web viewadding naf to a solution of hf causes more hf to be produced. major species are hf, na+, f-,...

APPLICATIONS OF AQUEOUS EQUILIBRIAChapter 15

COMMON ION EFFECTThe addition of an ion already present (common) in a system causes equilibrium to shift away from the common ion.

Example: The addition of concentrated HCl to a saturated solution of NaCl will cause some solid NaCl to precipitate out of solution. The NaCl has become less soluble because the addition of additional chloride ion. This can be explained by the use of Le Chatelier’s Principle. In a solution of NaCl and HCl, the major species are Cl-, H+, H2O, and Na+. The common ion is Cl-.

NaCl + HCl → more NaCl due to increased Cl-

Example: The addition of a common ion to a weak acid solution makes the solution LESS acidic. HC2H3O2(aq) ↔ H+

(aq) + C2H3O2-(aq)

If NaC2H3O2 is added to the system, the equilibrium shifts to undissociated HC2H3O2 raising the pH. The new pH can be calculated by putting the concentration of the anion into the Ka equation and solving for the new [H+].

Example: Adding NaF to a solution of HF causes more HF to be produced. Major species are HF, Na+, F-, and H2O. Common ion is F-. In a 1.0M NaF and 1.0M HF solution, there is more HF in the presence of NaF.

HF(aq) ↔ H+(aq) + F-

(aq)

Le Chatelier’s indicates that additional F- due to additional NaF causes a shift to the left and thus generates more HF.

Finding the pH1. Always determine the major species.2. Write the equilibrium equation and expression.3. Determine the initial concentrations4. Do ICE chart and solve for x5. Once [H+] has been found, find pH

PRACTICE ONEThe equilibrium concentration of H+ in a 1.0M HF solution is 2.7 x 10-2M and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka = 7.2 x 10-4) and 1.0M NaF.

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BUFFERED SOLUTIONS

A buffer is a solution of a weak acid or base and its salt [which is its conjugate]. is just a case of the common ion effect. resists a change in pH. consists of both an acid or base and its conjugate which is its complement, so an acid and a base

are present in all buffer solutions. Usually consists of a solution of a weak acid and its salt or a weak base and its salt. If a small

amount of strong acid is added to the buffer, there is a base component ready and waiting to neutralize the “invader”.

System is like a net – it “catches” the acid or base but only for so long until it “breaks”.

You are always adding a strong acid or strong base to a buffer solution.

Example: HC2H3O2 / C2H3O2- buffer system (the “net”)

Add a strong acid: H+ + C2H3O2- → HC2H3O2 forms a weak acid

Add a strong base: OH- + HC2H3O2 → C2H3O2- + H2O forms a weak base

Example: NH3 / NH4+ buffer system (the “net”)

Add a strong acid: H+ + NH3 → NH4+ forms a weak acid

Add a strong base: OH- + NH4+ → NH3 + H2O forms a weak base

PRACTICE TWOA buffered solution contains .050M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Calculate the pH of the solution.

HELPFUL TIPS:1. Buffered solutions are simply solutions of weak acids and bases containing a common ion.2. The pH calculations for buffered solutions require exactly the same procedures as determining

the pH of weak acid or weak base solutions learned previously.

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3. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations.

PRACTICE THREECalculate the change in pH that occurs when 0.010mol solid NaOH is added to 1.0L of the buffered solution contains .050M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Compare this pH change with that which occurs when 0.010mol solid NaOH is added to 1.0L of water.

HOW DOES BUFFERING WORK?In the equation below, when hydroxide ions are added to the solution, the weak acid provides the source of protons. The OH- ions are not allowed to accumulate, but are replaced by A-.

OH- + HA → A- + H2O

pH can be understood by looking at the EQ expression.

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Ka = ¿¿ or [H+] = Ka [ HA ]¿ ¿

So [H+] (and thus pH) is determined by the ratio of [HA]/[A-]. When OH- ions are added, HA converts to A- and the ratio decreases. However, is the amounts of [HA] and [A-] are LARGE, then the change in the ratio will be small.Going back to Practice Two and Three, [HA]/[A-] = 0.50M / 0.50M = 1.0M initially. After adding 0.010M OH-, it becomes [[HA]/[A-] = 0.49M / 0.51M = 0.96M. Not much of a change. [H+] and pH are essentially constant.

It is similar when protons are added to a buffered solution of a weak acid and a salt of its conjugate base.

HENDERSON-HASSELBACH EQUATIONOne way to calculate the pH of a buffer system. This is on your equations (pink) sheet.

For a particular buffering system, all solutions that have the same ratio of [A-]/[HA] have the same pH. Optimum buffering occurs when [HA] = [A-] and the pKa of the weak acid used should be as close to possible to the desired pH of the buffer system.

The Henderson-Hasselbach (HH) equation needs to be used cautiously. It is sometimes used as a quick, easy equation in which to plug in numbers. A Ka or Kb problem requires a greater understanding pf the factors involved and can ALWAYS be used instead of the HH equation. However, at the halfway point (as in a titration), the HH is very useful.

PRACTICE FOURCalculate the pH of a solution containing 0.75M lactic acid (Ka = 1.0 x 10-4) and 0.25M sodium lactate. Lactic acid, HC3H5O3, is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion.

Hints for Solving Buffer Problems:1. Determine the major species involved.2. If a chemical reaction occurs, write the equation and solve stoichiometry.3. Write the EQ equation.

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4. Set up the equilibrium expression (Ka or Kb) of the HH equation.5. Solve.6. Check the logic of the answer.

PRACTICE FIVEA buffered solution contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl. Calculate the pH of this solution.

PRACTICE SIXCalculate the pH of the solution that results when 0.10mol gaseous HCl is added to 1.0Lof the buffered solution of contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl.

BUFFERING CAPACITY

This is the amount of acid or base that can be absorbed by a buffer system without a significant change in pH.

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In order to have a large buffer capacity, a solution should have large concentrations of both buffer components.

PRACTICE SEVENCalculate the change in pH that occurs when 0.010mol gaseous HCl is added to 1.0L of each of the following solutions (Ka for acetic acid = 1.8 x 10-5):

Solution A: 5.00M HC2H3O2 and 5.00M NaC2H3O2

Solution B: 0.050M HC2H3O2 and 0.050M NaC2H3O2

PRACTICE EIGHTA chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt:

a. chloroacetic acid Ka = 1.35 × 10-3

b. propanoic acid Ka = 1.3 × 10-5

c. benzoic acid Ka = 6.4 × 10-5

d. hypochlorus acid Ka = 3.5 × 10-8

Calculate the ratio of [HA] / [A-] required for each system to yield a pH of 4.30. Which system works best?

TITRATIONS AND pH CURVES

Only when the acid AND base are both strong is the pH at the equivalence point 7. Any other conditions and you get to do an equilibrium problem. It is really a stoichiometry problem with a limiting reactant. The “excess” is responsible for the pH

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Weak acid + strong base equivalence pt. > pH 7Strong acid + weak base equivalence pt. < pH 7

There is a distinction between the equivalence point and the end point. The end point is when the indicator changes color. If you’ve made a careful choice of indicators, the equivalence point, when the number of moles of acid = number of moles of base, will be achieved at the same time.

Titrant – solution of know concentration (usually in the buret). The titrant is added to a solution of unknown concentration until the substance being analyzed is just consumes (stoichiometric point or equivalence point).Titration or pH Curve – plot of pH as a function of the amount of titrant added.

1. STRONG ACID – STRONG BASE TITRATIONH+

(aq) + OH-(aq) → H2O(l)

To compute H+, we have to know how much H+ remains at that point in the titration. New unit: millimole, mmol – titrations usually involve small quantities.

This means Molarity = mol soluteL solution =

mmol solutemilliliters solution . So a 1.0M =

1.0 mol1.0 L =

1.0 mmol1.0 mL

Example - For the titration of 50.0 mL of 0.200M HNO3 with 0.100M NaOH, calculate the pH of the solution at the following selected points of the titration: a. 0.0 mL of 0.100M NaOH has been added.

Major Species: H+, NO3-, H2O HNO3 = strong acid

pH = -log[H+] = -log (0.200) = 0.699

b. 10.0 mL of 0.100M NaOH has been added.Major Species: H+, NO3

-, H2O, Na+, OH-

1.00mmol (10.0mL x 0.100M) OH- reacts with 1.00mmol H+. H+ + OH- → H2OBefore: H+ = 10.0mmol; OH- = 1.00mmolAfter: H+ = 10.0mmol – 1.00mmol = 9.0mmol; OH- = 1.00mmol – 1.00mmol = 0.00mmolSo after, Major Species: H+, NO3

-, H2O, Na+. Determine pH with H+ remaining.

pH = -log[H+] = -log (9.0 mmol

50.0 mL+10.0 mL ) = -log[0.15] = 0.82

c. 20.0 mL (total as opposed to additional) of 0.100M NaOH has been added.Major Species: H+, NO3

-, H2O, Na+, OH-




50.0 mL+20.0 mL ) = -log[0.11] = 0.94

d. 50.0 mL (total) of 0.100M NaOH has been added.Major Species: H+, NO3

-, H2O, Na+, OH-


-, H2O, Na+. Determine pH with H+ remaining.7


50.0 mL+50.0 mL ) = -log[0.050] = 1.30

e. 100.0 mL (total) of 0.100M NaOH has been added.Major Species: H+, NO3

-, H2O, Na+, OH-

2.00mmol (20.0mL x 0.100M) OH- reacts with 10.0mmol H+. H+ + OH- → H2OBefore: H+ = 10.0mmol; OH- = 10.0mmolAfter: H+ = 10.0mmol – 10.0mmol = 0.0mmol; OH- = 10.0mmol – 10.0mmol = 0.0mmolSo after, Major Species: NO3

-, H2O, Na+. This is the EQUIVALENCE POINT (or stoichiometric point).pH = 7.00, neutral

f. 150.0 mL (total) of 0.100M NaOH has been added.Major Species: H+, NO3

-, H2O, Na+, OH-



pOH = -log[OH-] = -log (5.0 mmol

50.0 mL+150.0mL ) = -log[0.025] = 1.60, so pH = 12.40

g. 200.0 mL (total) of 0.100M NaOH has been added.Major Species: H+, NO3

-, H2O, Na+, OH-



pOH = -log[OH-] = -log (10.0 mmol

50.0 mL+200.0mL ) = -log[0.040] = 1.40, so pH = 12.60.

The results of a-g are plotted. The pH changes gradually until the titration is close to the equivalence point when there is a dramatic change. Why is this?

Characteristics:1. At equivalence point, pH = 7.2. Before the equivalence point, [H+] and thus pH

can be calculated by dividing mmol H+ by total volume of solution.

3. After the equivalence point, [OH-] and thus pOH and then pH can be calculated by dividing mmol OH- by total volume of solution.

2. WEAK ACID – STRONG BASE TITRATION We have to do a series of buffer problems like we did earlier.

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Curve characteristic of a strong acid and strong base.

Remember that although the acid is weak, it reacts to completion with the OH- ion, a very strong base.

A two-step procedure:o A stoichiometry problem: The rxn of OH_ with the weak acid is assumed to run to

completion, and concentrations of the acid remaining and conjugate base formed are determined.

o An equilibrium problem: The position of the weak acid equilibrium is determined and pH is calculated.

At halfway to the equivalence point, pH = pKa. At the equivalence point, a basic salt is present and pH > 7 After the equivalence point, the strong base will be the dominant species and a simple pH

calculation can be done after stoichiometry.

Example - For the titration of 50.0 mL of 0.10M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10M NaOH, calculate the pH of the solution at the following selected points of the titration:

a. 0.0 mL of 0.10M NaOH has been added.Major Species: HC2H3O2, H2OHC2H3O2

→ C2H3O2- + H+

Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

Ka = 1.8 x 10-5 = (x)(x )

0.10−x = x2 / 0.10

x2 = 1.8 x 10-6 x = 0.0013pH = -log[H+] = -log (0.0013) = 2.89

b. 10.0 mL of 0.10M NaOH has been added.Major Species: HC2H3O2, H2O, Na+, OH-

HC2H3O2 + OH- → C2H3O2

- + H2OBefore: OH- = 1.0mmol; HC2H3O2 = 5.0mmol; C2H3O2

- = 0.0mmolAfter: OH- = 1.0mmol – 1.0mmol = 0.0mmol; HC2H3O2 = 5.0mmol – 1.0mmol = 4.0mmol; C2H3O2

- = 0.0mmol + 1.0mmol = 1.0mmol formed.So after, Major Species: HC2H3O2, C2H3O2

-, H2O, Na+. Determine pH with HC2H3O2 equilibrium

[Initial]: HC2H3O2 = 4.0mmol

50.0 mL+10.0mL

[Initial]: C2H3O2- =

1.0 mmol50.0 mL+10.0 mL

[Initial]: H+ ≈ 0Ka = [C2H3O2

-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

Ka = 1.8 x 10-5 = (x )(0.017+x)

0.067− x = x(0.017)/0.067 = 0.25x

x = 7.2 x 10-5

pH = -log[H+] = -log[7.2 x 10-5] = 4.14

c. 25.0 mL (total as opposed to additional) of 0.100M NaOH has been added.Major Species: HC2H3O2, H2O, Na+, OH-


- + H2O9

HC2H3O2 → C2H3O2

- + H+

I 0.10 0 0

C -x +x +x

E 0.10-x x x

HC2H3O2 → C2H3O2

- + H+

I 0.067 0.017 0

C -x +x +x

E 0.067 - x 0.017 + x x

Before: OH- = 2.5mmol; HC2H3O2 = 5.0mmol; C2H3O2- = 0.0mmol

After: OH- = 2.5mmol – 2.5mmol = 0.0mmol; HC2H3O2 = 5.0mmol – 2.5mmol = 2.5mmol; C2H3O2-

= 0.0mmol + 2.5mmol = 2.5mmol formed.So after, Major Species: HC2H3O2, C2H3O2

-, H2O, Na+. Determine pH with HC2H3O2 equilibrium

[Initial]: HC2H3O2 = 2.5mmol

50.0 mL+25 .0mL


5.0 mmol50.0 mL+25 .0 mL


-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

Ka = 1.8 x 10-5 = (x )(0.033+x)0.033−x

= x(0.033)/0.033 = x

x = 1.8 x 10-5

pH = -log[H+] = -log[1.8 x 10-5] = 4.74

This is halfway to the equivalence point. Half of the HC2H3O2 has been converted. [HC2H3O2]0 = [C2H3O2

-]0. Ka = [H+] and pH = pKa.

d. 40.0 mL (total) of 0.10M NaOH has been added.Major Species: HC2H3O2, H2O, Na+, OH-




- = 0.0mmol + 4.0mmol = 4.0mmol formed.So after, Major Species: HC2H3O2, C2H3O2

-, H2O, Na+.

[Initial]: HC2H3O2 = 1.0 mmol

50.0 mL+40 .0mL


4 .0mmol50.0 mL+40 .0 mL


-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

Ka = 1.8 x 10-5 = (x )(0. 011+x)

0.044−x = x(0.044)/0.011 = 4.0x

x = 4.5 x 10-6

pH = -log[H+] = -log[4.5 x 10-6] = 5.35

e. 50.0 mL (total) of 0.10M NaOH has been added.Major Species: HC2H3O2, H2O, Na+, OH-




- = 0.0mmol + 5.0mmol = 5.0mmol formed.So after, Major Species: C2H3O2

-, H2O, Na+. [Initial]: HC2H3O2 = 0


5.0 mmol50.0 mL+50 .0 mL

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HC2H3O2 → C2H3O2

- + H+

I 0.033 0.033 0

C -x +x +x

E 0.033 - x 0.033 + x x

HC2H3O2 → C2H3O2

- + H+

I 0.011 0.044 0

C -x +x +x

E 0.011 - x 0.044 + x x

C2H3O2- → HC2H3O2

+ OH-

I 0.050 0 0

C -x +x +x

E 0.050 - x x x

[Initial]: OH- ≈ 0Kb = 1.0 x 10-14/1.8 x10-5 = 5.6 x 10-10

Kb = [C2H3O2-][H+] / [HC2H3O2] = 5.6 x 10-10

Kb = 5.6 x 10-10 = (x)(x )0.050−x

= x2 / 0.050

x2 = 2.8 x 10-11 x = 5.3 x 10-6

pOH = -log[OH-] = -log[5.3 x 10-6] = 5.28, so pH = 8.72

f. 60.0 mL (total) of 0.100M NaOH has been added.Major Species: HC2H3O2, C2H3O2

-, H2O, Na+, OH-




- = 0.0mmol + 5.0mmol = 5.0mmol formed.After, Major Species: C2H3O2

-, H2O, Na+, OH-. [Initial]: HC2H3O2 = 0


5.0 mmol50.0 mL+60 .0 mL this is weak

[Initial]: OH- = 1.0 mmol

50.0 mL+60.0 mLpOH = -log[OH-] = -log[9.1 x 10-3] = 2.04, so pH = 11.96

g. 75.0 mL (total) of 0.100M NaOH has been added.Major Species: HC2H3O2, C2H3O2

-, H2O, Na+, OH-




- = 0.0mmol + 5.0mmol = 5.0mmol formed.After, Major Species: C2H3O2

-, H2O, Na+, OH-. [Initial]: HC2H3O2 = 0


5.0 mmol50.0 mL+60 .0 mL this is weak

[Initial]: OH- = 2.5 mmol

50.0 mL+75 .0 mLpOH = -log[OH-] = -log[2.0 x 10-2] = 1.70, so pH = 12.30

Be sure to note the difference between the curve on the left and the curve for a strong acid and strong base titration. The difference is noted to the right. Why is the shape the same after the equivalence point?

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Equivalence points for the titrations of weak and strong

acids

Where the amount of added OH-

exactly equals the original amount of acid.

Remember that the equivalence point is defined by stoichiometry NOT by pH. When enough titrant has been added to react exactly with all the acid or base being titrated is the equivalence point.PRACTICE NINEHydrogen cyanide gas, HCN, a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka = 6.2 x 10-10) when dissolved in water. If a 50.0mL sample of 0.100M HCn is titrated with 0.100M NaOH, calculate the pH of the solution

a. After 8.00mL of 0.100M NaOH has been added.b. At the halfway point of the titration.c. At the equivalence point of the titration.

1. It took the same amount of NaOH in both the example and Practice Nine to reach the equivalence point. It is the MOUNT of the acid, not its strength that determines the equivalence point.

2. The pH value at the equivalence point IS affected by acid strength.3. The strength of a weak acid has a significant effect on the shape of

its pH curve. The weaker the acid, the greater the pH at the 12

equivalence point. As the acid becomes weaker, the vertical region (shaded to the left) becomes shorter.

FIVE POINTS OF INTEREST ALONG A TITRATION CURVE for weak acids/bases: A. The pH before the titration begins. Treat as usual, the acid or base in the flask determines the pH. If

weak, an ICE chart is in order. B. The pH on the way to the equivalence point. You are in the “land of buffer” as soon as the first drop

from the buret makes a splash and reacts to form the salt. Whatever is in the buret is the “added” part. Use to solve for the hydrogen ion concentration and subsequently the pH. Either the acid or the base [whichever is in the buret] starts at ZERO. Do the stoichiometry and then an ICE chart.

C. The pH at the midpoint of the titration (½ equivalence point): once the midpoint is reached, [H+] = Ka since ½ of the acid or base has been neutralized, AND the resulting solution in the beaker is composed of the half that remains AND the salt. That means that pH = pKa.

D. The pH at the equivalence point.—you are simply calculating the pH of the salt, all the acid or base is now neutralized [to salt + water!]. Write the hydrolysis reaction for your ICE chart.

E. The pH beyond the equivalence point—it is stoichiometry again with a limiting reactant. Calculate the molarity of the EXCESS and solve for either pH directly (excess H+) or pOH (excess OH−) and subtract it from 14.00 to arrive at pH. Be sure to track the total volume when calculating the molarity!

3. WEAK BASE – STRONG ACID TITRATION We have to do a series of buffer problems like we did earlier. Remember that although the base is weak, it reacts to completion with the H+ ion, a very strong

acid. At the equivalence point, an acidic salt is present and pH < 7 After the equivalence point, the strong acid will be the dominant species and a simple pH

calculation can be done after stoichiometry.

PRACTICE TEN - For the titration of 100.0 mL of 0.050M NH3 (Ka = 1.8 x 10-5) with 0.10M HCl, calculate the pH of the solution at the following selected points of the titration:

a. 0.0 mL of 0.10M HCl has been added.b. Before the equivalence point.c. At the equivalence pointd. Beyond the equivalence point

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SUMMARY OF BUFFERED SOLUTIONS

1. Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid, HA, and the conjugate base, A-, or a weak base, B and the conjugate acid, BH+.

2. When H+ is added to a buffered solution, it reacts essentially to completion with the weak acid present. H+ + A- → HA or H+ + B → BH+

3. When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA → A- + H2O or OH- + BH+ → B + H2O

4. The pH of the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or B and BH+) are large compared with the amounts of H+ and OH- added.

ACID-BASE INDICATORS

Use of an acid-base indicator marks the end point of a titration by changing color. The end point is NOT the equivalence point though with careful selection of an indicator, it can be. When choosing an indicator, we want the indicator end point and the titration equivalence point to

be as close as possible. Since strong acid-strong base titrations have a large vertical area, color changes will be sharp and a

wide range of indicators can be used. For titrations involving weak acids and bases, we must be more careful in our choice of indicator.

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Indicators are usually weak acids, HIn. They have one color in their acidic form, HIn, and another color in their basic form, In-.

A very common indicator is phenolphthalein which is colorless in its HIn form and pink in its In-

form. It changes color in the pH range of 8-10. Usually 1/10 of the initial form f the indicator must be changed to the other form before a new color

is apparent. The following equations are used to determine the pH at which an indicator will change color:

o For titration of an ACID: pH = pKa + log1/10 = pKa – 1o For titration of a BASE: pH = pKa + log10/1 = pKa + 1

The useful range of an indicator is usually pKa + 1. When choosing an indicator, determine the pH at the equivalence point of the titration and then choose an indicator with a pKa close to that.

There is a table of indicator colors in your test, page 715.

PRACTICE ELEVENBromothymol Blue, an indicator with a Ka of 1.0 x10-7, is yellow in its HIn form and blue in its In- form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible?

SOLUBILITY EQUILIBRIA

Saturated solutions of salts are another type of chemical equilibria. For a saturated solution of AgCl, the equation would be: AgCl(s) ↔ Ag+

(aq) + Cl-(aq).

The SOLUBILITY PRODUCT expression would be Ksp = [Ag+][Cl-]. The AgCl(s) is left out of the expression because solids are left out of equilibrium expressions (have constant concentrations).

The Ksp for AgCl = 1.6 x 10-10. This means that is the product of [Ag+][Cl-] is < 1.6 x 10-10, then the solution is unsaturated and no solid would be present. If the product were 1.6 x 10-10, the product is exactly saturdated and no solid would be present. If the product > 1.6 x 10-10, the solution is saturated and a solid (precipitate) would form.

Calculating Concentrations For Ag2CO3: Ag2CO3 (s) ↔ 2Ag+

(aq) + CO3-2

(aq) and Ksp = [Ag+]2[CO3-2]. The Ksp = 8.1 x 10-12.

We can determine the concentrations of 2Ag+(aq) and CO3

-2(aq) using the equation Ksp = [Ag+]2[CO3

-2] = 8.1 x 10-12.Initial [Ag+]2 = 0 EQ [Ag+]2 = x Ksp = [Ag+]2[CO3

-2] = 8.1 x 10-12

Initial [CO3-2] = 0 EQ [CO3

-2] = 2x Ksp = (x)(2x)2 = 4x3 = 8.1 x 10-12

x3 = 2.0 x 10-12 x = 1.3 x 10-4

EQ [Ag+]2 = 1.3 x 10-4M EQ [CO3-2] = 2.6 x 10-4M

PRACTICE TWELVEThe Ksp value for copper(II) iodate, Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at 25°C.

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You can also calculate Ksp for given concentrations.

PRACTICE THIRTEENCopper(I) bromide has a measured solubility of 2.0 x 10-4M at 25°C. Calculate the Ksp value.

PRACTICE FOURTEENCalculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15M at 25°C.

COMMON ION EFECTWhat happens if we dissolve the substance into something other than pure water? What happens when the solution contains a common ion?

PRACTICE FIFTEEN

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What is the molar solubility of solid calcium fluoride, CaF2 in a 0.025M solution of sodium fluoride? Ksp = 4.0 x 10-11

PRECIPITATE and QUALITATIVE ANALYSIS

This is now considering the reverse of the above - forming a solid from solution. The product of the initial concentrations of the ions (raised to the power of their coefficients) is

called the ION PRODUCT or Q. If Q > Ksp, precipitate forms. If Q < Ksp, no precipitate forms. Q = [A+]0

x[B-]0y

PRACTICE SIXTEENA solution is prepared by adding 750.0mL of 4.00 x 10-3M Ce(NO3)3 to 300.0mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 with a Ksp = 1.9 x 10-10 precipitate from this solution? Justify your answer.

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web viewadding naf to a solution of hf causes more hf to be produced. major species are hf, na+, f-,...

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