Analytical chemistry

Analytical chemistry is the area of chemistry responsible for characterizing the composition of matter, both qualitatively (what is present) and quantitatively (how much is present).

Analytical chemistry is the study of the separation, identification, and quantification of the chemical components of natural and artificial materials.

Analytical chemistry is also focused on improvements in experimental design, chemometrics, and the creation of new measurement tools to provide better chemical information.

Analytical chemistry has applications in forensics, bioanalysis, clinical analysis, environmental analysis, and materials analysis.

qualitative analysisAn analysis in which we determine the identity of the constituent species in a sample.

quantitative analysisAn analysis in which we determine howmuch of a constituent species is presentin a sample.

characterization analysisAn analysis in which we evaluate asample’s chemical or physical properties.

fundamental analysisAn analysis whose purpose is to improvean analytical method’s capabilities.

Chemical Analysis : A process that provides chemical or physical information about the constituents in the sample or the sample itself.

Qualitative analysis Quantitative analysis

gives an indication of the identity of the chemical species in the sample.

determines the amount of one or more of these components

what is present? how much is present?

Analytical methods can be separated into classical and instrumental

Classical methods (also known as wet chemistry methods)

Instrumental methods

use only chemicals for separations such as precipitation, extraction and qualitative analysis by color, odor, or melting point. Quantitative analysis is achieved by measurement of weight or volume.

Gravimetric analysis

Volumetric analysis

(Chemical tests )

use an apparatus to measure physical quantities of the analyte such as light absorption, fluorescence, or conductivity.

Spectroscopy

Electrochemistry

Chromatography

a sample is a limited quantity of something which is intended to be similar to and represent a larger amount of that thing(s).

The components of interest in the sample are called analytes,

the remainder of the sample is the matrix.

Technique is a chemical or physical principle that can be used to analyze a sample.

A method is the application of a technique for the determination of a specific analyte in a specific matrix.

A procedure is a set of written directions detailing how to apply a method to a particular sample.

Constituent : A component of a sample; it may be further classified as:

A major constituent % 1-100

Minor constituent % 0,01-1

Trace constituent ˂ %0,01

Statistical Analysis of the Results

• Measurement : An experimental determination of an analyte’s chemical or physical properties.

• Measurement data give only an estimate of the ‘true’ value.

• Length : • Mass :• Time:• Temperature:

True Result The 'correct' value for a measurement which remains unknown except when a standard sample is being analysed. It can be estimated from the results with varying degrees of precision depending on the experimental method.

Mean The mean, X, is the numerical average obtained by dividing the sum of the individual measurements by the number of measurements. (The average value of a set of data -X-).

Median The median, Xmed, is the middle value when data are ordered from the smallest to the largest value. Or the value for a set of ordered data, for which half of the data is larger in value and half is smaller in value ( –Xmed).

Range (Spread )

The range, w, is the difference between the largest and smallest values in the data set. It is a measure of precision. Range = w = Xlargest – Xsmallest

N ∑ xi

i=1 X1 + X2 +X3 +......+ Xn X = ------------- = ---------------------------------------- N NMean

Precision is the closeness of data to other data that have been obtained in exactly the same way. Sometimes called the variability, it can be represented statistically by the standard deviation or relative standard deviation.

Accuracy The closeness of an experimental measurement or result to the true or accepted value. Accuracy is usually expressed in terms of error or absolute error. (Bias)

Error The difference between the true result and the measured value.. E = xi - xt

Absolute Error defined as the actual difference between the true result and the experimental value in the same units.

Relative Error

is the absolute error divided by the true value.

Uncertainty The range of possible values for a measurement.

xi - xt

Er = ----------- . %100 Xt

(percentage relative error)

xi - xt

Er = ------------- . ‰1000 Xt

(parts per thousond)

Calculate the mean, median, absolute and relative error for the following data set

Fe concentration

19,4ppm19,5ppm19,6ppm19,8ppm20,1ppm20,3ppm

Classification of Errors

On the basis of their origin, errors may usually be classified as determinate or indeterminate.

Systematic (determinate) Error

Are the errors which can be avoided or whose mahnitude can be determined. They are (in principle at least) measurable and for which a correction may be made.

-         operational and personal error-         instrumental and reagent error-         errors of methods- Sampling Errors

Random (indeterminate) Error

They are due to causes over which the analyst has no control and which in general are so intangible taht they are incapable of analysis. They do not have a definite measurable value.

Gross Error

Statical Evulation of the Random (indeterminate) Error

– (x-μ)2

/2 σ2

e y = --------------------------------------

σ (2 П)^ 1/2

x herbir ölçümün değeriμ bir çok ölçümün aritmetik ortalamasıx – μ ortalamadan sapma σ standart sapma y her bir x - μ değeri için bulunma frekansı

Deviation

Standard Deviation

is a measure of the precision of a population of data

Variance (σ2 or s2) The square of the standard deviation. The variance is a parameter describing in part either the actual probability distribution of an observed population of numbers, or the theoretical probability distribution of a sample of numbers.

Relative Standard Deviation (RSD)

coefficient of variation

is the percent relative standard deviation

Degree of Freedom An independent variable. For N measurements it is equal to N -1

Calculate the standard deviation of the following experimental resultsLead concentration in blood

0,752ppm0,756ppm0,752ppm0,751ppm0,760ppm

∑ (xi – x1 )2 + ∑ (xi – x2 )2 + ... Spool = ( --------------------------------------------)^1/2

N1 + N2 + N3 ... – Nt

• Calculate the stardard deviation of the method.

Sample No Number of measurements Hg concentration ppm

1 3 1,80 1,58 1,64

2 4 0,96 0,98 1,02 1,10

3 2 3,13 3,35

4 6 2,06 1,93 2,12 2,16 1,89 1,95

5 4 0,57 0,58 0,64 0,49

6 5 2,35 2,44 2,70 2,48 2,44

7 4 1,11 1,15 1,22 1,04

What are the standard deviation, the relative standard deviation, and thepercent relative standard deviation for the folowing data

3.056 - 3.080 - 3.094 - 3.107- 3.112 - 3.174- 3.198

What are the standard deviation, the relative standard deviation, and thepercent relative standard deviation for the following data

3.056 - 3.080 - 3.094 - 3.107- 3.112 - 3.174- 3.198

1,76 (±0,03) + 1,89 (±0,02 ) - 0,59(±0,02) = ?

[ 1,76 (±0,03) x 1,89 (±0,02 ) ] / 0,59(±0,02) =?

The pH of a solution is defined as pH = –log[H+]

where [H+] is the molar concentration of H+. If the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, what is the [H+] and its absolute uncertainty?

The pH of a solution is defined as pH = –log[H+]

where [H+] is the molar concentration of H+. If the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, what is the [H+] and its absolute uncertainty?

The Nature and Origin of Errors

where μ is the mean and s is the standard deviation. The width of the curve is determined by s, which is a useful measure of the spread or precision of a set of results, and is unique for that set of data. An interval of μ ± s will contain 68.3% of the statistical sample, whilst the intervals μ ± 2s and μ ± 3s will contain 95.5% and 99.7% respectively.

On the basis of their origin, errors may usually be classified as determinate or indeterminate. The first are those having a value which is (in principle at least) measurable and for which a correction may be made. The second fluctuate in a random manner and do not have a definite measurable value.

Indeterminate errors arise from the unpredictable minor inaccuracies of the individual manipulations in a procedure. A degree of uncertainty is introduced into the result which can be assessed only by statistical tests. The deviations of a number of measurements from the mean of the measurements should show a symmetrical or Gaussian distribution

Confidence Intervals for Populations

confidence interval: Range of results around a mean value that could be explained by random error.

there is a 68.26% probability that a member selected at random from a normally distributed population will have a value in the interval of ± 1.

In general, we can write

Xi = ± z

where the factor z accounts for the desired level of confidence. Values reported in this fashion are called confidence intervals.

Confidence intervals also can be reported using the mean for a sample of size n, drawn from a population of known . The standard deviation for the mean value, , which also is known as the standard error of the mean

What is the 95% confidence interval for the analgesic tablets if an analysis of five tablets yields a mean of 245 mg of aspirin? the population standard deviation for the amount of aspirin in a batch of analgesic tablets is known to be 7 mg of aspirin.

What is the 95% confidence interval for the analgesic tablets if an analysis of five tablets yields a mean of 245 mg of aspirin? the population standard deviation for the amount of aspirin in a batch of analgesic tablets is known to be 7 mg of aspirin.

Confidence Intervals for Samples

Once the reliability of a replicate set of measurements has been established the mean of the set may be computed as a measure of the true mean. Unless an infinite number of measurements is made this true mean will always remain unknown. However, the t-factor may be used to calculate a confidence interval about the experimental mean,

the confidence interval as a way to report the most probable value for a population’s mean, , when the population’s standard deviation, , is known.

Since s2 is an unbiased estimator of 2, the term z inThe confidence interval equation is replaced with the variable t, where t is defined such that t z at allconfidence levels.

t is a statistical factor derived from the normal error curve s is the estimated standard deviation and N is the number of results.

t is a statistical factor derived from the normal error curve s is the estimated standard deviation and n is the number of results.

What is the 95% confidence interval for the folowing data?

3.056 - 3.080 - 3.094 - 3.107- 3.112 - 3.174- 3.198

The accepted value for the chloride content of a standard sample obtained from extensive previous analysis is 54.20%.

Five analyses of the same sample are carried out by a new instrumental procedure,

54.01, 54.24, 54.05, 54.27, 54.11%

being the results obtained. For the 95% confidence interval, is the new method giving results consistent with the accepted value?

The accepted value for the chloride content of a standard sample obtained from extensive previous analysis is 54.20%.

Five analyses of the same sample are carried out by a new instrumental procedure,

54.01, 54.24, 54.05, 54.27, 54.11%

being the results obtained. For the 95% confidence interval, is the new method giving results consistent with the accepted value?

Dixon’s Q-test. Statistical test for deciding if an outlier can be removed from a set of data. The Q-test compares the difference between the suspected outlier and its nearest numerical neighbor to the range of the entire data set.

outlier: Data point whose value is much larger or smaller than the remaining data.

differences between outlier and Nearest value to it Qexp = -------------------------------------------------------------------------

Range (spread)

Qexp is greater than Q(, n), then the null hypothesis is rejected and the outlier may be rejected. When Qexp isless than or equal to Q(, n) the suspected outlier must be retained.

Q (% 90 confidence interval)

0,94 0,76 0,64 0,56 0,51 0,47 0,44 0,41

Number of measurements 3 4 5 6 7 8 9 10

The following masses, in grams, were recorded in an experiment to determine the average mass of a U.S. penny.

3.067 3.049 3.039 2.514 3.048 3.079 3.094 3.109 3.102

Determine if the value of 2.514 g is an outlier at % 90 confidence level.

Comparing Two Sample Means

Tables below show results for two separate experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the means of these analyses at a %95 confidence level.

Tables below show results for two separate experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the means of these analyses at a %95 confidence level.

The %w/w Na2CO3 in soda ash can be determined by an acid–base titration.The results obtained by two analysts are shown here. Determine whether thedifference in their mean values is significant at a %95 confidence level

Comparing Two Sample Variances

F-test Statistical test for comparing two variances to see if their difference is too large to be explained by indeterminate error. The F-test can be extended to the comparison of variances for two samples

If Fexp is greater than Fcri, the analysis is not under statistical control.

• Tables below show results for two separate experiments to determine the mass of a circulating U.S. penny. Determine whether there is a difference in the precisions of these analyses at %95 confidence level. the variances are s2A = 0.00259 and s2B= 0.00138.

A new method for Ni analysis in the stainless stel is being compared with another method, is the new method is acceptable?

the results of the analysis of vitamin B are shown below,. 81,10 79,30 80,40 79,70 Calculate confidence interval for 95% confindence level

If the true value 80,00, is there any significant difference between mean values at a %95 confidence level

Sistematik Belirli Hata

1- Sabit hatalar : analiz edilen numune miktarından bağımsızdır.2- Orantılı hatalar : numunenin miktarı ile orantılı olarak azalır veya

artar

F Testi (standart sapmaların karşılaştırılması)

• İki farklı kişinin yaptığı analizlerin veya iki farklı metotla elde edilen sonuçların standart sapmalarının karşılaştırılmasında kullanılır. Farklı sonuçların standart sapmalarının karesi birbirine bölünür ve sonuç tablodaki F değeri ile karşılaştırılır.

Aqueous Solution Chemistry

Precipitation Reactions

Acids, Bases, and Neutralization Reactions

Oxidation–Reduction (Redox) Reactions

Complex formation reactions

Solutions

Homogeneous mixtures are called solutions.

Solvent solute

is the component present in greatest amount or determines the state of solution exist

is a component present in a lesser quantity and it is said to be dissolved in the solvent.

Solvent is the component present in greatest amount or determines the state of solution exist.

A solute is a component present in a lesser quantity and it is said to be dissolved in the solvent.

Solubility is the maximum amount of solute that dissolves in a given quantity of solvent at a specific temperature to produce a saturated solution

saturated solution.

The solution in which no more solute can be dissolved is said to be saturated solution.

Super saturated solution.

is a solution in which has higher amount of solute than saturated solution

Unsaturated solution

is a solution in which contain less amount of solute than saturated solution

nonelectrolyte Substances such as sucrose or ethyl alcohol, which do not produce ions in aqueous solution, are called nonelectrolytes.

electrolytes Substances such as NaCl or KBr, which dissolve in water to produce conducting solutions of ions, are called electrolytes.

strong electrolytes, weak electrolytes.

Compounds that dissociate to a large extent (~100%) into ions when dissolved in water are said to be strong electrolytes,

compounds that dissociateto only a small extent are weak electrolytes.

Mass of solute

Percent by mass = ------------------------ x 100

Mass of solution

Volume of solute

Percent by volume = ------------------------- x 100

Volume of solution

Mass of solute

Mass / volume percent = --------------------- x 100 (gr/100ml)

Volume of solution

Solution Concentrations

Solution Concentrations

Number of moles of soluteMolarity = ----------------------------------------------- Number of liters of solution

Number of moles of soluteMolality = -------------------------------------------------- Number of kilograms of solvent

Moles of component iMole fraction = --------------------------------------------------------- total moles of all solution components

Parts per million (ppm), Parts Per Billion(ppb) mg solute g solute1 ppm = -------------------- 1 ppb = ----------------------- kg solution kg solution

Mol number

mn = --------- MA

Molarity nM= ------- mol/L V

density md = ---------- gr/mL V

Diluting Concentrated Solutions

Minitial x Vinitial = mol number = Mfinal x Vfinal

Minitial x Vinitial = Mfinal x Vfinal

Dilution The addition of solvent to solution is called dilution. it does not change the amount of solute in a solution but changes the solution concentration..

Find the molarity of a solution that 23.4g of Na2SO4 was dissolved in water and diluted to

250.0ml Na2SO4: 142

calculate the molarity of H2SO4 solution when we dilute 50.0 mL of a solution of 2.00 M H2SO4 to a volume of 200.0 mL.

There is a 2000ml of 8M H2SO4 solution in the lab. A student needs 800ml of 2M

H2SO4 solution to carry out an experiment. In order to prepare this solution a-

how many ml of 8 M H2SO4 solution should the student use? b- how many ml of

distilled water should be added?

Calulate the molarity of the solution prepared by mixing 400ml of 0.6M HNO3 solution with 200ml water?

Q1 How can you prepare 500ml 0.10 M H2SO4 solution from 3.0M H2SO4?

Q2 How would you prepare 500.0 mL of 0.2500 M NaOH solution starting from a concentration of 1.000 M?

Q3 What is the final concentration if 75.0 mL of a 3.50 M glucose solution is diluted to a volume of 400.0 mL?

• question.

A solution has a density of 1.235g/ml and contains 90.0% glycerol C3H8O3 and 10% H2O by mass. Determine

a- the molarity of C3H8O3

b- the molarity of H2O

c- mole fraction of C3H8O3

d- mole percent of H2O

e- the molality of H2O

Reactions in Aqueous Solution

Precipitation reactions

Are the reactions, an insoluble solid is formed, the solid product is called as a precipitate.

Ag+(aq) + Cl-(aq) → AgCl(s)

Solubility is the maximum amount of solute that dissolves in a given quantity of solvent at a specific temperature to produce a saturated solution

saturated solution.

The solution in which no more solute can be dissolved is said to be saturated solution.

Super saturated solution.

is a solution in which has higher amount of solute than saturated solution

Unsaturated solution

is a solution in which contain less amount of solute than saturated solution

• Solubility of most of solids increase with temperature.

• Solubility of gases decrease with increased temperature

• Solubility of a gas increases as the gas pressure is increased.

• Types of solvent and solute is important,

like dissolves like, generally nonpolar substances are soluble in nonpolar substances, polar solvents dissolves ionic and polar substances.

solubility rules

Soluble salts

salts of 1A groups (Na, K, Li,) and NH4+ are soluble

all nitrates, acetates and perchlorates are solubleNO3 (Nitrat), CHCOO- (asetat) ,ClO4 (perklorat)

all chlorides (halogens) are soluble except (AgCl, Hg2Cl2, PbCl2)

most sulfates (SO42+ ) are soluble except (Sr SO4, Ca SO4 , Ba SO4 , Pb SO4 )

Cl-, Br-, I- SO42-

Li+, Na+, K+, Rb+, Cs+ NH4+ NO3- ClO4- CH3CO2-

solubility rules

Slightly soluble

all OH- hydroxides are insoluble except those of 1A group (Na, K, Li,) and NH4+

all sulfides S2- are insoluble except those of 1A group (Na, K, Li,) and NH4+

all carbonates are insoluble except those of 1A group (Na, K, Li,) and NH4+

All PO43- are insoluble except those of 1A group (Na, K, Li,) and NH4

+

CO32-, PO4

3- OH-, S2-

• What will happen if Na2CO3 and CaCl2 solutions are mixed ?

• What will happen if CuSO4 and NaNO3 solutions are mixed ?

solubility product constant

SrCrO4(s) SrCrO4(aq)

SrCrO4(aq) Sr2+(aq) + CrO42-(aq)

Ksp = [Sr2+] [CrO42-]

Ionic compounds of alkali metals (like NaCl, KBr,

LiNO3 ..) generally dissolve completely in water.

But many ionic compounds are only slightly

soluble in water. We use the term insoluble for such

compound.

Ksp is the solubility product constant and is equal

to the product of the concentration of the ions

involved in the equilibrium, each raised to the

power of its coefficient in the equilibrium equation.

Ksp has a fixed value for a given system at a

particular temperature.

Solubility equilibrium KspSolubility equilibrium Ksp

Al(OH)3 Al3+ + OH- 1.3x10-33 ZnS Zn2+ +S2- 1.6x10-24

BaCO3 Ba2+ +CO32-

5.1x10-9

MgCO3 Mg2+ +CO32-

3.5x10-8

PbCl2 Pb2+ + 2Cl-1.6x10-5

AgCl Ag+ + Cl- 1.8x10-10

question: write the solubility product constants

for

Ca3(PO4)2, Ag2CrO4, CaF2, PbSO4

Solubility and Kspquestion: solid barium sulfate is shaken in

contact with pure water at 25°C for several days.

At the equilibrium saturated barium sulfate

solution contains 1.04x10-5M Ba2+, what is the Ksp

for BaSO4

 

 

question: Ksp for CaF2 is 3.9x10-11, what is the

solubility of CaF2 in water in grams per liter.

 

Solubility is the quantity of

substance that dissolves in a given

quantity of water. It is often

expressed as grams of solid per

100g of water. Ksp and solubility are

related and one can be calculated

from other.

Common ion effect

question: calculate the molar solubility of

Ag2SO4 in 1.0 M Na2SO4. Ksp for Ag2SO4 is

1.4x10-5

The solubility of a slightly soluble ionic

compound is lowered when a second

soluble that furnishes a common ion is

added to solution.

Ksp and Precipitate

formationQip is generally called ion product, it is the

product of ion concentrations raised to

appropriate powers.

SrCrO4(aq) Sr2+(aq) + CrO42-(aq)

Qip = [Sr2+] [CrO42-]

Qip < Ksp no precipitate occur

Qip = Ksp solution just saturated

Qip > Ksp precipitation should occur

Selective prepicitation or fractional precipitation

One ion is precipitated while the other(s) remains in solution. For this purpose there

must be significant differences in their Ksp.

 

question: 0.10 M AgNO3 is adding slowly to a solution that has [CrO42-] = 0.010M

and [Br-]= 0.10M ( Ksp for Ag2CrO4 is 1.4x10-5, Ksp for AgBr is 5.0 x10-

13 )

show that AgBr (s) precipitates before than Ag2CrO4(s) at the point where Ag2CrO4 begins to precipitate, what is the [Br-] remaining in solution? is separation of Br- and CrO4

2- by fractional precipitation feasible?

Activity Effects

activityTrue thermodynamic constants use a species activity in place of its molar concentration (a).

activity coefficientThe number that when multiplied by a species’ concentration gives that species’ activity ().

ionic strengthA quantitative method for reporting the ionic composition of a solution that takes into account the greater effect of more highly charged ions ().

Calculate the ionic strength of 0.10 M NaCl. Repeat the calculation for a solution of 0.10 M Na2SO4.

Calculate the solubility of Pb(IO3)2 in a matrix of 0.020 M Mg(NO3)2.

Quantitatif Analysis

Titrimetry and Gravimetry

Gravimetry: is a method in which the signal is a mass or change in mass to find the amount (or concentration) of analyte in the sample. .

2 H2 (g) + O2 (g) 2 H2O (s)

2 molecules of H2 1molecules O2

two moles of H2 reacts with one mole O2 and two moles H2O is produced.

4 grams of H2 reacts with 32 g O2 and 36 g H2O is produced.

Reaksiyon Stokiometrisi

2 H2 (g) + O2 (g) 2 H2O (s)

2 molekül H2 1 molekül O2 ile reaksiyona girer 2 molekül H2O oluşur

2 mol H2 1 mol O2 ile reaksiyona girer ve 2 mol H2O oluşur

4 gram H2 32 g O2 ile reaksiyona girer ve 36 g H2O oluşur

what mass of AgBr is formed when a solution containing 3.45 g of KBr is mixed with a solution containing 7.28 g AgNO3?

KBr(aq) + AgNO3 (aq) AgBr(s) + K+ (aq) + NO3- (aq)

Gravimetry: is a method in which the signal is a mass or change in mass to find the amount (or concentration) of analyte in the sample.

Types of Gravimetric Methods

 precipitation gravimetry

A gravimetric method in which the signal is the mass of a precipitate.

Fe3+ + OH- → Fe(OH)3 →Fe2O3

volatilization gravimetry

A gravimetric method in which the loss of a volatile species gives rise to thesignal. In determining the moisture content of food, thermal energy vaporizes the H2O. The amount of carbon in an organic compound may be determined by using the chemical energy of combustion to convert C to CO2.

CaCO3 (k)→CaO (k) + CO2(g)

electrogravimetry A gravimetric method in which the signal is the mass of an electrodeposit onthe cathode or anode in an electrochemical cell.

 

 particulate gravimetry

 

 the analyte is determined following its removal from the sample matrix by filtration or extraction. The determination of suspended solids is one example of particulate gravimetry.

Quantitative Calculations In precipitation gravimetry

the relationship between the analyte and the precipitate is determined by the stoichiometry of the relevant reactions.

Fe3+ + OH- → Fe(OH)3 →Fe2O3

Grams of analyte = k grams of precipitate

Gravimetric factor : is a stochiometric ratio between the analyte and its precipitate

2 Formula weight of FeGF = ---- x -------------------------------------- 1 molecular weight of Fe2O3

2x 56Mass of the Fe = the mass of the precipitate x -------------------------- 160

Quantitative Calculations In precipitation gravimetry

the relationship between the analyte and the precipitate is determined by the stoichiometry of the relevant reactions.

mass of the analyte X% X (the percentage of analyte) = --------------------------------------- x 100 sample mass

Grams of analyte = k grams of precipitate

Gravimetric factor : is a stochiometric ratio between the analyte and its precipitate

a Formula weight of the analyte GF = ---- x ---------------------------------------------------- b molecular weight of the final precipitate

Mass of X analyte = the mass of the precipitate x GF

An ore containing magnetite, Fe3O4, was analyzed by dissolving a 1.5419 –g sample in concentrated HCl, giving a mixture of Fe2+ and Fe3+ . After adding HNO3 to oxidize any Fe2+ to Fe3+ the resulting solution was diluted with water and the Fe3+ precipitated as Fe(OH)3 by adding NH3. After filtering and rinsing, the residue was ignited, giving 0.8525 g of pure Fe2O3. Calculate the

%w/w Fe3O4, in the sample.

An impure sample of Na3PO3 weighing 0.1392 g was dissolved in 25 mL of water. A solution containing 50 mL of 3% w/v mercury(II) chloride, 20 mL of 10% w/v sodium acetate and 5 mL of glacial acetic acid was then prepared. The solution containing the phosphite was added dropwise to the second solution, oxidizing PO3 3– to PO4 3– and precipitating Hg2Cl2. After digesting, filtering, and rinsing, the precipitated Hg2Cl2 was found to weigh 0.4320 g. Report the purity of the original sample as %w/w Na3PO3

Phosphorus has an oxidation state of +3 in PO3 3– and +5 in PO4 3–; thus,oxidizing PO3 3– to PO4 3– requires two electrons.

The formation of Hg2Cl2 by reduction of HgCl2 requires 2 electrons as the oxidation state of each mercury changes from +2 to +1.

Since the oxidation of PO3 3– and the formation of Hg2Cl2 both require two electrons, we have Moles Na3PO3 = moles Hg2Cl2

An impure sample of Na3PO3 weighing 0.1392 g was dissolved in 25 mL of water. A solution containing 50 mL of 3% w/v mercury(II) chloride, 20 mL of 10% w/v sodium acetate and 5 mL of glacial acetic acid was then prepared. The solution containing the phosphite was added dropwise to the second solution, oxidizing PO3 3– to PO4 3– and precipitating Hg2Cl2. After digesting, filtering, and rinsing, the precipitated Hg2Cl2 was found to weigh 0.4320 g. Report the purity of the original sample as %w/w Na3PO3

Moles Na3PO3 = moles Hg2Cl2

Soru: to find the %w/w of NaCl in a mixture, 0,9532 g sample is disolved in pure water and precipitated with AgNO3. After drying the AgCl precipitate ie weighed 0,7033g. What is the %w/w of NaCl in a mixture? (NaCl: 58,44 AgCl: 143,32)

A typical gravimetric analysis procedure may be divided into five stages: sample pretreatment; precipitation; filtration; drying and ignition; weighing. 

1- sample pretreatment

2 - precipitation;

3- filtration;

5- drying and ignition;

6- weighing.

Çökelek ve çöktürücülerin özellikleri.

Precipitant A reagent that causes the precipitation of a soluble species. Must be selevtive for one analyte.

Precipitate: is the product of a simple reaction between the analyte and precipitant. The precipitate must be formed quantitatively and within a reasonable time. Its solubility should be low enough for a quantitative separation to be made. It must be readily filterable and, if possible, have a known and stable stoichiometric composition when dried so that its weight can be related to the amount of analyte present. Failing this, it must be possible to convert the precipitate to a stoichiometric weighable form (usually by ignition).

The theoretical and experimental details of precipitation gravimetry

Precipitates form in two ways; by nucleation and by particle growth.

Nucleation : A few ions, atoms or molecules come together to form a stable solid.

Particle growth is the addition of new preciptates onto existing nucleus..

If nucleation predominates, a large number of very small particles results, if particle growth predominates a smaller number of larger particles is obtained.

Solutions with a large, positive value of relative supersaturation (RSS) show high rates of nucleation, producing a precipitate consisting of numerous small particles.

When RSS is small, precipitation is more likely to occur by particle growth than by nucleation.( smaller number of larger particles is obtained)

Controlling Particle SizeFollowing precipitation and digestion, the precipitate must be separated from the supernatant solution and freed of any remaining impurities, including residual solvent. These tasks are accomplished by filtering, rinsing,and drying the precipitate. The size of the precipitate’s particles determines the easeand success of filtration. Smaller, colloidal particles are difficult to filter because they may readily pass through the pores of the filtering device. Large, crystalline particles, however, are easily filtered.

A solute’s relative supersaturation, RSS, can be expressed as

Q is the solute’s actual concentration, S is the solute’s expected concentration at equilibrium, Q – S is a measure of the solute’s supersaturation when precipitation begins.3

A large, positive value of RSS indicates that a solution is highly supersaturated. Such solutions are unstable and show high rates of nucleation, producing a precipitate consisting of numerous small particles. When RSS is small, precipitation is more likely to occur by particle growth than bynucleation.

Colloidal solution: the particle size are in the range of 10-7-10-4 cm. colloidal solutions are genereally not filterable.

Crystalline suspensions. Particle size are around 10-1 mm, they have the tendency to precipitate.

A precipitate’s solubility usually increases at higher temperatures, and adjusting pH may affect a precipitate’s solubility if it contains an acidic or basic anion.

Temperature and pH, therefore, are useful ways to increase the value of S. Conducting the precipitation in a dilute solution of analyte, or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of Q.

When RSS is higher, colloidal solutions will form, when RSS is lower crystalline solutions will form

Colloidal solution:  

coagulation

The process of smaller particles of precipitate clumping together to form larger particles.

 

Heating and string.

Peptization

İs a process by which a coagulated colloid returns to smaller particles. It could be happen when preciptate is being washed. HNO3, HCl, NH4Cl gibi elektrolitler eklenir.

 

 

 

 

 

A 101.3-mg sample of an organic compound known to contain Cl is burned in pure O2 and the combustion gases collected in absorbent tubes. The tube used to trap CO2 increases in mass by 167.6 mg, and the tube for trapping H2O shows a 13.7-mg increase. A second sample of 121.8 mg is treated with concentrated HNO3 producing Cl2, which subsequently reacts with Ag+, forming 262.7 mg of AgCl. Determine the compound’s composition, as well as its empirical formula.

Çökeltinin Saflığı

Normal olarak çözeltide kalması istenen bileşiklerin çökeltiye geçmesi safsızlıklara, ürün kütlesinin yanlış tartılmasına dolayısıyla analizde hataya neden olur.

Yüzey Adsorpsiyonu normal koşullarda çözünen bileşiğin çöken teneciklerin yüzeyinde tutunmasıdır. Koloidal çökeleklerde görülür. Yıkama, Yeniden çötürme ile azaltılabilir.

Karışık kristal oluşumu

kristalin yapısında bulunan bir iyonun yerine ortamda bulunan başka bir iyonun yer değiştirerek çökmesidir.

Hapsetme ve Mekanik

sürüklenme

kristaller arasındaki boşluklarda bir bileşiğin kalmasıdır. Yavaş çötürme ve bağıl aşırı doygunluk kontrol edilerek azaltılır.

Titration

• Titration is a procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a standard solution of another substance, whose concentration is known. By finding the volume of the standard solution that reacts with the measured volume of the first solution, the concentration of the first solution can be calculated.

• equivalent point The point that all reactants are consumed, Stoichiometric mol numbers of both

reactants are equal.

• an indicator a compound that change its color around equivalent point. such as phenolphthalein, is colorless in acidic solution but turns pink in basic solution.

Calculations in the volumetric titrimetry

Mol number

mn = --------- MA

Molarity nM= ------- mol/L V

aA + bB ↔ cC + dD

a mol A react with b mol B

When A is a standard (titrant) the volume of A used in the titration and its molarity can be used to calculate the mol number of B

Calculations in the volumetric titrimetry

A 25.0 mL sample of vinegar (dilute acetic acid, ) is titrated and found to react with 94.7 mL of 0.200 M NaOH. What is the molarity of the acetic acid solution?

Soru: antiasit ilaç tabletlerinin anabileşeni CaCO3 tür. 0.542 gr olarak tartılan bir tablet HCl ile titre edildiğinde, reaksiyonun tamamlanması için 38.5 ml 0.200M HCl harcandığına göre tablet içindeki CaCO3 yüzdesi nedir?

CaCO3 + HCl → Ca2+ + CO2 + Cl- + H2O

soru: 0,2879g sodyum okzalatı (Na2C2O4) asidik çözeltide aşağıdaki redoks tepkimesine göre titre etmek için 25,12mL KMnO4 çözeltisi harcandığına göre KMnO4 ün molaritesi nedir?

C2O42- + MnO4- → Mn2+ + CO2

Titrimetric Analysiskonsantrasyonu bilinen bir çözeltinin analit ile reaksiyona giren miktarının ölçümüne dayanan kantitatif analiz metotlarıdır. Standart çözelti reaksiyon tamamlanıncaya dek ortama yavaş yavaş eklenir.

 

Volumetry   Gravimetric Titrimetry   Coulometric titrimetri

 Standart reaktifin hacminin ölçülmesi temeline dayanan bir titrimetrik metottur

 Standart reaktifin kütlesinin ölçülmesine dayanan titrimetrik metottur

 Analit ile bir reaksiyonun tamamlanması için gerekli kulon cinsinden yük miktarının ölçüldüğü titrimetrik metottur.

- İndirgenme-Yükseltgenme reaksiyonları- Asit Baz reaksiyonları- Kompleks oluşum reaksiyonları- Çöktürme titrasyonları

Titrant: The reagent added to a solution containing the analyte and whose volume is the signal. 

titration: is a process in which a standard reagent is added to a solution of an analyte until the reaction between the analyte and reagent is completed.  

equivalence point The point in a titration where stoichiometrically equivalent amounts of analyte and titrant react. 

Indicator A colored compound whose change in color signals the end point of a titration.

 

end point The point in a titration where we stop adding titrant.

 

titration error The determinate error in a titration due to the difference between the end point and the equivalence point. 

Veq:

 

Venepoint:

Acids and Bases

Acids and Bases

Acids;Have a sour taste,dissolve metals such as zinc

and carbonate minerals change color of litmus to red

Acetic acid in vinegar,Citric acid in lemons

BasesHave a bitter taste,Have a slippery feelchange color of litmus to blue,React with dissolved metal to form prepiciate

Household cleaning products

Acid-Base Definitions

Arrhenius Acid-BaseDefinition

Brønsted-Lowry Acid-BaseDefinition

Lewis acid-baseDefinition

Arrhenius Acid-Base Definition (1884)

An acid is a substance that contains hydrogen and dissociates to produce Hydrogen ion : H+

HCl(aq) H+(aq) + Cl-(aq)

A base is a substance that contains the hydroxyl group and dissociates to produce Hydroxide ion : OH –

NaOH (aq) Na+ (aq) + OH -(aq)

Neutralization is the reaction of an H+ ion from the acid and the OH - ion from the base to form water, H2O

H+(aq) + OH-(aq) <=> H2O(l)

Brønsted-Lowry Acid-Base Definition (1923)

An acid is a species having a tendency to donate an H+ ion.

HCl + H2O Cl– + H3O+

A base is a species having tendency to accept an H+ ion.

NH3 + H2O NH4+ + OH-

The Conjugate Pairs in Some Acid-Base Reactions

Conjugate Pair

Acid + Base Base + Acid

Conjugate Pair

In the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process.

question : show the conjugated acid base pairs in the reactions below

HF + H2O F– + H3O+

H2PO4– + OH– HPO42– + H2O

NH4+ + CO32– NH3 + HCO3–

The Lewis acid-base definition :

• A base is any species that donates an electron pair

• An acid is any species that accepts an electron pair.

Strengths of Acids

• Strong Acids; An acid that completely ionized in water, is called as a strong acid

• HCl(aq) + H2O H3O+(aq) + Cl-(aq)

• Weak acid is an acid that partly ionized in water.

• CH3COOH (aq) + H2O CH3COO-(aq) + H3O+(aq)

Percent Ionization

HA + H2O ↔ H3O+ + A-

Degree of ionization =[H3O+] from HA

[HA] originally

Percent ionization =[H3O+] from HA

[HA] originally x 100%

Strong acids. 1. The hydrohalic acids HCl, HBr, and HI2. Oxoacids in which the number of O atoms exceeds the number of ionizable H atoms by two or more, such as HNO3, H2SO4, HClO4

Weak acids. 1. The hydrohalic acid HF2. Those acids in which H is bounded to O or to halogen, such as HCN and H2S 3. Oxoacids in which the number of O atoms equals or exceeds by one the number of ionizable H atoms, such as HClO, HNO2, and H3PO4 4. Organic acids (general formula RCOOH), such as CH3COOH and C6H5COOH

Strong bases. A base that completely ionized in water, is called as a strong base. Soluble compounds containing O2- or OH- ions are strong bases.

NaOH (aq) Na+ (aq) + OH -(aq)

1) M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs)2) MO or M(OH)2, where M = Group 2A(2) metals (Ca, Sr, Ba) [MgO and Mg(OH)2

are only slightly soluble, but the soluble portion dissociates completely.]

A weak base is a base that partly ionized in water. Many compounds with an electron-rich nitrogen are weak bases (none are Arrhenius bases). The common structural feature is an N atom that has a lone electron pair in its Lewis structure

NH3(g) +H20(l) NH4+ (aq) + OH -(aq)

1) Ammonia (:NH3)2) Amines (general formula RNH2, R2NH, R3N), such as CH3CH2NH2, (CH3)2NH,

(C3H7)3N, and C5H5N

The Meaning of Ka, the Acid Dissociation ConstantKa is the equilibrium constant of the ionization reaction of a weak acid.

Kb represents the ionization constant of a base.

Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C)

The pH Scale

• The pH is defined as the negative logarithm in base 10, of the hydronium ion concentration

• pH = - log[H3O+]

• The pOH is defined as the negative logarithm in base 10, of the hydroxyl ion concentration

• pOH = - log[OH-]

pH of an acidic solution < 7.00

pH of a neutral solution = 7.00

pH of a basic solution > 7.00

Calculating [H3O+], pH, [OH-], and pOH

Strong acid (or base)

Weak acid (or base)

Strong and weak acid (or base)

[H+] = acid concentration

calculate [H+] from Ka

[H+] is generally equal to strong acid (effect of weak acid neglect unless strong acid is too dilute)

question : What is the pH of a solution that contains 10-2 M hydronium ion ?

question : Calculate the [H3O+], pH, [OH-], and pOH of 0.0024 M hydrochloric acid solutions at 25°C.

• question : The weak acid hypochlorous acid is formed in bleach solutions. What is the [H3O+] of a 0.125 M HClO solution? Ka = 3.5 x 10^-8

Buffer solution

• is a solution whose pH changes only very slightly upon the addition of small of either an acid or a base.

• Buffer solutions contain a weak acid and its conjugate base ( its salt) or a weak base and its conjugate acid ( its salt).

• question : calculate the the pH of a buffer solution containing 0.246 M NH3 and 0.0954 M NH4Cl (for NH3 Kb = 1.74x 10-5)

• NH3 +H2O NH4+ + OH-

Hidrolysis

Salts of the weak acids (or weak bases) react with water to form the weak acid (or weak base).

Salt + Water ↔ weak acid + OH-

equilibrium constant of this reaction is Kh=Kw/Ka the pH of the solution can be calculated from Kh.

question: calculate the ph of the 0.01 M sodium acetat solution. For acetic acid, Ka= 1.8x10-5

Polyprotic Acids

H3PO4 + H2O ↔ H3O+ + H2PO4-

H2PO4- + H2O ↔ H3O+ + HPO4

2-

HPO42- + H2O ↔ H3O+ + PO4

3-

Phosphoric acid:

A triprotic acid.

Ka = 7.1x10-3

Ka = 6.3x10-8

Ka = 4.2x10-13

Ionization Constants of Some Polyprotic Acids

Phosphoric Acid

• Ka1 >> Ka2

• All H3O+ is formed in the first ionization step.

• H2PO4- essentially does not ionize further.

• Assume [H2PO4-] = [H3O+].

• [HPO42-] ≈ Ka2

regardless of solution molarity.

Calculating Ion Concentrations in a Polyprotic Acid Solution.

For a 3.0 M H3PO4 solution, calculate:

(a) [H3O+]; (b) [H2PO4-]; [HPO42-] (d)

[PO43-]

H3PO4 + H2O ↔ H2PO4- + H3O+

Initial conc. 3.0 M 0 0

Changes -x M +x M +x M

Eqlbrm conc. (3.0-x) M x M x M

Acid Base titrations

A titration in which the reaction between the analyte and titrant is an acid–base reaction.

Standard reactants:

acid–base titrations were conducted using H2SO4, HCl, and HNO3 as acidic titrants, and K2CO3 and Na2CO3 as basic titrants.

Indicator : A colored compound whose change in color signals the end point of a titration.End points were determined using visual indicators such as litmus, which is red in acidic solutions and blue in basic solutions,

Acidimetry: Determination of the Base content of a sample by titrating strong acid solution. Diluted solutions of HCl, H2SO4, HClO4 are used as a titrant. Since HNO3

gives some extra reactions, it can not be used for acidimetry.

Alkalimetry: Determination of the acid contents of a sample by titrating with a strong base solutions like NaOH, KOH ve Ba(OH)2

Visual Indicators for Acid-base Titrationsan acid–base indicator are the weak organic acids. the weak acids and bases derivatives of organic dyes can serve as a useful means for determining the end point of a titration. Because such compounds have at least one conjugate acid–base species that is highly colored,

Indicator A colored compound whose change in color signals the end point of a titration.

an acid–base indicator changes color with in changing the pH of the solution.

In general terms a visual indicator is a compound which changes from one colour to another as its chemical form changes with its chemical environment

If the indicator is present in an environment where a titration reaction generates or consumes the X species, the indicator will change with the concentration of X in the solution and the colour of the solution will be determined by the ratio [InB]/[InA]. As a general guide, the eye will register a complete change from one colour to the other when this ratio changes from 10:1 to 1:10.

The pH at which an acid–base indicator changes color is determined byits acid dissociation constant. For an indicator that is a monoprotic weakacid, HIn, the following dissociation reaction occurs

HIn(aq) + H2O(l)↔H3O+(aq) + In–(aq)

For example, the acid-base indicator methyl orange has a pKIn of 3.7 and will thus change colour over the pH range 2.7–4.7.

The ultimate sharpness of the end point will further depend upon the rate at which pX is changing at the end point of the titration.

Acid-base Titration Curves

titration curve: A graph showing the progress of a titration as a function of the volume of titrant added.

A titration curve provides us with a visual picture of how a property, such as pH, changes as we add titrantTitrating Strong Acids and Strong Bases

Titrating a Strong Base with a Strong Acid

Titrating a StrongAcid with a Strong Base

Titrating a Weak Acid with a Strong Base

Titrating a Weak Base with a Strong Acid

Titration Curves

Titrating a Strong Acid with a Strong Base

starts with a low pH, then pH changes less till equivalent point then there will be sudden change (increase) at the equivalent point.

Titrating a Strong Base with a Strong Acid

The curve starts at the basic region, till equivalent point pH decrease less but at the equivalent point sudden change (decrease) will occur.

Titration Curves

Titrating a Weak Acid with a Strong Base

Titrating a Weak Base with a Strong Acid

the titration curve when titrating a diprotic weak acid, H2A, with a strong base.

pH Calculations in the Titrating of a Strong Acid with a Strong Base

Before equivalence point

At the equivalence point

After equivalence point

question: 50.0ml 0.050 M HCl solutrion is being titrated with 0.100 M NaOH, calculate the pH when

0 ml base added, 10 ml base added, 20ml base added, 25ml base added, 30 ml base added, and 40 base added,

Draw the titration curve.

pH Calculations in the Titrating of a weak Acid with a Strong Base

Buffer solution A solution containing a conjugate weak acid/weak base pair that is resistant to a change in pH when a strong acid orstrong base is added.

Buffer capacity is the mol number of a stronge acid or base needed to add for changing the pH 1,0L solution in 1 unit

The pH of a buffer changes within the range of pH = pKa ± 1

Question: calculate the the pH of a solution containing 0.246 M NH3 and 0.0954

M NH4Cl (for NH3 Kb = 1.74x 10-5)

NH3 +H2O NH4+ + OH-

Titrating of a weak Acid with a Strong Base

Hidrolysis

Salts of the weak acids (or weak bases) react with water to form the weak acid (or weak base).

Salt + Water ↔ weak acid + OH-

equilibrium constant of this reaction is Kh=Kw/Ka the pH of the solution can be calculated from Kh.

Question : calculate the pH of a 0.01 M sodium acetat solution. Ka= 1.8x10-5

pH Calculations in the Titrating of a weak Acid with a Strong Base

.

Adding 0ml base Calculate the pH from Ka of the weak acid

Before the Equivalence point

Calculet the pH of the buffer solution

At the Equivalence point

Calculate pH of the salt of the weak acid (Hydrolysis)

After the Equivalence point

Calculate the pH from the concentraion of the base (not consumed in the reaction)

Soru: 50ml 0.10 M acetic acid is being titrated with 0.10 M NaOH. calculate the pH when.

0ml base added , 10 ml base added 45 ml base added 50 ml baz ilavesi 51 base added 55 ml base added

Ka= 1.8x10-5

Titrasyonda İndikatörden gelen Hatalar

Belirli hatalar Rastgele hatalar

indikatörün renk değişim pH aralığının titrasyonun eşdeğerlik noktasındaki pH ının faklı olması sonucu olur. Uygun indikatör seçimi, tanık deney düzeltmesi ile azaltılır.

renk değişimi izlenirken göz yanılgılarından kaynaklanır.

complexometric titrations

A complexometric titration is based on the essentially stoichiometric reaction of a complexing agent with another species to form a complex species that is only slightly dissociated and that is soluble in the titration medium.

Complex The product of the complexation reaction is called a metal–ligand complex.

Ligant A Lewis base that binds with a metal ion

Chelat bir metalin bir ligantla bağlanarak oluşturduğu halkalı kompleks tir

formation constant

The equilibrium constant for a reaction in which a metal and a ligand bind to form a metal–ligand complex (Kf).

EDTA The most widely use ligands is ethylenediaminetetraacetic acid, EDTA, forms strong 1:1 complexes with many metal ions.

Dissosiation constants of EDTA: K1= 1.02x10-2 K2= 2.14x10-3 K3= 6.92x10-7 K4= 5.50x10-11

EDTA çözeltilerinde α değerlerinin hesaplanması

EDTA Hidrojen iyonlarını kaybederek şu türlere iyonlaşır

[Y4- ]+[HY3-]+[H2Y2-]+[H3Y

-]+[H4Y]

K1= 1.02x10-2 K2= 2.14x10-3 K3= 6.92x10-7 K4= 5.50x10-11

Y4- nin çözeltideki mol kesri

[Y4-] α4= -------------- CEDTA

CEDTA= [Y4- ]+[HY3-]+[H2Y2-]+[H3Y

-]+[H4Y]

Buradan K sabitleri kullanılarak

K1K2K3K4

α4= ------------------------------------------------------------------------- [H+]4 + K1[H

+]3-+ K1K2[H+]2- + K1K2K3[H

+] + K1K2K3K4

K1K2K3K4

α4 = ------------------------------ olarak bulunur CEDTA

EDTA Metal İyonları Kompleksleri

Mn+ + Y4- ↔ MYn-4 [MYn-4 ] KMY = ------------------

[Mn+] [Y4-]

K’MY (durum sabiti)

[MYn-4 ] K’MY = ----------------- = α4 KMY

[Mn+] CEDTA

Calculate the Concentration of Ni2+ in a soltion which prepared as mixing 50.0ml 0.0300 M Ni2+ and 50.0mlml 0.0500 M EDTA at pH= 3At pH=3 , α4= 2.5x10-11 KNiY=4.2x1018

EDTA titrasyon eğrileri

Verilen pH değeri için durum sabiti değeri bulunur hesaplamalarda kullanılır.

Eşdeğerlik noktası öncesinde

ortamda metal iyonu fazlası vardır ve harcanmadan kalan metal iyonu konsantrasyonu hesaplanır.

Eşdeğerlik noktasında

ortamda sadece kompleks vardır ve bir miktar ayrışır, ayrışan metal iyonu konsantrasyonu durum sabitinden hesaplanır.

Eşdeğerlik noktası sonrasında

EDTA fazlası vardır ve artan EDTA konsantrasyonu durum sabitinde yerine konarak metal iyonu konsantrasyonu hesaplanır.

50.0ml 0.0500 M Mg2+ nın 0.0500 M EDTA ile titrasyonunda pH=10 na tamponlanıyor,0ml, 5.0ml, 50.0ml ve 51.0 ml EDTA ilaveleri içinMg2+ konsantrasyonunu hesaplayınız ve titrasyon eğrisini çiziniz.PH=10 için α4= 3.5x10-1 Kol= 6.2 x 108

EDTA titrasyonları için İndikatörler

Genel olarak metal iyon indikatörleri kullanılır. Bunlar metal iyonlarına bağlandığında renk değiştiren organik boyalardır.MetalIn + EDTA → MetalEDTA + In Renk 1 renksiz renksiz renk 2

Chelometric titrations may be classified according to their manner of performance: direct titrations, back titrations, substitution titrations, or indirect methods.

direct titrations

back titrations substitution titrations

indirect methods

0,300 g metal wire is being dissolved in nitric acid and diluted to 100,0ml with pure water. 25.00ml sample is taken from this solution and added 0.052 M EDTA and then neutralized with NaOH. After that the pH of the solution is buffered to pH=5.5. and excess EDTA is back titrated with 17.61ml 0.02299M Zn2+. Calculate the Ni2+ percent in the metal wire. (Ni: 58,71g/mol EDTA:372,0g/mol)

Application of Equilibrium Calculations to Complex Systems

when aqueous solutions contain several species that interact with one another and water that yield two or more simultaneous equilibria.

PbS ↔ Pb2+ + S2-

S2- + H2O ↔ HS- + OH-

HS- + H2O ↔ H2S + OH-

2 H2O ↔ H3O+ + OH-

Steps for solving problems involving several equilibria

1. Write all chemical equations and balance them2. Write all equilibrium constant wxpressions. 3. State (undrline) which quantity (unknown) you want to find

(solve).4. Write mass balance expression for the system. 5. Write the charge balance equation (IF possible)6. Compare the number of equations and the number of

unknowns. 7. Make suitable approximations to simplify the algebra. 8. Solve the algebraic equations, find the unknown defined at

step 3. 9. Check the validity of approximations.

mass balance equation: (which is simply a statement of the conservation of matter.)

An equation stating that matter is conserved, and that the total amount of a species added to a solution must equal the sum of the amount of each of its possible forms present in solution.

charge balance equation: (is a statement of solution electroneutrality.) An equation stating that the total concentration of

positive charge in a solution must equal the total concentration of negative charge. Total positive charge from cations = total negative charge from anions.

Write the mass balance equation for 1 liter of 0.250mol H3PO4

aqueous solution.  

Write charge balanse equation for a solution which contains H+, OH-, K+, H2PO4

-, HPO42- and PO4

3-

Calculate the Hg22+ cocentration in a saturated Hg2Cl2 solution.

Ksp= 1.2x10-18 Kw=1x10-14

 

Calculate the molar solubility of Mg(OH)2 in water.

Ksp Mg(OH)2 = 7.1x10-12 Kw=1x10-14

Calculate the molar solubility of Fe(OH)3 in water.

Ksp Fe(OH)3 = 2.0x10-39 Kw=1x10-14

Calculate the solubility of calcium oxalate in a solution which the Hydronium ion cocentration is 1.0 x 10-4

For CaC2O4 Ksp= 2.3 x 10-9

for H2C2O4 Ka1= 5.36 x 10-2, Ka2= 5.42 x 10-8

Kw=1x10-14

Calculate the solubility of AgBr in 0.1 M NH3 solution.

For AgBr Ksp= 5.2 x 10-13 for Ag(NH3)

+ Kformation oluşum= 2 x 103

for Ag(NH3)2 Kformation= 6.9 x 103

for NH3 Kb= 1.76 x 10-5

Soru: 0.1 M Pb2+ ve 0.1 M Fe2+ içeren bir çözeltiye H2S eklenerek Pb2+ ve Fe2+ iyonları birbirinden ayrılması için

gerekli koşulları belirleyiniz. Kçç(PbS)=3x10-28, Kçç(FeS)=8x10-19 H2S için K1=9.6x10-8 K2=1.3x10-14

Calculate the solubilty of HgS in water by using all equations which occur.

for HgS Ksp = 5 x 10-54 Kw= 1.0 x 10-14, for H2S Ka1= 9.6x10-8 Ka2= 1.3x10-14