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EXPERIMENT 4

Rates of Chemical Reactions A Clock Reaction

OBJECTIVE To determine the order of a reaction with respect to the reactant concentrations. To obtain the rate law for a chemical reaction.

APPARATUS & CHEMICALS burets (2)1-mL pipets (2)clock or watch with second hand125-mL Florence flask400-mL beaker test tubepipet bulb250-mL Erlenmeyer flasks (4)buret clampring stand

25-mL pipet50-mL pipetthermometer0.2 M KI (200-mL)0.4 M Na2S2O3 (100- mL) (freshlyprepared)1%percent starch solution, boiled0.2 M KNO3 (300- mL)0.1 M solution of Na2H2EDTA0.2M (NH4)2S2O8, (200- mL)(prepared from fresh solid)

DISCUSSIONKinetics is the study of how fast chemical reactions occur. Among the important factors that affect the rates of chemical reactions are:1.Reactant concentration2. Temperature

3. Catalyst Before a reaction can occur, the reactants must come into direct contact via collisions of the reacting particles. The reacting particles (ions or molecules) must have the right orientation and must collide with sufficient energy to result in a reaction. With these considerations in mind, we can qualitatively explain how the various factors influence the rates of reactions.

Concentration Changing the concentration of a solution alters the number of particles per unit volume. The more particles present in a given volume, the greater the probability of their collision. Therefore, increasing the concentration of a solution increases the number of collisions per unit time and therefore the rate of reaction.

Temperature Since temperature is a measure of the average kinetic energy, an increase in temperature increases the kinetic energy of the particles. An increase in kinetic energy increases the velocity of the particles and therefore the number of collisions between them in a given period of time. Thus, the rate of reaction increases. Also, an increase in kinetic energy results in a greater proportion of the collisions having the required energy for reaction.

Catalyst Catalysts, in some cases, are believed to increase reaction rates by bringing particles into close juxtaposition in the correct geometrical arrangement for reaction to occur. In other instances, catalysts offer an alternative

route to the reaction, one that requires less energetic collisions between reactant particles. If less energy is required for a successful collision, a larger percentage of the collisions will have the requisite energy, and the reaction will occur faster. Actually, the catalyst may take an active part in the reaction, but at the end of the reaction, the catalyst can be recovered chemically unchanged.

Let's examine now precisely what is meant by the expression rate of reaction. Consider the hypothetical reaction

A+B C+D [1]

Speed of a reaction is measured by the change in concentration of reactants or products with time.Suppose A reacts with B to form C and D. For the reaction A and B there are two ways of measuring rate:1. the speed at which the products appear (i.e. change in concentration of C or D per unit time), or2. the speed at which the reactants disappear (i.e. the change in concentration of A or B per unit time).

average rate = change in concentration - [ A] = - [ B] = [ C ] = [ D] time required for this change t t t t

The units for average rate are mol/L.s or M/s.

In general, the rate of the reaction will depend upon the concentration of the reactants. Thus, the rate of our hypothetical reaction may be expressed as

Rate = k[A]x[B]y [2]

where [A] and [B] are the molar concentrations of A and B, x and y are the powers to which the respective concentrations must be raised to describe the rate, and k is the specific rate constant. One of the objectives of chemical kinetics is to determine the rate law. Stated slightly differently, one goal of measuring the rate of the reaction is to determine the numerical values of x and y. Suppose that we found x = 2 and y = 1 for this reaction. Then

Rate = k[A]2[B] [3]

would be the rate law. It should be evident from Equation [3] that doubling the concentration of B (keeping [A] the same) would cause the reaction rate to double. On the other hand, doubling the concentration of A (keeping [B] the same) would cause the rate to increase by a factor of 4, because the rate of the reaction is proportional to the square of the concentration of A. The powers to which the concentrations in the rate law are raised are termed the order of the reaction . In this case, the reaction is said to be second order in A and first order in B. The overall order of the reaction is the sum of the exponents, 2 + 1 = 3, or a third-order reaction. It is possible to determine the order of the reaction by noting the effects of changing reagent concentrations on the rate of the reaction.It should be emphasized that k, the specific rate constant, has a definite value that is independent of the concentration. And depends only on temperature. Once the rate is known, the value of k can be calculated.

Reaction of Peroxydisulfate Ion with Iodide IonIn this experiment you will measure the rate of the reaction

S2O82- + 2I- I2 + 2SO4

2- [4]

and you will determine the rate law by measuring the amount of Peroxydisulfate, S 2O82-, that reacts as a

function of time. The rate law to be determined is of the form:

Rate of disappearance of S2O82- = k [S2O8

2-]x[I-]y [5]

or [ S 2O8

2- ] = k[S2O82-]x[I-]y

tYour goal will be to determine the values of x and y as well as the specific rate constant, k.You will add to the solution a small amount of another reagent (sodium thiosulfate, Na2S2O3 which will cause a change in the color of the solution. The amount is such that the color change will occur when 2 × 10 -4 mol of S2O8

2- has reacted. For reasons to be explained shortly, the solution will turn blue-black when 2 × 10-4 mol of S2O8

2- has reacted. You will quickly add another portion of Na2S2O3 after the appearance of the color, and the blue-black color will disappear. When the blue-black color reappears the second time, another 2 × 10-4 mol of S2O8

2- has reacted, making a total of 2(2 × 10-4) mole of S2O82 that has reacted. You will

repeat this procedure several times, keeping careful note of the time for the appearance of the blue-black colors. By graphing the amount of S2O8

2- consumed versus time, you will be able to determine the rate of the reaction. By changing the initial concentrations of S2O8

2- and I- and observing the effects upon the rate of the reaction, you will determine the order of the reaction with respect to S2O8

2- and I-.The blue-black color that will appear in the reaction is due to the presence of a starch-iodine complex that is formed from iodine, I2, and starch in the solution. The color therefore will not appear until a detectable amount of I2 is formed according to Equation [4]. The thiosulfate that is added to the solution reacts extremely rapidly with the iodine, as follows:

I2 + S3O32- 2I- + S4O6

2- [6]

Consequently, until the same amount of S2O32- that is added is all consumed, there will not be a sufficient

amount of I2 in the solution to yield the blue-black color. You will add 4 x 10 -4 mol of S2O32- each time (these

equal portions are termed aliquots), and from the stoichiometry of Equations [4] and [6] you can verify that when this quantity of S2O3

2- has reacted, 2 X 10-4 mol of S2O32- has reacted. Note also that although iodide, I-,

is consumed according to Equation [4], it is rapidly regenerated according to Equation [6] and therefore its concentration does not change during a given experiment.

Graphical Determination of RateThe more rapidly the 2 X 10-4 mol of S2O8

2- is consumed, the faster is the reaction. To determine the rate of the reaction, a plot of moles of S2O8

2- that have reacted versus the time required for the reaction is made, as shown in Figure 4.1The best straight line passing through the origin is drawn, and the slope is determined. The slope, [S2O8

2-]/t, corresponds to the moles of S2O82- that have been consumed per second and is

proportional to the rate. Since the rate corresponds to the change in the concentration of S 2O82- per second,

dividing the slope by the volume of the solution yields the rate of disappearance of S2O82-, that is,

[S2O82-] /t.

Helpful Comments

1. According to the procedure of this experiment, the solution will turn blue-black when exactly 2 ×10-4 mol of S2O8

2- has reacted.2. The purpose of the KNO3 solution in this reaction is to keep the reaction medium the same in each run in terms of the concentration of ions; it does not enter into the reaction in any way.3. The reaction studied in this experiment is catalyzed by metal ions. The purpose of the drop of the EDTA

solution is to minimize the effects of trace quantities of metal ion impurities that would cause spurious effects on the reaction.

4. You will perform a few preliminary experiments to become acquainted with the observations in this experiment so that you will know what to expect in the reactions.

5. The initial concentrations of the reactants have been provided for you on the report sheet.

PROCEDUREA. Preliminary Experiments

1. Dilute 5 mL of 0.2 M KI solution with 10 mL of distilled water in a test tube, add 3 drops of starch solution and mix thoroughly, and then add 5 mL of 0.2 M (NH4)2S2O8. solution. Mix and wait a while and observe color changes.2. Repeat the procedure in (1), but when the solution changes color add 4 drops of 0.4 M Na2S2O3, mix the solution, and note the effect that the addition of Na2S2O3 has on the color.

B. Kinetics ExperimentSolution Preparation Prepare four reaction solutions as follows (one at a time):

Solution 1: 25.0 mL KI solution 1.0 mL starch solution

1.0 mL Na2S2O3 solution 48.0 mL KNO3 solution

1 drop EDTA solution Total volume = 75.0 mL


1.0 mL Na2S2O3 Solution 23.0 mL KNO3 solution

1 drop EDTA solution Total volume = 50.0 mL


1.0 ml Na2S2O3

23.0 mL KNO3 solution 1 drop EDTA solution Total volume = 75.0 mL


1.0 mL Na2S2O3 solution 35.5 mL KNO3 solution 1 drop EDTA solution

Total volume = 50.0 mL

Equipment SetupSet up two burets held by a clamp on a ring stand as shown in Figure 4.2. Use these burets to measure accurately the volumes of the KI and KNO3 solutions. Use two separate 1-mL pipets for measuring the volumes of the Na2S2O3 and starch solutions and use 25-mL and 50-mL pipets to measure the volumes of the (NH4)2S2O8. Solutions. Each solution must be freshly prepared to begin the rate study-that is, prepare solutions 1, 2, 3, and 4 one at a time as you make your measurements.Rate Measurements Prepare solution 1 in a 250-mL Erlenmeyer flask that has been scrupulously cleaned and dried. Pipet 25.0 mL of (NH4)2S2O8 solution into a clean, dry 100-mL beaker. Be ready to begin timing the reaction when the solutions are mixed (READ AHEAD). The reaction starts the moment the solutions are mixed! BE PREPARED! ZERO TIME! Quickly pour the 25.0 mL of (NH 4)2S2O8, solution into solution 1 and swirl vigorously; note the time you begin mixing to the nearest second. At the instant when the blue-black color appears, 2 × 10-4 mol of S2O8

2- has reacted. IMMEDIATELY (be prepared!) add a 1-mL aliquot of Na2S2O3 solution from the pipet and swirl the solution; the color will disappear. If the students fill each of seven clean, dry test tubes with 1mL of Na2S2O3 solution, they then can add these aliquots to their reactions at the appearance of the blue color without loss of time.Record the time for the reappearance of the blue-black color. Add another 1-mL aliquot of Na 2S2O3 solution and note the time for the reappearance of the color. The time interval being measured is that between the appearance of the blue-black color. For good results, these aliquots of Na 2S2O3 must be measured as quickly, accurately, and reproducibly as possible. Continue this procedure until you have added seven (7) aliquots to solution 1.You are finished with solution 1 when you have recorded all your times on the report sheet. (The time intervals are cumulative.) Solutions 2, 3, and 4 should be treated in exactly the same manner except that 50.0-mL portions of (NH4)2S2O8, solutions should be added to solutions 2 and 4 and 25 mL of (NH4)2S2O8

solution should be added to solution 3.

(CAUTION: Be on guard-solution 3 will react much more rapidly than solution l.). In each of these reactions the final total solution volume is 100 mL.

Time saving Hint to InstructorsTime may be saved in this experiment by setting up several burets on side tables in the laboratory filled with (NH4)2S2

O8, KNO3, KI, and Na2S2O3

solutions. The

students may obtain the solution from these burets and refill the burets from bottles of stock solutions also kept there.

CalculationsTabulate on the data sheet for each aliquot of Na2S2O3 added to each of the four solutions:1. The time interval from the start of the reaction (addition of S2O8

2-) to the appearance of color for the first aliquot of S2O8

2- and the time interval from the preceding color appearance for each succeeding aliquot (column2);2. The cumulative time from the start of the reaction to each appearance of color (column 3);3. The corresponding numbers of moles S2O3

2- consumed (column 4) For each solution plot on the graph paper provided the moles of S2O8

2- consumed (as the ordinate, vertical axis) versus time in seconds (as the abscissa, horizontal axis), using the data in columns 3 and 4. Calculate the slope of each plot, and from these calculations answer the questions on your report sheet. REVIEW QUESTIONS Before beginning this experiment in the laboratory, you should be able to answer the following questions:

1. What factors influence the rate of a chemical reaction?

2. What is the general form of a rate law?

1. What is the order of reaction with respect to A and B for a reaction that obeys the rate law

rate = k[A]2[B]3?

4. Write the chemical equations involved in this experiment and show that the rate of disappearance of [S2O8

2-] is proportional to the rate of appearance of the blue-black color of the starch-iodine complex.

5. It is found for the reaction A + B C that doubling the concentration of either A or B quadruples the rate of the reaction. Write the rate law for this reaction.

6. If 2 X 10-4 mol of S2O82- in 50 mL of solution is consumed in 188 s, what is the rate of consumption of

S2O82-?

7. Why are chemists concerned with the rates of chemical reactions? What possible practical value does this type of information have? 8. Suppose you were dissolving a metal such as zinc with hydrochloric acid. How would the particle size of the zinc affect the rate of its dissolution?

9. Assuming that a chemical reaction doubles in rate for each 100 temperature increase, by what factor would the rate increase if the temperature were increased 40°C?

Experiment 4

Rates of Chemical Reactions A Clock Reaction

REPORT SHEET

Name Desk #

Date Laboratory Instructor

A. Preliminary Experiment1. What are the colors of the following ions: K+ ; I- 2. The color of the starch I2 complex is

B. Kinetics ExperimentSolution 1. Initial [S2O8

2-] = 0.05 M; initial [I-] = 0.05 M. Time experiment started

Time (s) between Total moles of S2O82-

Aliquot no. appearances of color Cumulative time(s) consumed 1 2.0 × 10-4

2 4.0 × 10-4

3 6.0 × 10-4

4 8.0 × 10-4

5 10 × 10-4

6 12 × 10-4

7 14 × 10-4

Solution 2. Initial [S2O82-] = 0.10 M; initial [I-] = 0.05 M. Time experiment started



2 4.0 × 10-4

3 6.0 × 10-4

4 8.0 × 10-4

5 10 × 10-4

6 12 × 10-4

7 14 × 10-4




2 4.0 × 10-4

3 6.0 × 10-4

4 8.0 × 10-4

5 10 × 10-4

6 12 × 10-4

7 14 × 10-4




2 4.0 × 10-4

3 6.0 × 10-4

4 8.0 × 10-4

5 10 × 10-4

6 12 × 10-4

7 14 × 10-4

CALCULATIONS

1. Rate of reaction,, [S2O82-]/T, as calculated from graphs (that is, from slopes of lines):

Solution 1 Solution 3

Solution 2 Solution 4 2. What effect does doubling the concentration of I- have on the rate of this reaction?

3. What effect does changing the [S2O82-] have on the reaction?

4. Write the rate law for this reaction that is consistent with your data.

5. From your knowledge of x and y in the equation (as well as the rate in a given experiment from your graph), calculate k from your data. Rate = k[S2O8

2-]x[l-]y

EXPERIMENT 5An Equilibrium

Constant

OBJECTIVETo determine the equilibrium constant of a chemical reaction by spectrophotometric measurementsTo use graphing techniques and data analysis to evaluate data

DISCUSSIONA spectrophotometric method of analysis involves the interaction of electromagnetic (E" radiation with matter. The most common regions of the EM spectrum used for analyses are the ultraviolet, visible, and the infrared regions. We are most familiar with the visible region of the spectrum, having a wavelength range from 400 to 800. The colors of nature, the bluebonnet flowers, the red rocks, the green grass, and the changing colors of the leaves in Fall are a consequence of visible light interacting with the compounds that are present in the material.Every chemical substance possesses its own unique set of electronic, vibrational, and rotational energy states . When EM radiation falls incident upon an atom or molecule, the radiation absorbed (the absorbed light) is an energy equal to the difference between two energy states in the atom or molecule, placing the atom or molecule in an "excited state.' lle remainder of the EM radiation passes through the sample (the transmitted light) and an EM radiation detector detects it.As the energy absorbed (and transmitted) equals the energy difference between the unique sets of energy states in an atom or molecule, absorption and emission spectrophotometry methods are used to detect its presence in a sample.The energy absorbed, E, by an atom or molecule is related to the wavelength, .X, of the EM radiation.

E = h

h is Planck's constant ,and is the frequency of light

When a substance absorbs EM radiation from the visible region of the spectrum, it is usually an electron that is excited from a lower to a higher energy state. When white light (EM radiation containing all wavelengths of visible light) passes through the sample, our eyes detect the wavelengths of visible light not absorbed, i.e., the light transmitted. The light that is absorbed excites electrons of the sample. Therefore, the color we see is complementary to the one absorbed . If, for example, the atom or molecule absorbs energy from the violet region of the visible spectrum, the transmitted light (and the substance) appears yellow (violet's complementary color)-the higher the concentration of violet absorbing atoms or molecules, the more intense is the yellow.

E2 (excited state

incident light transmitted light

E1 (ground state)

Thus the eye, one kind of EM radiation detector, detects only the transmitted light. Table 1 lists the colors corresponding to wavelength regions of light (and their complements) in the visible region of the EM spectrum.

Table 1. Color and Wavelengths in the Visible Region of the Electromagnetic Spectrum

Color Absorbed Wavelength(nm) Color Transmitted

red 750-610 green-blue

orange 610-595 blue-green

yellow 595-580 violet

green 580-500 red-violet

blue 500-435 orange-yellow

violet 435-380 yellow

In this experiment, EM radiation is used to determine the concentration of an absorbing substance in an aqueous solution. the amount of transmitted light is measured using an instrument called a spectrophotometer, an instrument that measures light intensities with a photosensitive detector at specific (but variable) wavelengths . The wavelength at which maximum absorption of the EM radiation by the absorbing substance occurs is determined and set on the spectrophotometer.The visible light path through the spectrophotometer from the light source through the sample to the photosensitive detector.Several factors control the amount of EM radiation (light energy) that a sample absorbs: -Concentration of the absorbing substance -Thickness of the sample containing the absorbing substance (determined by the width of the cuvet) -Probability of light absorption by the absorbing substance (called the molar absorptivity coefficient or extinction coefficient)The ratio of the intensity of the transmitted light to that of the incident light IO , I, is called the transmittance , TThis ratio, expressed as percent, is

I/Io x 100% = %T [2]

The spectrophotometer has a %T (percent transmittance of light) scale. Because it is linear, the %T scale is easy to read and interpolate. Chemists often perform calculations based on the amount of light absorbed by the sample, rather than the amount of light transmitted, because the extent of absorption is directly proportional to the concentration of the absorbing substance. The absorbance, A, of the substance is related to the intensity of the incident and transmitted light and the percent transmittance by the equation A= Log I/Io = Log 1/T = Log 100/ %T [3] Io I t

b

Figure 4. 2 Incident light, Io, and transmitted light, I t, for a sample of concentration c in a cuvet of thikness b,

A = abc This equation is commonly referred to as Beer's law.A is the absorbance, a (the molar absorptivity coefficient ) constant at any given wavelength for a thickness b, and c is the molar concentration of the absorbing substance.

The absorbance value is directly proportional to the molar concentration of the absorbing substance, if the same (or a matched) cuvet and a set wavelength are used for all measurements. A plot of absorbance versus concentration data is linear; a calculated slope and absorbance data can be used to determine the molar concentration of the same absorbing specie in an unknown solution .

Measuring an Equilibrium constant

The magnitude of an equilibrium constant, Ke, expresses the equilibrium position fora chemical system. For the reaction,

aA + bB xX + yY the mass action expression, [XI x IYI y equals the equilibrium constant, Kc, when a dynamic equilibrium

[A]a [B]b

has been established between reactants and products. The brackets in the mass action expression denote the molar concentration of the respective substance.the magnitude of the equilibrium constant indicates the principal species, products or reactants, that exist in the chemical system at equilibrium. For example, a large equilibrium constant indicates that the equilibrium lies to the right with a high concentration of products and correspondingly low concentration of reactants. The value of Kc is constant for a chemical system at a given temperature.

This experiment determines Kc for a chemical system in which all species are soluble. The chemical system involves the equilibrium between iron(III) ion, Fe3+,thiocyanate ion, SCN-, and thiocyanatoiron(III) ion, FeNCS2+ : Fe (H2O)6

3+ (aq) + SCN-(aq) Fe(H2O)5NCS2+(aq) + H2O(l) [4] Because the concentration of water is essentially constant in dilute aqueous solutions, we omit the waters of hydration and simplify the equation to read

Fe3+ (aq) + SCN-(aq) FeNCS2+(aq) [5]

The mass action expression for the equilibrium system, equal to the equilibrium constant, is Kc = [FeNCS 2+ ] [Fe2+][SCN-l]

In Part A you will prepare a set of standard solutions of the FeNCS2+ ion. As FeNCS2+ ion is a deep, blood-red complex with an absorption maximum at 447 nm, its concentration is determined spectrophotometrically. the absorbance for each solution is plotted vesus the molar concentration of FeNCS 2+; this establishes a

calibration curve from which the concentrations of FeNCS2+, are determined for the chemical systems in Part B.

In preparing the standard solutions, the Fe3+ concentration far exceeds the SCN-concentration. This huge excess of Fe3+ pushes the equilibrium (Equation 5)far to the right, nearly consuming all of the SCN - placed in the system. As a result the FeNCS2+ concentration at equilibrium approximate the original concentration. In other words, we assume that the position of the equilibrium is driven so far to the right by the excess Fe3+ that all of the SCN- is complexed, forming FeNCS2+

In Part B, the concentrations of the Fe3+ and SCN- ions in the various tests solutions are nearly the same, thus creating equilibrium systems in which there is an appreciable amount of each of the species in the equilibrium. The chemical system is prepared by mixing known molar concentrations of Fe3+ and SCN-. By knowing the initial concentrations of Fe3+ and SCN-, and by measuring the equilibrium concentration of FeNCS 2+

spectrophotometrically, the equilibrium concentrations of Fe3+ and SCN- are calculated. Using these equilibrium concentrations, the Kc for the system is calculated. [Fe3+]equilibrium = [Fe3+] initial _ [FeNCS2+]equilibrium

PROCEDUREOne set of solutions having known molar concentrations of FeNCS2+ is prepared for a plot of absorbance versus concentration. A second set of PROCEDURE solutions is prepared to determine unknown molar concentrations of FeNCS2+. By carefully measuring the initial amounts of reactants placed in the reaction systems, the mass action expression at equilibrium can be solved; this equals Kc.

A. A Set of StandardSolutions to Estabfish a Calibration Curve

These solutions are used to determine the percent transmittance (and absorbance) of known molar concentrations of FeNCS2+. A plot of the data, known as a calibration curve, is used to determine the equilibrium molar concentrations of FeNCS2, in Part B.Record on the Report Sheet the exact molar concentrations of the Fe(NO3)3 and NaSCN reagent solutions.

1.Prepare a Set of the Standard Solutions. Pipet 0, 2, 5, 10, and 15 mL of 0.002 M NaSCN into separate, labeled, and clean 100-mL volumetric flasks (Table 2). Pipet 25 mL of 0.2 M Fe(NO3)3 into each flask and dilute to the 100 mL "mark" with 0.2 M Fe(NO3)3. Stir each solution thoroughly to ensure that equilibrium is established.

Table 2 Compositios of Standard Solutions for Preparing the Calibration Curve

0.002 M NaSCN 0.2 M Fe(NO3)3

Solution (in 0.25 M HNO3) (in 0.25 M HNO3) 0.25 M HNO3 1(blank) 0 ml 25 ml dilute to 100 ml2 2 ml 25 ml dilute to 100 ml3 5 ml 25 ml dilute to 100ml4 10 ml 25 ml dilute to 100 ml5 15 ml 25 ml dilute to 100 ml 2.Prepare the Blank Solution. Solution I is called the blank solution. After the instrument has been turned on for 10 minutes and the wavelength scale has been set at 447 nm, rinse a cuvet with several portions of Solution 1. Dry the outside of the cuvet with a clean Kimwipe or Joseph paper, removing water and fingerprints.3 Handle the lip of the cuvet thereafter.

3.Calibrate the Spectrophotometer. Place the cuvet containing the blank solution into the sample compartment and set the meter on the spectrophotometer to read 100%Twith the 100%Tadjust knob. Remove the cuvet and set the meter at 0%T with the zero adjust knob. Repeat this procedure until the 100% T and 0% T readings are reproducible. Once they are set, do not touch the knobs for the remainder of the experiment. If you accidentally move them, merely repeat this calibration procedure.

4.Record the Percent Transmittance of the Standard Solutions. Empty the cuvet and rinse it thoroughly with several portions of Solution 2. Fill it approximately three- fourths full. Carefully dry the outside of the cuvet with a clean Kim wipe. Remember, handle only the lip of the cuvet. Place the cuvet into the sample compartment; read as accurately as possible the percent transmittance and record. Repeat with Solutions 3, 4, and 5.Disposal. Discard each test solution and each rinse into the "Waste Salts" container.

5. Graph the Data. convert the % T readings for the four solutions to absorbance values. Plot, on linear graph paper, A (ordinate) versus [FeNCS2+] (abscissa) for the five solutions. Draw the best straight line through the five (or seven) points and the origin . Ask your instructor to approve your graph.

Record on the Report Sheet the exact molar concentrations of the Fe(NO3)3 and NaSCN reagent solutions.

1. Prepare the Test Solutions. In clean 150-mm test tubes (or 10-mL volumetric flasks)-5 prepare the test solutions in Table 25.3. Use pipets for the volumetric measurements. Be careful not to "mix' pipets to avoid contamination of the reagents prior to the preparation. Also note that the molar concentration of Fe(NO3)3 for this set of solutions is 0.002 mol/L, not the 0.2 mol/L solution used in Part A.

Table 3 Compositions of Test Solutions for Determination of Kc

0.002 M NASCN 0.002 M Fe(NO3)3+

Solution (in 0.25 M HNO3) (in 0.25 M HNO3) 0.25 M HNO3

1 1 mL 4 ml 5 ml 2 2 ml 5 ml 3 ml3 3 ml 5 ml 2 ml4 4 ml 5 ml 1 ml5 5 ml 5 ml -

If 0.002 M Fe(NO3)3 is not available, dilute 1.0 ml (measured with 1.0 ml pipet) of 0.2 M Fe(NO3)3 used in Part A to 100 mL with 0.25 M HNO3 in a volumetric flask.

2.Recalibrate the Spectrophotometer. Use the blank solution (Solution 1) from Part A to check the 100% T and 0% T readings on the spectrophotometer scale.

3.Determine the Percent Transmittance of the Test Solutions. Stir each test solution until equilibrium is reached (approximately 1 minute). Rinse the cuvet thoroughly with several portions of the test solution and fill it three-fourths full. Clean and dry the outside of the cuvet.6. Be cautious in handling the cuvets. Record the percent transmittance of each test solution.

Disposal.. Dispose of the waste thiocyanatoiron(III) ion solutions from the 100-ml volumetric flasks and the cuvets in the ‘Waste Salts' container.

CLEANUP: Rinse the volumetric flasks, the pipets, and the cuvets twice with tap water and once with deionized water. Discard each rinse in the sink.

4. Use Data to Determine Equilibrium Concentrations. Convert the percent transmittance readings from Part B.3 to absorbance values. Using the calibration curve prepared in Part A.5 determine the equilibrium molar concentration of FeNCS2+, for each test solution.

5.Do the Calculations. Complete the calculations as outlined on the Report Sheet and described in the I discussion. Complete an entire Kc calculation for Test Solution 1 (Part B) before attempting the calculations for the remaining solutions.The equilibrium constant varies from solution to solution and from chemist to chemist in this experiment, depending on chemical technique and the accumulation and interpretation of the data. Consequently, it is beneficial to work with other colleagues through your own calculations, and then "pool" your final, experimental K, values to determine an accumulated "probable" value .

REVIEW QUESTIONS

Before beginning this experiment in the laboratory, you should be able to answer the following questions: 1. What effect does a dirty cuvet (caused by fingerprints, water spots, or lint) have on the percent transmittance reading for a FeNCS2+ solution? Does this error cause the Kc to be reported as being too high or too low? Explain. 2.In our calculations, the thickness of the solution (the cuvet) and the molar absorptivity of FeNCS2+ are not considered. Explain. 3. If the percent transmittance reads less than 3.0%T on the spectrophotometer, how can the procedure be modified to obtain a higher %Treading? 4.Over a period of time the 0 %T (no sample in the sample compartment) and the l00%T(blank solution in the sample compartment) may drift from the initial calibration of the spectrophotometer. If the 100% T reading drifts downward (less than 100%1), how does his error affect, in Part B, the :a. absorbance readings? b. [FeNCS2+], equilibrium? c. [Fe3+], equilibrium? d. [SCN], equilibrium?

Experiment 5

An Equilibrium Constant

REPORT SHEET

Name Desk #


Solutions 1 2 3 4 5

Volume of Fe(NO3)3 (ml) -------- -------- -------- -------- --------

Moles of Fe3+ -------- -------- -------- -------- --------

Volume of NaSCN(ml) + -------- -------- -------- -------- --------

Moles of SCN- -------- -------- -------- -------- -------

Absorbance -------- -------- -------- -------- -------

[Fe(NCS2+], equilibrium -------- -------- -------- -------- -------

Moles of Fe3+, equilibrium -------- -------- -------- -------- --------

Moles of Fe3+, reacted -------- -------- -------- -------- --------

[ Fe3+], equilibrium -------- -------- -------- -------- --------(unreacted )

Moles of SCN- , reacted -------- -------- -------- -------- -------

Moles of SCN- , unreacted -------- -------- -------- -------- -------

[SCN-], equilibrium -------- -------- -------- -------- --------(unreacted )

Kc = [FeNCS2+] -------- -------- -------- -------- -------- [Fe3+] [SCN-]

EXPERIMENT 6

Titration Curves ofPolyprotic Acids

OBJECTIVEDetermination of dissociation constants of polyprotic acid

APPARATUS AND CHEMICALSpH meter with electrodes weighing bottlepotassium acid phthalate (KHP) buret clamp, and ring stand sodium hydroxide solution about 0.1 M 25-mL pipet100 mL of approx. 0.1M phosphoric acid 150-mL beaker (3) standard buffer solution 250-mL beaker phenolphthalein indicator solution

DISCUSSION Consider the triprotic acid (H3PO4) .It undergoes the following dissociations in aqueous solution:

H3PO4 (aq) H2PO4 -(aq) + H+(aq) Ka1 = [H2PO4

- ][H+] [1]

[H3PO4] H2PO4

- (aq) HPO4 2-(aq) + H+(aq) Ka2 = [HPO4

2- ][H+] [2]

[H2PO4--]

HPO42- (aq) PO4

3-(aq) + H+(aq) Ka3 = [PO43-

][H+] [3] [HPO4

2--]

The acid H3PO4 possesses three dissociable protons, and for this reason it is termed a triprotic acid. If you

were to perform a titration of H3PO4 with NaOH , the following reactions would occur :

H3PO4+ NaOH NaH2PO4 + H2O [4]

NaH2PO4 + NaOH Na2HPO4 + H2O [5]

Na2HPO4 + NaOH Na3PO4 + H2O [6]

The resultant titration curve, when plotted as pH versus milliliters of NaOH added, would be similar to that shown in Figure 6.1. At the point at which one-half of the protons in the first dissociation step of H 3PO4 have been titrated with NaOH, the H3PO4 concentration is equal to the H2PO4

- concentration. Substituting [H3PO4] = [H2PO4

-] into Equation [1] yields Ka1 = [H+], or pH = pKa1 at this point.

Similarly, at one-half the second equivalence point, one-half of the H2PO4- has been neutralized and [H2PO4

-] = [HPO42-] . Substituting this into Equation [2] yields Ka2 = [H+] or pKa2 = pH at this point.In the same manner, at one-half the third equivalence point, [HPO4

2-] = [PO43-]. Substituting this into Equation

[3], we obtain the expression Ka3 = [H+], or pKa3 = pH.The same type of result is obtained for any polyprotic acid. If a titration of the acid is performed with a pH meter, the dissociation constants may be obtained from titration curves as long as the dissociation constants exceed the ion product of water, which you should recall is 10-14 for the reaction

2H2O H3O+ + OH-

In practice, if the acidity of the acid being studied approaches that of water, as in the case for the third proton of H3PO4 for which Ka3, is 4.2 x 10-13, it is difficult to determine the dissociation constant in this manner. Thus for H3PO4, both Ka1 and Ka2 readily obtained in this way, but Ka3 is not.In this experiment you will determine the dissociation constants K,a1 and K.a2 of a phosphoric acid

PROCEDURE 1 . Standardize the pH meter .2. Prepare 200 mL approximately 0.1 M NaOH solution 3. Titrate three separate 25-mL aliquots of the phosphoric acid in three separate 250-mL beakers and plot the

titration curves.4. Using the relations given above, and the relevant equations from this experiment, determine the pK an and

Kan values for phosphoric acid .5. Determine the volumes necessary to reach the equivalence points.

(HINT): You can save time if you do your first titration rapidly so that you know the approximate volumes of the equivalence points; then you can do the next two titrations with large-volume increment away from the equivalence points and small-volume increments near the equivalence points.)

REVIEW QUESTIONS

Before beginning this experiment in the laboratory, you should be able to answer the following questions:

1. What is a polyprotic acid?

2. If 20.2 mL of 0.122 M NaOH is required to reach the first equivalence point of a solution of citric acid (H3C6H5O7), how many mL of NaOH are required to completely neutralize this solution? 3. How many mmol of NaOH will react with 50 mL of 6.2 M H2C2O4?

4. How many moles of H3O+ are present in 50 mL of a 0.3 M solution of H2SO4? 5. Why is it necessary to standardize a pH meter?

6. If the pH at one-half the first and second equivalence points of a dibasic acid is 3.52 and 6.31, respectively, what are the values for pKa1 and pKa2? From pKa1 and pKa2 calculate the Ka1 and Ka2.

7. Derive the relationship between pH and pKa at one-half the equivalence point for the titration of a weak acid with a strong base.

8. Could Kb for a weak base be determined in the same way that Ka for a weak acid is determined in this experiment? 9. What is the relationship of the successive equivalence-point volumes in the titration of a polyprotic acid? 10. If the Ka1 of a diprotic acid is 3.20, what is the pH of a 0.10 M solution of this acid.

Experiment 6 Titration Curves

of Polyprotic Acids

REPORT SHEET

Name Desk #


Determination of pKa values of H3PO4 mL NAOH pH mL NaOH pH mL NaOH pH _________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_______ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------_________ _______ ------------ -------- ----------- -------- ________ _______ ------------ -------- ----------- --------

PKa1 ------------- pK.a2 ------------

K.a1----------- Ka2--------------

EXPERIMENT 7

Determination of the Solubility-Product Constant for a Slightly Soluble Salt

OBJECTIVEDetermination of the value of the solubility-product constant for a slightly soluble salt.

APPARATUS AND CHEMICALS Buret 0.0024 M K2CrO4

Ring stand and buret clamp 0.004 M AgNO3

centrifuge 0.25 M NaNO3

75-mm test tubes (3) 100-mL volumetric flasks(4) Spectrophotometer and cuvettes no. 1 corks (3)

5-mL pipets (2)

DISCUSSION Acids, bases, and salts are classified as inorganic substances . When an acid reacts with a base in aqueous solution, the products are a salt and water, as illustrated by the reaction of H2SO4 and Ba(OH)2:

H2SO4(aq) + Ba(OH)2(aq) BaSO4(s) + 2H2O(1) [1]

In general most common salts are strong electrolytes. The solubilities of salts span a broad spectrum, ranging from slightly soluble to very soluble. This experiment is concerned with heterogeneous equilibria of slightly soluble salts. In order for a true equilibrium to exist between a solid and solution, the solution must be saturated. For instance barium sulfate is a slightly soluble salt, and in a saturated solution this equilibrium may be rewritten as : BaSO4(s) Ba2+(aq) + SO42-(aq) [2] The equilibrium constant for Equation [2] is: Ksp= [Ba2+] [SO4 2-] [3] The value of Ksp is a constant at constant temperature. The solubility product for a slightly soluble salt can easily be calculated by determining the solubility of the substance in water. Suppose, for example, we determined that 2.42 x 10-4 g of BaSO4 dissolves in 100 ml of water. The molar solubility of this solution (that is, the molarity of the solution) is 1.4 x 10-5 M In a saturated solution the product of the molar concentrations of Ba2+ and SO4

2- cannot exceed 1.08 x 10-10. If the ion product [Ba2+][SO4

2-] exceeds 1.08 x 10-10, precipitation of BaSO4 would occur until this product is reduced to the value of Ksp or if, for example, a solution of Na2SO4 is added to a solution of Ba(NO3)2, BaSO4

would precipitate if the ion product [Ba2+1[SO42-] is greater than Ksp, if we determined that the solubility of

Ag2CO3 were 3.49 x 10-3 g/ 100 mL, we could calculate the solubility-product constant for Ag2CO3 as follows:

The solubility equilibrium involved is Ag2CO3 2 Ag+(aq) + CO3

2- (aq) [4]

and the corresponding solubility-product expression isKsp,= [Ag+]2 [CO3

2-]The solubility of Ag2CO3 in moles per liter is: 3.49 /100 x 10-3 x 278 = 1.27 x 10-4 M [CO3

2-] = 1.27 x 10-4 M ; [Ag+]= 2 x 1.27 x 10-4 = 2.54 x 10-4 M

Ksp,= [Ag+]2 [CO32-] = [2.54 x 10-4 ]2 [1.27 x 10-4 ] =8.19 x 10-12 M3

To determine the solubility-product constant for a slightly soluble substance, we need only to determine the concentration of one of the ions since the concentration of the other ion is related to the first ion's concentration by a simple stoichiometric relationship

In this experiment you will determine the solubility-product constant for Ag2CrO4. This substance contains the yellow chromate ion, CrO2- that should be determined experimentally by spectrophotometric measurement at 375 nm

To determine the solubility of AgCrO4, you will first prepare it by the reaction of AgNO3 with K2CrO4: 2AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2KNO3(aq)

If a solution of AgNO3 is added to a solution of K2CrO4, precipitation will occur if the ion product [Ag2+][CrO4

2-] numerically exceeds the value of Ksp if not, no precipitation will occur.

PROCEDUREA. Preparation of a Calibration Curve Using a buret, add 1, 5, 10, and 15 mL of standardized 0.0024 M K 2CrO4 to each of four clean, dry 100mL volumetric flasks and dilute to the 100 mL mark with 0.25 M NaNO3.1.Calculate the CrO4

2-concentration in each of these solutions. 2. Measure the absorbance of these solutions at 375 mn 3. Plot the absorbance versus concentration to construct your calibration curve.

B. Determination of the Solubility-Product Constant Accurately prepare three separate solutions in separate 150-mm test tubes by adding 5 mL of 0.004 M AgNO3

to 5 ml of 0.0024 M K2CrO4. Stopper each test tube and shake the solutions thoroughly at periodic intervals for about 15 min to establish equilibrium between the solid phase and the ions in solution. Transfer the contents of each test tubs into 75-mm test tubes and centrifuge. Discard the supernatant liquid and retain the precipitate. To each of the test tubes add 2 ml of 0.25 M NaNO3. Shake each test tube thoroughly for about 15 minutes to establish equilibrium between the solid and the solution and centrifuge again. There must be some solid Ag2CrO4 remaining in these test tubes. If there is not, start over again.. Transfer the clear, pale yellow supernatant liquid from each of the three test tubes to a clean, dry cuvette. Measure and record the absorbance of the three solutions.

Using your calibration curve, calculate the molar concentration of CrO42- in each solution.

Note on Calculations:You are determining the Ksp of AgCrO4 in this experiment. The equilibrium reaction for the dissolution of Ag2CrO4 is:

AgCrO4 (s) 2Ag+(aq) + CrO4 2-(aq)

for which Ksp = [Ag+]2 [CrO4 2-].

You should note that at equilibrium [Ag+1 = 2[CrO42-]. Therefore, having determined the concentration of

chromate ions, you know the silver-ion concentration REVIEW QUESTIONS Before beginning this experiment in the laboratory, you should be able to answer the following questions:

1 Write the solubility equilibrium and the solubility-product constant expression for the slightly soluble salt CaF2,

2 Calculate the number of moles of Ag+ in 5 mL of 0.004 M AgNO3 and the number of moles of CrO42- in 5 mL

of 0,0024 M K2CrO4,

3 If 10 ml of 0,004 M AgNO3 is added to 10 mL of 0,0024 M K2CrO4' is either Ag+ or CrO42- in stoichiometric

excess? If so, which is in excess? 4. The Ksp for BaCrO4 is 1.2 x 10-10. Will BaCrO4 precipitate upon mixing 10 mL of 1 x 10-4 M Ba(NO3)2 with 10 mL of 1 x 10-4 M K2CrO4?

5. The Ksp for BaCO3 is 5.1 x 10-9. How many grams of BaCO3 will dissolve in 100 mL of water?

6. Distinguish between the equilibrium-constant expression and Ksp for the dissolution of a sparingly soluble salt.

7. List as many experimental techniques as you can that may be used to determine Ksp for a sparingly soluble salt.

8. Why must some solid remain in contact with a solution of a sparingly soluble salt in order to ensure equilibrium?

9. In general, when will a sparingly soluble salt precipitate from solution?

10. Sparingly soluble bases and salts, such as Fe(OH)2 and FeCO3 are more soluble in acidic than in neutral solutions. Why?

Experiment 7 Determination of the Solubility-Product

Constant for a Slightly Soluble Salt

REPORT SHEET

Name Desk #


A. Calibration Curve

Initial [CrO42-] : 0.0024 M

Volume of0.0024 M K2CrO4 Total volume [CrO4 2-] Absorbance

1. 1ml 100 ml -----------------2. 5ml 100 ml -----------------3. 10 ml 100 ml -----------------4. 15 ml 100ml -----------------

Molar absorption coefficient for [CrO4 2-]

1. ---------- 2. ---------- 3. ---------- 4.----------

Average molar absorption coefficient ---------------

B. Determination of Ksp

Absorbance [CrO4 2-] [Ag+] Ksp of Ag2CrO4

1. -------- ----------- ---------- -----------

2. -------- ----------- ---------- ----------

3. -------- ----------- ---------- ----------

Average Ksp -----------

EXPERIMENT 8Molar Solubility,

Common-ion Effect

OBJECTIVETo determine the molar solubility and the solubility constant of calcium hydroxide.To study the effect of a common ion on the molar solubility of calcium hydroxide.

DISCUSSIONSalts that have a very limited solubility in water are called slightly soluble (or "insoluble') salts. A saturated solution of a slightly soluble salt is a result of a dynamic equilibrium between the salt and its ions in solution; however, because the salt is only slightly soluble, the concentrations of the ions are low. For example, in a saturated silver sulfate, Ag2SO4, solution, the dynamic equilibrium between solid

Ag2SO4 , the Ag+ and SO42- ions in solution lies far to the left because of the low solubility of silver sulfate.

Ag2SO4(s) 2 Ag+(aq) + SO42-(aq) [1]

For which the solubility product Ksp is: Ksp = [Ag+]2 [SO4

2-]. What happens to the molar solubility of a salt when an ion, common to the salt, is added to the saturated solution? According to Le Chatelier's principle, the equilibrium for the salt shifts to compensate for the added ions; that is, it shifts left to favor the formation of more of the solid salt. This effect, the addition of an ion common to an existing equilibrium, is called the common-ion effect. As a result of the common ion addition and the corresponding shift in the equilibrium, fewer moles of the salt dissolve in solution, lowering the molar solubility of the salt.The molar solubility of a salt is the number of moles of that salt that dissolves per liter of (aqueous) solution. In this experiment, you will determine the molar solubility and the solubility constant for calcium hydroxide, Ca(OH)2. A saturated Ca(OH)2 solution is prepared; after an equilibrium is established between the solid Ca(OH)2 and the Ca 2+ and OH- ions in solution, the decanted solution is analyzed. The hydroxide ion, OH-, in the solution is titrated with a standardized HCI solution to determine its molar concentration.

According to the following equation:

Ca(OH)2(s) Ca2+(aq) + 2 OH-(aq) [2]

for each mole of Ca(OH)2 that dissolves, I mol of Ca2+ and 2 mol of OH- are present in solution. Therefore, by determining the molar concentration of hydroxide ion, the |Ca2+], the Ksp, and the molar solubility of Ca(OH)2(s) can be calculated [Ca2+]=1/2 [OH-] ; Ksp = [Ca2+][OH-]2

Likewise, we can use the same procedure to determine the molar solubility of Ca(OH)2 in the presence of added calcium ion, an ion common to the slightly soluble salt equilibrium.

PROCEDUREThe decante from a saturated calcium hydroxide solution is titrated with a standardized hydrochloric acid solution to the methyl orange endpoint. An analysis of the data results in determination of the molar solubility and solubility constant of calcium hydroxide. The procedure is repeated on a decantate from a saturated calcium hydroxide solution containing added calcium ion.

A. Three analyses are to be completed. To hasten the analyses in Parts A.4 and A.5, prepare three, clean labeled 125- or 250-mL Erlemneyer flasks and pipet 25 mL of the saturated Ca(OH)2 solution into each flask before you begin any titration.

1. Prepare the Stock Calcium Hydroxide Solution. Prepare a saturated Ca(OH)2 solution 1 week before the experiment by adding approximately 3 g of Ca(OH)2 to 120 mL of boiled, deionized water in a 125-mL Erlenmeyer flask. Stir the solution and stopper.A saturated solution of calcium hydroxide is called limewater. This solution may have been prepared for you. Ask your instructor.

2. Set Up the Titration Apparatus. Prepare a clean, 50-mL buret for titration. Rinse the clean buret and tip with three 5-mL portions of the standard 0.05 M HCI solution and discard. Fill the buret with standardized 0.05 M HCI, remove the air bubbles in the buret tip, and, after 30 seconds, read and record the initial volume

(±0.02 mL). Record the actual concentration of the 0.05 M HCI on the Report Sheet. Place a sheet of white paper beneath the receiving flask.

3.Transfer the Saturated Calcium Hydroxide Solution. Allow the solid Ca(OH)2 to remain settled on the bottom of the flask (in Part A. 1). Carefu lly ( try not to disturb the finely divided Ca(OH)2 solid) decant about 90 mL of the saturated Ca(OH)2 solution into a second 125-mL flask. 4.Prepare a Sample for Analysis. Rinse a 25-niL pipet twice with 1- to 2-mL portions of the saturated Ca(OH)2 solution and discard. Pipet 25 mL of the saturated Ca(OH)2 solution into a 125-mL flask and add 2 drops of methyl orange indicator.

5.Titrate. Titrate the Ca(OH)2 solution to the methyl orange endpoint (see color plate), where the color changes from yellow to red. Remember the addition of HCI should stop within one-half drop of the endpoint. Read (±0.02 mL) and record the final volume of standard HCI in the buret.

6.Repeat. Titrate two additional samples of the saturated Ca(OH)2 solution until I % reproducibility is achieved

7.Do the Calculations. The reported values for the Ksp of Ca(OH)2 will vary from chemist to chemist. Complete your calculations as outlined on the Report Sheet.

Again, three analyses are to be completed. Clean and label three 125- or 250-mL Erlenmeyer flasks. Prepare all three of the Ca(OH)2-CaCI2 samples at the same time.

1. Prepare the Stock Solution. Mix 3 g of Ca(OH)2 and 1 g of CaCI2.2H2O with 120 mL of boiled, deionized water in a 125-mL flask 1 weeks before the experiment. Stir and stopper the flask. Prepare a Buret for Analysis, Prepare the Sample, and titrate. Repeat Parts A.2-A.6.

2. Prepare a Buret for Analysis, Prepare the Sample, and titrate. Repeat Parts A.2-A.6. Disposal: Discard all of the reaction mixtures in the sink, followed by a generous supply of water.

CLEANUP: Discard the HCI solution in the buret into the sink. Rinse the buret twice with tap water and twice with deionized water.

REVIEW QUESTIONS 1. How did the addition of CaCl2 affect the molar solubility of Ca(OH)2? 2. a. In Part A.3, suppose that some solid Ca(OH)2 was accidentally transferred to the titrating flask. What effect does this error have on the reported Ks value? b. As a result of the inadvertent transfer, will the calculated molar solubility for Ca(OH)2 be too high or too low? Explain. 3. If the endpoint in the titration is surpassed in Part A.5, will the reported Ks value be too high or too low? Explain.. 4. Does adding boiled, deionized water to the titrating flask to wash the wall of the flask and the buret tip affect the Ks value of the Ca(OH)2? Explain. 5. How will tap water instead of boiled, deionized water affect the Ks value of Ca(OH)2 in Part A? Hint: How will the minerals in the water affect the solubility of Ca(OH)2 ?

Experiment 8Molar Solubility,

Common-ion Effect

Report Sheet

Name Desk #


A. Molar Solubility and Solubility Constant of Calcium Hydroxide

Trial 1 Trial 2 Trial 3

1.Concentration of standard HCI solution ----------- ---------- -----------(Mol/L)

2. Buret reading,flnal (mL) ----------- ---------- -----------

3. Buret reading, initial (mL) ----------- ---------- -----------

4. Volume of HCI used (mL) ----------- ---------- -----------

5. Amount of HCI added (mol) ----------- ---------- -----------

6. Amount of OH- in satd solution (mol) ----------- ---------- -----------

7. Volume of satd Ca(OH)2 solution (mL) 25.0 25.0 25.0

8. [OH], equilibrium (moL/L) ----------- ---------- ----------

9. [Ca2+] equilibrium (mol/L) ----------- ---------- ----------

10. Molar solubility of Ca(OH)2 ----------- ---------- -----------

1. Average molar solubility of Ca(OH)2 ----------- ---------- -----------

12. KSP of Ca(OH)2 ----------- ---------- -----------

13. Average KSP ---------------------------------------------------------------

B. Molar Solubility of Calcium Hydroxide in the Presence of a Common Ion

Trial 1 Trial 2 Trial 3

1. Concentration of standardized HCI solution (mol,/L) ----------- ---------- ---------

2. Buret reading,.final (mL) ----------- ---------- -----------

3. Buret reading, initial (mL) ----------- ---------- -----------

4. Volume of HCI used (mL) ----------- ---------- -----------

5. Amount of HCI added (mol) ----------- ---------- -----------

6. Amount of OH- in satd solution (mol) ----------- ---------- -----------

7. Volume of satd Ca(OH)2/CaCl2 solution (mL) 25.0 25.0 25.0

8. [OH-], equilibrium (mol/L) ----------- ---------- -----------

3. Molar solubility of Ca(OH)2/CaCI2 ----------- ---------- -----------

10. Average molar solubility of Ca(OH)2/CaCI2 -----------------------------------------------------

Experiment 9

Galvanic Cells, the Nernst Equation

OBJECTIVE To measure the relative reduction potentials for a number of redox couples.

To develop an understanding of the movement of electrons, anions, and cations in a galvanic cell. To study factors affecting cell potentials. To estimate the concentration of ions in solution using the Nemst equation.

When iron corrodes, a change in the oxidation number of the iron atoms occurs; when gasoline bums, a change in the oxidation number of the carbon atoms occurs. A change in the oxidation number of an atom is the result of an exchange of electrons between the reactants of the reaction. A decrease in an oxidation number requires a gain of electrons, whereas an increase in an oxidation number requires a loss of electrons. The substance that lowers its oxidation number gains electrons and is reduced; the substance that increases its oxidation number loses electrons and is oxidized. A chemical reaction that involves the transfer of electrons from one substance to another is an oxidation-reduction (redox) reaction .Experimentally, when copper metal is placed into a silver ion solution, copper atoms spontaneously lose electrons (copper atoms are oxidized) to the silver ions (which are reduced). Silver ions migrate to the copper atoms to pick up electrons and form silver atoms at the copper metal solution interface; the copper ions that form then move into the solution away from the interface. The overall reaction that occurs at the interface is

Cu(s) + 2 Ag+(aq) 2 Ag(s) + Cu+(aq) [1]

This redox reaction can be divided into an oxidation and a reduction half-reaction. Each half-reaction, called a redox couple, consists of the reduced state and the oxidized state of the substance.

Cu(s) Cu2+(aq) + 2 e- oxidation half-reaction [2]

2 Ag+(aq) + 2 e- 2 Ag(s) reduction half-reaction [3]

A galvanic cell is designed to take advantage of this spontaneous transfer of electrons. Instead of electrons being transferred at the interface between the copper metal and the silver ions in solution, a galvanic cell separates the copper metal from the silver ions and forces the electrons to pass externally through a wire, an external circuit. Figure 1 is a schematic diagram of a galvanic cell setup for these two redox couples.

Figure 9.1

The two redox couples are placed in separate compartments, called half-cells. Each half-cell consists of an

electrode, usually the metal (reduced state) of the redox couple, and a solution containing the corresponding cation (oxidized state) of the redox couple. The electrodes of the half-cells are connected by a wire; this is where the electrons flow, providing current for the external circuit.A salt bridge, which connects the two half-cells, completes the construction of the galvanic cell (and the circuit). The salt bridge permits limited movement of ions from one half-cell to the other so that when the cell operates, electrical neutrality is maintained in each half-cell. For example, when copper metal oxidizes in the Cu 2+/Cu half-cell, either anions must enter or copper ions must leave the half-cell to maintain neutrality.

Similarly, when silver ions form silver metal in its half cell, either anions must leave or cations must enter its half-cell to maintain neutrality.The electrode at which reduction occurs is called the cathode; the electrode at which oxidation occurs is called the anode. As oxidation releases electrons to the electrode to provide a current in the external circuit, the anode is designated the negative electrode in a galvanic cell. The reduction process draws electrons from the circuit and supplies them to the ions in solution; the cathode is the positive electrode. This sign designation allows us to distinguish the anode from the cathode in a galvanic cell.Different metals, such as copper and silver, have different tendencies to oxidize; similarly, their ions have different tendencies to undergo reduction. The cell potential of a galvanic cell is due to the difference in tendencies of the two metals to oxidize (lose electrons) or of their ions to reduce (gain electrons). Commonly, a measured reduction potential, the tendency for the metal ion to gain electrons, is the value used to identify the relative ease with which a given metal ion undergoes reduction.A potentiometer, placed in the external circuit between the two electrodes, measures this relative tendency of the electrons to be transferred between the two redox couples. The potentiometer measures a cell potential, Ecell, a value that represents the difference between the tendencies of the metal ions in their respective half-cells to undergo reduction (i.e., the difference between the reduction potentials of the two redox couples).For the copper and silver redox couples, we can represent their reduction potentials as Ecu2+,cu and EAg+.Ag , respectively. The cell potential is therefore

Ecell = Ecu2+,cu - EAg+.Ag [ 4]

Experimentally, silver ion has a greater tendency than does copper ion to be in the reduced (metallic) state; therefore, Ag+ has a greater reduction potential. As the cell potential, Ecell, is measured positive, EAg+, Ag is placed before ECu2+,Cu in Equation 4.The measured cell potential corresponds to the standard cell potential when the concentrations of all ions are 1mol/L and the temperature of the solutions is 25°C.The standard reduction potential for the Ag+(l M)/Ag redox couple, Eo

Ag+.Ag, is +0.80 V, and the standard reduction potential for the Cu2+(l A.I)/Cu redox couple Eo

Cu2+.Cu is +0.34 V. Theoretically, a potentiometer should show the difference between these two potentials, or, at standard conditions,

Ecell = Ecu2+,cu - EAg+.Ag = 0.80 V - (+0.34 V) = +0.46 V [5]

Deviation from the theoretical value may be the result of surface activity at the electrodes or activity of the ions in solution.

In Part A of this experiment, several cells are 'built' from a selection of redox couples and data are collected. From an analysis of the data, the relative reduction potentials for the redox couples are determined and placed in an order of decreasing reduction potentials.

In Part B, the effects of concentration changes and the formation of a complex on the reduction potential of a redox couple are determined. The presence of ammonia in a Cu2+ solution affects the potential of the Cu2+/Cu redox couple by theequation

Cu2+(aq) + 4 NH3(aq) Cu(NH3)4 (aq)

The Nemst equation is applicable to redox systems that are not at standard conditions, most often when the concentrations of the ions in solution are not 1 mol/L. At 25'C, the measured cell potential, Ecell, is related to EO cell and ionic concentrations by

Ecell - Eo cell = 0.0592 logQ [6] n

where n represents the moles of electrons exchanged according to the cell equation. For the copper/silver cell, n - 2; two electrons are lost per copper atom and two electrons are gained per two silver ions (see Equations 1-.3). For dilute ionic concentrations, the reaction quotient, Q, equals the mass action expression for the cell reaction-the product of the molar concentrations of the products divided by the product of the molar concentrations of the reactants, each concentration raised to the power of its coefficient in the balanced cell equation. For the copper/silver cell (see Equation .1)

Q = [ Cu 2+ ] [ Ag+]

In Part C of this experiment, we study, in depth, the effect of that changes in concentration of an ion have on the potential of the cell. The cell potentials for a number of zinc/copper redox couples are measured in which the copper ion concentrations are varied but the zinc ion concentration is maintained constant.

Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)

The Nemst equation for this reaction is Ecell - EO

cell - = 0.0592 Log[ Zn 2 +] [7] 2 [ Cu2+]

Rearrangement of equation 7 yields an equation for a straight line:

Ecell = EOcell + 0.0592 Log[ Zn2+] - 0.0592 Log[ Cu2+] [8]

2 2 if EO

cell + 0.0592 Log[ Zn2+] = b and - 0.0592 = a then 2 2

y = b + axA plot of Ecell versus log[Cu2+] for solutions of known copper ion concentrations is a straight line that has a slope of 2 and an intercept b that leads to the determination of the concentration of the zinc ion .

PROCEDURE The cell potentials for a number of galvanic cells are measured, and the redox couples are placed in order of decreasing reduction potentials, the effects of changes in ion concentrations on cell potentials are observed and analyzed.

PART A: Reduction Potential of Several Redox Couples 1. Collect the Electrodes, Solutions, and Equipment. Obtain a 24-test tubes. Place about 2 ml (three-fourths full) of the solutions in test tubs A1, A2, B 1, and B2 Polish 2-cm strips of copper, zinc, lead, and iron metal with steel wool, rinse briefly with dilute (-1 M HNO3) , and rinse with deionized water. These polished metals are used as electrodes. Check out a potentiometer (or a voltmeter) with two electrical wires (different colors) attached to "alligator" clips.

2.Set Up the Copper/Zinc Cell. Place a Cu strip (electrode) in test tube 1 A1 and a Zn strip (electrode) in tube B2. Roll and flatten a 2 x 2-cm piece of filter paper; wet the filter paper with a 0. 1M KNO3 solution. Fold and insert the ends of the filter paper into test tubes A1 and B2; this is the salt bridge. Connect one electrode to the negative terminal of the potentiometer and the other to the positive terminal.1

3.Determine the Copper/Zinc Cell Potential. If the needle on the potentiometer swings the 'wrong way,' reverse the connections. Read and record the (positive) cell potential. Identify the cathode (positive terminal) and the anode. Write an equation for the reaction occurring at each electrode.

4.Repeat for the Remaining Cells. Determine the cell potentials for all possible galvanic cells that can be constructed from these four redox couples. Be sure to prepare a 'new" salt bridge for each galvanic cell.

You have now combined two half-cells to form a galvanic cell.5. Determine the Relative Reduction Potentials. Assuming the reduction potential of the Zn2+(0. 1 M)/Zn redox couple is -0.79 V, determine the reduction potentials of all other redox couples.2

6.Determine the Reduction Potential of the Unknown Redox Couple. Place a 0.1 M solution and electrode obtained from your instructor in well A3. Determine the reduction potential, relative to the Zn 2+(0.1 M/Zn) redox couple, for your unknown redox couple.

a. Effect of Concentration. Place about 2 mL of 1 M CuSO4 in test tube Dl and 0.001 M CuSO4 in test tube D2. Immerse a polished copper electrode in each solution. Prepare a new salt bridge to connect the two redox couples. Measure the cell potential. Determine the anode and the cathode. Write an equation for the reaction occurring at each electrode.

b. Effect of Complex Formation. Add 5-10 drops of 6 M NH3 to the 0.001 M CuSO4 solution.3 (Caution: do not inhale NH3) Observe any changes in the cell potential. If an expanded scale potentiometer (0-2 or 0-1.5 V) is not available, only calculate the theoretical potentials for each solution in Parts C. 1 -C.3 Complete Parts C.4 and C.5 as suggested.

c. Prepare the Diluted Solutions. Prepare Solutions 1 through 4 as outlined using a 1-mL pipet and 100-mL volumetric flasks. Be sure to rinse the pipet with the more concentrated solution before making the transfer. Use deionized water for dilution "to the mark' in the volumetric flasks. Calculate the molar concentration of each solution. Solution: 10 ml of 0.1 M Cu(NO3)2

Solution 2 :Take 10 ml of solution 1 and diluted to 100 ml Solution 3 Take 1. ml of solution (2) and diluted to 100 ml Solution 4 Take 1 ml of (solution 3) and diluted to 100 ml

d.Measure and Calculate the Cell Potential for Solution 2. The Zn2+/Zn redox couple in test tube B2 is the reference half-cell for this part of the experiment. Place about 2 mL of Solution 2 in test tube C1 and a polished Cu strip. Connect the two half-cells with a "new' salt bridge. Connect the electrodes to the potentiometer and record the potential difference, Ecell. Calculate the theoretical cell potential. (Use a table of standard reduction potentials and the Nemst equation.)

e.Measure and Calculate the Cell Potentials for Solutions 3 and 4. Repeat Part C.2 with Solutions 3 and 4 in wells C2 and C3, respectively. A freshly prepared salt bridge is required for each cell.

f. Plot the Data. Plot Ecell, expt and Ecell, calc (ordinate) versus log[Cu2+) abcissa)on the same piece of linear graph paper for the four concentrations of Cu(NO3)2 solutions in wells A1, Cl, C2, and C3. Have your instructor approve your graph.

g. Determine the Concentration of the Unknown. Obtain a Cu(NO3)2 solution with an "unknown" copper ion concentration from your instructor and place the unknown solution in test tube B3. Determine Ecell as in Part C.2. Using the graph, determine the unknown copper ion concentration in the solution.

Disposal. Dispose of the waste zinc, copper, lead, and iron solutions in the 'Waste Metal Solutions' container. Return the metals to appropriately marked containers.

CLEANUP: Rinse the 24-test tubes plate twice with tap water and once with deionized water. Discard the first rinse in the "Waste Metal Solutions" container and the other rinses into the sink.

REVIEW QUESTIONS 1. Identify the oxidizing and reducing agents in each cell in Part A. a. Cu-Zn d. Fe-Pb b. Cu-Pb e. Zn-Fe c. Zn-Pb 2. List two reasons why an experimentally measured reduction potential would not be equal to the theoretical reduction potential. See Part A.3 of the Report Sheet. 3. Why is it not possible to use a nonmetal as an electrode in a nonmetal-nonmetal ionredox couple,such as the Cl2/Cl- redox couple? Explain.

Experiment 9

Galvanic Cells, the Nernst Equation

REPORT SHEET

Name Desk #


A. Reduction Potentials of Several Redox Couples

Fill in the following table with your observations and interpretations from the galvanic cells.

Galvanic MeasuredCell Ecell Anode Anode Reaction Cathode Cathode Reaction

Cu-Zn ------- ------------ -------------------------------- ----------- ----------------------------

Cu-Pb ------- ------------ -------------------------------- ----------- ----------------------------

Zn-Pb ------- ------------ -------------------------------- ----------- ----------------------------

Fe-Pb ------- ------------ -------------------------------- ----------- ----------------------------

Zn-Fe ------- ------------ -------------------------------- ----------- ----------------------------

1. Compare the sum of the Zn~Pb and Cu-Pb cell potentials with the Zn-Cu cell potential.

2. Compare the sum of the Zn~Fe and Fe-Pb cell potentials with the Zn-Pb cell potential.

3.Arrange the four redox couples in order of decreasing (measured) reduction potentials. List the reduction potential for each redox couple relative to that of the Zn"'(0.1 M/Zn couple, which is -0.79 V. Use a table of standard reduction potentials and the Nemst equation to calculate the reduction potentials for each of these redox couples.

Reduction Potential Reduction Potential

Redox Couple (measured) (calculated) % Error------------------------------- ---------------------------- ----------------------------- ------------------------------------------- ---------------------------- ----------------------------- ------------------------------------------- ---------------------------- ----------------------------- ------------------------------------------- ---------------------------- ----------------------------- ------------

4.Reduction potential of the unknown redox couple: ----------------

B. Effect of Concentration and Complex Formation on Cell Potential

1.Cell potential of "concentration cell":

Anode reaction: ----------------------------------------

Cathode reaction: ----------------------------------------

2.Cell potential from complex formation:

Explain why the potential changes with the addition of NH3(aq).

C. The Nernst Fquation and an Unknown Concentration Complete the following table with the concentrations of the Cu(N03)2 solutions and the measured cell potentials.Use a table of standard reduction potentials and the Nemst equation to calculate the Ecell.

SolutionNumber Concentration of Cu(NO3)2 Log [Cu 2+] Ecell measured Ecell (calculated)

1 0. 1 mol/L -1 ----------------------- -------------------------

2 ---------------------- --------------- -------------------------- -----------------------

3 ---------------------- --------------- -------------------------- -----------------------

4 ---------------------- --------------- -------------------------- -----------------------

Account for any significant difference between the measured and calculated Ecell values. 3. Ecell for the solution of unknown concentration:

Molar concentration of Cu2+ in the unknown:

EXPERIMENT 10

Determination of Orthophosphate

in Water

OBJECTIVETo gain some familiarity with the techniques of spectrophotometric analysis by analysing a water solution for its phosphate content.

APPARUTUS & CHEMICALSwater sample spectrophotometerKH2PO4 (oven-dried) 1-L volumetric flaskconc. H2SO4 l-, 5-, and 10-mL pipets

ammonium vanadomolybdate 100-mL volumetric flasks (6)solution graduated 5-mL pipet

cuvettes CHCI3

balance

DISCUSSIONTripolyphosphates have been found to be extremely effective in enhancing the cleansing ability of detergents; they also are very inexpensive. Their aid in cleaning is probably due, in part, to the stable complexes that they form with Ca2+ and Mg2+, thus softening the water. Their extensive use, however, has very serious side effects on nature.The accelerated eutrophication, or overfertilization, of our lakes has aroused a great deal of ecological concern. Nutrient enrichment enhances the growth of algae and other microscopic organisms. This produces the green scum of an algal bloom on the water surface, masses of waterweeds, and a depletion of dissolved oxygen; it also kills fish and other aquatic organisms and produces malodorous water systems.When the photosynthetically active algae population near a lake's surface rapidly expands, most of the oxygen produced escapes to the atmosphere. After the algae die, they sink to the lake bottom, where they are biochemically oxidized. This depletes the dissolved oxygen needed to support aquatic life. When oxygen is removed, anaerobic decomposition of the algae continues, producing foul odors.Although many factors affect algae growth, the only one that is readily subject to preventive control is the supply of nutrients. The many nutrients important to the growth of algae include phosphorus, carbon, nitrogen, sulfur, potassium, calcium, and magnesium. Many environmentalists have accepted the idea that phosphorus is generally the key nutrient that limits the plant growth that a body of water can support.There are at least four major sources of phosphorus associated with human activity: human and food wastes, fertilizers, industrial wastes, and detergents. Although detergent products contribute only about one-third of the phosphates entering our water systems, curtailing this particular source is a logical place to begin to combat eutrophication.

The phosphate found in natural waters is present as orthophosphate, PO43-, as well as the polyphosphates P2O7

-4

and P3O105-. The species present, PO4

3-, HP O42-, H2PO4

-, or H3PO4, depend on the pH. Trace amounts are also present as organophosphorus compounds. Detergents usually contain triphosphate, P3O10

5- which slowly hydrolyses to produce orthophosphate, PO43- , according to the following reaction:

P3O105- + 2H2O 3PO4

3- + 4H+

In this experiment you will determine the amount of orthophosphate present in a sample from a natural body of water, whose source will be given to you by the laboratory instructor.

Analytical MethodIn dilute phosphate solutions ammonium metavanadate, (NH4VO3), molybdate (MoO4

2-), and phosphate (PO43-)

condense to form an intensely yellow colored compound called a heteropoly acid, whose formula is thought to be (NH4)3PO4 .NH4VO3.16MoO3. The intensity of the yellow color is directly proportional to the concentration of phosphate. The relative amount of color developed is measured with a spectrophotometer. The amount of light absorbed by the sample is directly proportional to the concentration of the colored substance. This is stated by the Beer-Lambert law,

A = abc

where A is absorbance, b is solution path length, c is concentration, and a is absorptivity or extinction coefficient. The colored solutions that you study in this experiment have been found to obey the Beer-Lambert law in the region of wavelengths ranging from 350 nm to 410 nrn. It is convenient to run this experiment at 400 nm. The

amount of phosphate in the unknown sample of interest is determined by comparison with a calibration curve constructed by using a distilled-water reference solution and solutions of known phosphate concentration. The minimum detectable concentration of phosphate is about 0.01 mg/L (10 ppb). The usual experimental precision will lie within about ±l percent of the result obtained by an experienced analyst.

Comparative Phosphate Levels in Water SystemsLimiting nutrients and their critical concentrations are likely to differ in different bodies of water. Analysis of the waters of 17 Wisconsin lakes has led to the suggestion that an annual average concentration of 0.015 mg/L of inorganic phosphorus (0.05 mg phosphate/L) is the critical level above which algal blooms can be expected if other nutrients, such as nitrogen, are in sufficient supply. During the 1968-69 period, Lake Tahoe in Nevada had an average phosphate level of 0.006 mg/L, while its tributaries averaged 0.08mg/L. Lake Tahoe is one of the two purest lakes in the world.

The other is Lake Baikal in Russia. This figure thus represents the lowest natural-water value of phosphate one is likely to find. In July of 1969 the phosphate level of Lahontan Reservoir (about 55 km east of Reno, Nevada) was 0.52mg/L. By comparison, Lake Erie's phosphate level increased from 0.014mg/L in 1942 to 0.40 mg/L in 1967-68. The U.S. Public Health Service has set 0.1mg/L of phosphorus (0.3 mg phosphate/L) as the maximum value allowable for drinking water. Raw sewage contains an average of about 30 mg/L of orthophosphate, of which about 25 percent is removed by most secondary sewage-treatment plants.

PROCEDUREA. Preparation of Calibration CurveDissolve about 136 mg of oven-dried KH2PO4 (weigh accurately) in about 500 mL of water. Quantitatively transfer this solution to a I-L volumetric flask, add 0.5 mL of 98 percent H 2SO4, and dilute to the mark with distilled water. This yields a stock solution that is about I x10 -3 M in various phosphate species. From this stock solution prepare a series of six solutions with phosphate concentrations 2 x 10 -5 , 5 x10-5 , 1 x 10-4 2 x10-4

, 5 x 10-4 , and 7.5 x 10-4 M by appropriate dilution of the stock solution. You must know the precise concentrations of these solutions.Each point on the calibration curve is obtained by mixing 10 mL of the phosphate solution with 5 mL of the ammonium vanadomolybdate solution (see note 1) and measuring the absorbance on the spectrophotometer at 400 nm. Your curve is constructed by plotting absorbance as the ordinate versus concentrations of phosphate as the abscissa. A straight line passing through the origin should be obtained. Your calibration curve should be handed in with your report sheet.

B. Analysis of Water SampleThe unknown samples (A, B, and C) to be analyzed are stored in three test tubes and tightly stoppered. Add 5 mL of the ammonium vanadomolybdate solution to 10 mL of the unknown and measure the absorbances at 400 nm. Note: Preparation of Ammonium vanadomolybdate solution: Dissolve 40 g of ammonium molybdate (molybdic acid, 85 percent MoO3) in about 400 mL of distilled water. Dissolve 1 g of ammonium metavanadate, NH4VO2 in about 300 mL of distilled water and add 200 mL of concentrated nitric acid. Mix the two solutions and dilute to 1 L. This solution is stable for about 90 days and will be provided for your analysis.

REVIEW QUESTIONS

Before beginning this experiment in the laboratory, you should be able to answer the following questions:

1. What volume of 1 x 10-3 M solution is required to make 50 mL of solution with the following concentrations: 2 x 10-5,5 x 10-5, 1 x 10-4,2 x 10-4,5 x 10-4, and 7.5 x 10-4 M?

2. What species is thought to be the light-absorbing species in this experiment?

3. Write a balanced chemical equation for the formation of the light- absorbing species that results from reaction of phosphate and ammonium vanadomolybdate.

4. State the Beer-Lambert law and define all terms in it. 5. What are the five fundamental components of a spectrophotometer? 6. Why is a calibration curve constructed? How? 7. How do you know whether to measure the absorbance of a more dilute or more concentrated solution if the absorbance of your unknown solution is not within the limits of your calibration curve?

8. A 0.0750 M sample of CO(NO3)2 gave an absorbance of 0.38 at 505 nm in a l-cm cell. What is the cobalt concentration of a solution giving an absorbance of 0.52 in the same cellat the same wavelength?

9. A 17.28-ppm (1 ppm = 1 mg/L) solution of FeSCN2+ has a transmittance of 0.59 when

measured in a 1.00-cm cell at 580 nm. Calculate the extinction coefficient for FeSCN2+ at this wavelength. 10. Define eutrophication.

11. If raw sewage contains 30 mg/L phosphate and a secondary sewage- treatment plant removes 25 percent of the phosphate, would a secondary treatment plant provide potable water if 0.3 mg/L is the maximum phosphate concentration allowable in drinking water?

12. Write a balanced chemical equation for the hydrolysis of triphosphate, P3O105-, to orthophosphate, PO4

3-.

Experiment 10

Determination of Orthophosphatein Water

REPORT SHEET

Name Desk #


A.Preparation of Calibration Curve Concentration Absorbance

mol/L PO43-

A-------------- ----------------B -------------- ----------------C -------------- ----------------D -------------- ----------------E-------------- ----------------F -------------- ----------------

B. Determination of the concentration of PO43- in the unknown

Absorbance Concentration of PO4

3- A-------------- ----------------

B -------------- ----------------

C -------------- ----------------

EXPERIMENT 11 Water Analysis

OBJECTIVETo learn how water sample is collected and storedPerform some chemical analyses on water sample.

APPARATUS AND CHEMICALSconductivity devices (Figure 11.1) hot plates150-mL beakers pH meter with electrodes 100 mL, graduate cylinders 100 mL, 25-mL pipets Buffer solutions weiging papersvacuum filtration apparatus filter paper analytical balance

INTRODUCTIONWater is the universal solvent, many contaminants (impurities) are easily dissolved upon contact. They may give water a bad taste, color, odor, or cloudy appearance (turbidity), and cause hardness, corrosiveness, or staining. They can also damage growing plants and transmit disease. At low levels, impurities generally are not harmful in water. Removing all contaminants would be extremely expensive and in nearly all cases would not provide greater protection of health. At high level (waste water) many of these impurities are treated and removed or rendered harmless. Chemists are concerned with the purity of water but regulatory agencies are concerned with setting standards to protect the environment and public health. One mean of establishing and assuring the purity and safety of water is to meet standards for various contaminants found in water. In this project you be able to get some practice on how water is collected and stored and to perform some routine chemical analysis. a. Sample Collection and StorageSite data should be recorded for all sampling locations. The information generally required includes time, date, grid references of site, weather. Temperature, method of collection and information about any local activities that might influence the results. Some of these data may only be applicable for certain classes of water samples. A number of sampling devices are available for taking water samples from small ponds and from different depths in large stratified lakes. The simplest system uses a weighted bottle which is suspended at the required depth. The stopper is then removed by a sharp pull on a separate line.

Water samples are especially subject to alteration in chemical composition due to microbiological activity and chemical reactions. Heavily polluted waters can undergo changes in composition within an hour of collection, and most natural waters are affected to some degree. Some tests (particularly pH and dissolved gases) should, if possible, be carried out in the field. Glass sample containers are frequently recommended for storage since polythene vessels can be porous to gaseous constituents and have been round to absorb phosphorus. In general, to minimize possible bias of results caused by any changes occurring during storage it is important to: 1 Analyze the samples as soon as possible. Ensure that solution collectors at sites are emptied regularly. 2 Fill containers to exclude air. 3 Keep sample cool, but do not freeze.

Physical preservation methods Fine filtration If the interest is only in the dissolved fraction then fine filtration, which removes many of the microorganisms, can be applied. It will also remove fine mineral matter and any traces of turbidity which could affect a later analytical stage. An alternative approach is centrifuging which can be used to separate various particle sizes.

Temperature reduction Although some microbial activity appears to continue even at 0 °C the rapid cooling of samples after collection is generally to be recommended.

Preliminary and general tests For reasons given in the previous section tests on waters should be made as soon as possible after sampling and in some cases in the field. This particularly applies to labile and gaseous constituents.

Odor, turbidity, and color Odor can serve as a guide to gross pollution of water. For example, characteristic odors are associated with chlorination plants, untreated sewage and chemical industry effluents. Color in water may be a true color due to dissolved material or an apparent color when suspended material is present. The latter is quite common in natural waters, seen for example when algal blooms impart a greenish tinge.

Turbidity may be used as estimate of undissolved substances in the sample. It is generally measured by visual comparison with standards or photometrically, using a neophelometer or spectrophotometer. Turbidity and color control light penetration in lake which in turn affect phytoplankton population.

SOLIDTotal suspended solids (TSS)The finer suspended matter in natural waters is usually of an organic nature representing colloidal matter, which has been flocculated under the influence of bacteria and protozoa. Inorganic suspended matter is chiefly restricted to siliceous material resulting from the erosion of mineral soils.

Total dissolved solids (TDS) It is often convenient to determine the dissolved solids in the filtrate remaining from the TSS determination.

Total organic matter (TOM)TOM is all organic matter that can be found in a given water sample

Alkalinity (and acidity) The alkalinity of water is its capacity to neutralize a strong acid, and the values obtained will depend on the pH of the titration end-point. In practice it is the bicaronate, carbonates and hydroxides in solution that largely determine the alkalinity although there are minor contributions from silicates and other anions.Total alkalinity is determined by titration to the equivalent point of carbonic acid which occurs between pH 4.2 and 5.4 depending on the carbon dioxide content of water.It bas long been the practice in water analysis to determine solids as dissolved, suspended and organic.

ConductivityConductivity is a property of water governed by the total ionic content. Although it is nn specific and varies with the proportion of species presents, it is often measured, because of its value in characterising waters. It expresses the resistance of 1 cm cube of water to the passage of a current, usually at 25°C (specific resistance).

PROCEDURE

1. TSS determination:Filtrate 100 ml of your sample and the determine the weight of the solid in the filter paper

2. TDS determinationEvaporate the filtrate (liquid) to a small volume (from 100 to 50 ml)Transfer to a weighed 100 ml beaker for evaporation. Dry at 105°C to a constant weight. Cool and weigh. Express the result as mg/l

3. TOM determination Transfer to a small pre-weighed evaporating beaker 50ml of your sample. Evaporate to dryness at constant temperature. and weigh beaker plus contents. Ash in a muffle furnace, leaving at 500°C for 1 hour. The loss in weight of the residue gives the TOM in the sample. The method is only approximate and estimates of total organic carbon are preferable when organic contents are low.

4. Alkalinity (and acidity) Measure pH of water by a pH meter If pH> 8.3 a dd 3-4 drops of phenolphthalein indicator and titrate against 0.01 M HCI.

5.Conductivity Add the unfiltered sample into a two beakers and bring to the required temperature (preferably 25 °C) by immersion in a water bath. Immerse the electrodes in each beaker that contain water samples . Record the conductivity in the second tube (having used the first as a rinse). Check the sample temperature just after immersion of the electrode.

From the book:

A.P. Rowland & H.M. Grimshaw, in ChemicalAnalysis ofEcologicalMaterials (S.T. Allen, Editor) 2nd Edition, Oxford, UK: Blackwell, 1989, p. 62.

Experiment 11Water Analysis (I)

.REPORT SHEET

Chemistry 1402 Field Trip

Student Names: ------------------------------------------------------------------------

Observations (in the field)

Water temperature pH Conductivity Amount of plants, or living things you observe

Think about it and ANSWER these questions in the field

Areas with large amount of water insects and underwater plants are warmer because these living things produce heat -Compare your results with other groups that collected water from areas with and without plants/insects, is there a temperature difference?

Plants need CO2 from the water as food. When CO2 dissolves in water, an acid is produced, if CO2 is removed from water by the plants. What would happen to the pH in areas where there is plenty of plants? Does this agree with your observations of pH

The conductivity is a measure of the amount of ions in water. Plants need CO3

-, NO3- and PO4

3-ions and sunlight to produce food. Given the conductivity measurements that you have made water that you took and combining it with the pH measurements, do you think that there can be life in thc area where you took the water?

The solubility of ionic compounds in water depends on the temperature. Higher temperatures dissolve more of the compound to produce more ions. Does your measurement of conductivity and temperature respectively, when compared to that of other groups, agree with this observation?

Experiment 11

Water Analysis (II)

REPORT SHEET

(Water analysis will be performed in AUI Chemistry laboratory)

SOLIDMass of total suspended solids (TSS) in mg --------------- Mass of total dissolved solids (TDS) in mg ---------------- Mass of total organic matter (TOM) in mg -----------------

Alkalinity (and acidity) pH ------------------- If pH is larger than 8.3 continue otherwise skip to conductivity measurements

Concentration of standard HCI solution

(0.01mol/L) Trial 1 Trial 21. Buret reading,flnal (mL) ----------- ----------

2. Buret reading, initial (mL) ----------- ----------

3. Volume of HCI used (mL) ----------- ----------

4. Amount of HCI added (mol) ----------- ---------- -

5. Amount of OH- in satd solution (mol) ----------- ----------

6. Volume of sample solution (mL) 25.0 25.0

7. [OH], equilibrium (mol/L) ----------- ----------

Conductivity in µ MHOS -------------

EXPERIMENT 12

Determination of Chemical Oxygen Demand (COD) in Water From Ifrane Lake

OBJECTIVEThe objective of this experiment is to determine the chemical oxygen demand (COD) of a water sample using a standard method of Open Reflux Method ( potassium dichromate digestion in an open-reflux condenser followed by titration with ferrous ammonium sulfate).

APPARATUS AND CHEMICALS COD apparatus (Figure 12.1) 2 x buret clamps, and ring stands 0.04167 M K2Cr2O7 (standard solution) 3 x 25-mL pipets0.25 M Fe(NH4)2(SO4)2·6H2O (standard) 3 x 150-mL beakers solution) 100 mL, graduate cylinderHOOCC6H4COOK (standard solution) magnet stirrer Ferroin indicator

DISCUSSION the COD test is used as a measure of the oxygen equivalent of the organic matter content of a sample that is susceptible to oxidation by a strong chemical oxidant The analysis of the COD value in waste and surface waters is one of the most expressing factor to determine the degree of pollution of the water. COD reflects the total quantity of oxidizable components whether it is carbon (C) and hydrogen (H) from hydrocarbons, nitrogen (N) from e.g. proteins, or sulfur (S) and phosphorus (P) from detergents. In a COD test, results can be obtained in 2 hours or less when the sample allows (e.g. time for water sample with a predictable quickly oxidized component (like sugar) can be set to 15 min). Also the method is simple and inexpensive. Unlike the TOC value (Total Organic Carbon), which express only the carbon coming from pollutants. On the other hand BOD (biological oxygen demand) test provides the closest measure of the processes actually occurring in the natural water system. However, BOD is very time-consuming (5 days) and involves many uncertain factors such as the origin, concentration, pollutants, and the number and viability of active microorganisms present to affect the oxidation of all pollutants. One disadvantage of the COD method is that dichromate can oxidize materials that would not ordinarily be oxidized in nature. Therefore COD test is unable to differentiate between biologically oxidizable and biologically inert organic matter. The COD test can also generate a large volume of liquid hazardous waste (acid, chromium, silver, and mercury).

The COD test is used extensively in the analysis of industrial wastes. It is particularly valuable in surveys designed to determine and control losses to sewer systems. Results may be obtained within a relatively short time and measures taken to correct errors on the day they occur. In conjunction with the BOD test, the COD test is helpful in indicating toxic conditions and the presence of biologically resistant organic substances. The test is widely used in the operation of treatment facilities because of the speed with which results can be obtained.

The most stable oxidizer agent is a solution of potassium dichromate (K 2C12O7) mixed with sulfuric acid (H2SO4). The dichromate oxidation is normally 95-100% complete for most organic substance. However, it will not oxidize a number of refractory molecules including aromatic hydrocarbons, pyridine and related

compounds, and straight-chain aliphatics. Volatile organic compounds are oxidized only to the extent that they remain in contact with the oxidant. Another very suitable oxidizer agent is a solution of potassium permanganate (KMnO4). This is often used in the potable water industry. The permanganate test has the advantage to release a non-toxic waste.

METHODE PRINCIPLEThe water sample being measured is refluxed with excess of potassium dichromate in concentrated sulfuric acid for ~2 hrs. The reaction involved in the usual case, where organic nitrogen is all in a reduced state (oxidation number of –3), may be represented in a general way as follows:

CnHaObNc + dCr2O72- + (8d + c) H+ nCO2 + (a + 8d –3c)/2 H2O + cNH4

+ + 2dCr3+ [1]

Where d = 2n/3 + a/6 – b/3 – c/2. The reaction requires strong acidic conditions and elevated temperature (150°C). During the COD determination, organic matter is converted to CO2 and water. Organic nitrogen in a reduced state will be converted to NH4

+, while higher oxidation states will be converted to nitrates. Silver sulfate (Ag2SO4) may be included to catalyze the oxidation process if samples contain alcohols or low molecular weight fatty acids.

Reduced inorganic species such as Fe2+, S2-, Mn2+ are oxidized under the test conditions. Chloride ions give a positive interference by the reaction:

Cr2O72- + 6Cl- + 14H+ 2Cr3+ + 3Cl2 + 7H2O [2]

The interference is reduced by addition of mercury sulfate (HgSO4). Mercuric ion combines with the chloride ions to form a poorly ionized mercuric chloride complex:

Hg2+ + 2Cl- HgCl2 [3]

In the presence of excess mercuric ions the chloride ion concentration is so small that it is not oxidized to any extent by dichromate.

After dichromate digestion, the excess of dichromate is titrated with ammonium iron (II) sulfate:

6Fe2+ + Cr2O72- + 14H+ 6Fe3+ + 2Cr3+ + 7H2O [4]

The indicator used for the above equation is a chelating agent 1,10-phenanthroline (ferroin). When all Cr2O7

2- is reduced, ferrous ions (Fe2+) react with ferroin to form a red-colored complex:

Fe2+ + 3C12H8N2 Fe{C12H8N2}3 [5]Note that Cr2O7

2- has a yellow to orange brown color depending on the concentration, and Cr3+ has a blue to green color. So the color of the solution during the titration starts with an orange brown, and then a sharp change from blue-green to reddish brown, which corresponds to the color of Cr2O7

2-, Cr3+, and Fe{C12H8N2}3, respectively.

REAGENT PREPARATIONS

a. Dichromate standard solution 0.250 N:Standard potassium dichromate solution, o.250 or 0.04167 M : Dissolve 12.259 g K2Cr2O7 (MW = 294.2) in 1 L distilled water. (What is the concentration in normality? You need this number for later calculation.)

b. Standard ferrous ammonium sulfate solution (FAS) 0.125 NStandard ferrous ammonium sulfate (FAS) titrant (approximately 0.125N or 0.25 M): 98 g Fe(NH4)2(SO4)2·6H2O (MW = 392.14) in 1 L of distilled water. However, the morality of FAS solution is to be controlled each day of analysis, it decreases with time since FAS solution can be slowly oxidized by oxygen, hence standardization is required.. The variation of the molarity must be considered when computing COD. This is obtained by introducing in one DCO reaction tube the reagents given in the following table:

V(0.250 NK2Cr2O7) in ml V(H2O) in ml

V(H2SO2) 96 °/° in ml

10 100 30

Follow the experimental procedure given herewith. c. Potassium hydrogen phtalate standard solution (PHF)Using hydrogen phtalate (HOOCC6H4COOK)dried at 120°C, a solution containing 425 mg in 100 ml of H2O is prepared. The theoretical value of COD of this solution is 500 mg/l. d. Sulfuric acid and silver sulfate solution:5.4 g of silver sulfate analytical grade (Ag2SO4) are added to 1 Kg of H2SO2 96 °/° analytical grade (d=1.835) corresponding to 545 ml. Complete solution requires two days. The reagent is preserved in a well stoppered and dark bottle.

PROCEDURE A. Calibration of standard ferrous ammonium sulfate (FAS) solution 1. Transfer proper volume 100 ml of distilled water into DCO reaction tube. Label your DCO reaction tube and add few boiling chips. Use a pipet to transfer accurately 10 mL 0.250 N K2Cr2O7 solution into DCO reaction tube and diluted to 100 ml with H2O and slowly add 30 mL of sulfuric acid 96 °/° and mix the reactive slowly. Wipe the outside of the reaction tube and make sure there are no spills. Mix reflux mixture thoroughly before applying heat to prevent local heating of flask bottom and a possible blowout of flask contents.

follow the instructions of the professor on how to operate a DCO machinethe time for digestion for this experiment is about 1 hour.

2. Add 2-3 drops ferroin indicator; Titrate with FAS titrant. You should expect the color change at the endpoint from blue-green to reddish brown. Record the volume (V2) of FAS used.

Molarity of FAS solution = Volume of 0.250 N K2Cr2O7 solution (mL) * 0.2500/Volume FAS used in titration (mL)

Calculation the Normality of FAS: Normality (N2) of FAS solution = N1V1/V2, where

N1 = normality of K2Cr2O7

V1 = volume of K2Cr2O7 (V1 = 10.0 mL)V2 = volume of FAS used in titration (mL)

B. COD of Water Sample and Blank

1. Wash the reaction tubs thoroughly to exclude extraneous contamination of organic compounds. Each group should work with two water samples: one blank sample (Sidi Ali) and water sample from Ifrane lake. 2. Transfer proper volume (e.g., 20, 20, or 10 ml) for blank, water sample, or COD calibration standard, respectively) into two or three DCO reaction tubs. Add few boiling chips.3. Carefully and very slowly add 30.0 ml of solution d (sulfuric acid and silver sulfate solution) into each flask.4. Cool while mixing to avoid possible loss of volatile organics.5. Use a pipet to transfer accurately 10.0 mL of solution a (0.250 N K2Cr2O7 solution) into each reaction tube . Swirl and mix while adding the concentrated sulfuric acid reagent. Wipe the outside of the flasks and make sure there are no spills. Mix reflux mixture thoroughly before applying heat to prevent local heating of flask bottom and a possible blowout of flask contents.6. Follow the professor instructions on how to operate a DCO machinethe time for digestion for this experiment is about 1 hour since the water sample is not polluted.

7. Add 2 to 3 drops of ferroin indicator. Titrate the excess K2Cr2O7 with standard ferrous ammonium sulfate (FAS) solution. Your sample should have an initial orange brown color, the titration should have a sharp color change from blue-green to reddish brown at the endpoints. Record the volumes for the blank (A), your water sample (B) and the COD standard sample (C).8. Calculation:COD as mg O2/L = (A-B) x N x 8000 / mL water sample, orCOD as mg O2/L = (A-C) x N x 8000 / mL COD standard sample A = mL FAS used for blank,B = mL FAS used for your water sample, C = mL FAS used for your COD standard sample, and N = normality of FAS as determined in part A. 9. Dispose of your wastes according to the instruction.

Safety Issues1) Safety glasses must be worn at all times in this experiment, especially during digestion. Please bring your own goggle. 2) Do not exceed its recommended temperature (150oC). Shield the block and keep the door of the hood closed during digestion. 3) The reaction tubes contain concentrated sulfuric acid, potassium chromate (a strong oxidizing agent and possible carcinogen), and a very toxic mercury salt (if your sample contains Cl -). Use gloves when handling these solutions.4) Drain disposal of the waste (acid, chromium, silver and mercury) is prohibited. The spent liquid waste must be disposed into waste storage tanks.

Experiment 12

Determination of Chemical Oxygen Demand (COD) in Water Sample

REPORT SHEET

Name Desk #


Part ANormality of FAS: V2 = volume of FAS used in titration (mL) --------------

Normality (N2) of FAS solution = N1V1/V2 --------------N1 = normality of K2Cr2O7 (0.250N)

V1 = volume of K2Cr2O7 (V1 = 10.0 mL)

Part B: A = mL FAS used for blank --------------B = mL FAS used for your water sample --------------C = mL FAS used for your COD standard sample --------------N = normality of FAS as determined in part A. --------------

COD as mg O2/L = (A-B) x N x 8000 / mL water sample --------------COD as mg O2/L = (A-C) x N x 8000 / mL COD standard sample --------------

experiment 4 - al akhawayn university in ifrane - homea.ouardaoui/experim4-12.doc · web viewuse...

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