Chapter Seventeen: Additional Aspects of Aqueous Equilibria Water is the most important solvent on this planet. Aqueous solutions encountered in nature contain many solutes Many equilibria take place in these solutions

The Common Ion Effect Common Ion Effect is still important in Acid/Base equilibria. Using a salt that contains a common ion will cause an acid base equilibria to shift just as we saw in Chapter 15, using Le Chateliers Principle

The Common Ion Effect Consider: HC2H3O2(aq)  H+(aq) + C2H3O2-(aq) - if adding NaC2H3O2, complete dissociation into Na+ and C2H3O2- ions will occur. - additional C2H3O2- ions causes the reaction to shift left, using up H+ ions, and reducing acidity

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2.

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. HNO2  H+ + NO2- Ka = 4.5 x 10-4 = [H+][NO2-] [HNO2]

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. HNO2  H+ + NO2- I 0.085 0 0.10 D -x +x +x E 0.085-x x 0.10+x

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. HNO2  H+ + NO2- Ka = 4.5 x 10-4 = [x][0.10+x] [0.085-x]

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. HNO2  H+ + NO2- Ka = 4.5 x 10-4 = [x][0.10+x] [0.085-x] *make assumption

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. HNO2  H+ + NO2- Ka = 4.5 x 10-4 = [x][0.10] [0.085]

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. HNO2  H+ + NO2- Ka = 4.5 x 10-4 = [x][0.10] [0.085] x = 3.83 x 10-4

The Common Ion Effect Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, HNO2 (Ka = 4.5 x 10-4), and 0.10 M potassium nitrite, KNO2. [H+] = 3.83 x 10-4 pH = -log [H+] pH = 3.42

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3.

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. HCHO2  CHO2- + H+ Ka = 1.8 x 10-4 = [CHO2-][H+] [HCHO2]

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. HCHO2  CHO2- + H+ I 0.050 0 0.10 D -x +x +x E 0.050-x x 0.10+x

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. HCHO2  CHO2- + H+ Ka = 1.8 x 10-4 = [x][0.10+x] [0.050-x]

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. HCHO2  CHO2- + H+ Ka = 1.8 x 10-4 = [x][0.10+x] [0.050-x] * make assumption

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. HCHO2  CHO2- + H+ Ka = 1.8 x 10-4 = [x][0.10] [0.050]

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. HCHO2  CHO2- + H+ Ka = 1.8 x 10-4 = [x][0.10] [0.050] x = 9.0 x 10-5

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. pH will depend on nitric acid, which is a strong acid with 100% dissociation.

The Common Ion Effect Sample exercise: Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 (Ka = 1.8 x 10-4), and 0.10 M in HNO3. pH = -log[H+] pH = -log[0.10] pH = 1.0

The Common Ion Effect The common ion effect is equally important in the consideration of a basic solution! NH3(aq) + H2O  NH4+(aq) + OH-(aq) - adding NH4Cl will cause a shift to the left and a decrease in OH- ion, increasing acidity.

Buffered Solutions Solutions like those just discussed, containing weak conjugate acid-base pairs, resist drastic changes in their pH levels. These solutions are called buffers human blood is an extremely important buffer system

Buffered Solutions Buffers resist changes in pH because they contain both an acidic species to neutralize OH- and a basic species to neutralize H+. The species must not actually react with each other in a neutralization reaction, and that requirement is only fulfilled by weak conjugate acid/base pairs.

Buffered Solutions Buffers can be created by dissolving a salt in its common ion acidic solution. HC2H3O2/ NaC2H3O2 To better understand, consider the following: HX(aq)  H+(aq) + X-(aq) Ka = [H+][X-] [HX]

Buffered Solutions Solving for [H+]: Ka[HX] = [H+] [X-] Examining this we can see that the pH will be dependent on two factors: the value of Ka and the ratio of the conjugate acid base pair.

Buffered Solutions Adding minute amounts of base will alter the proportions of the acid/base concentrations, but not enough to affect the pH. Fig. 17.2

Buffered Solutions Two important characteristics of a buffer are its capacity and its pH. Buffer capacity is the amount of acid or base a buffer can neutralize before the pH begins to change to an appreciable degree The pH depends on the Ka of the solution and the relative amounts of acid/base pair

Buffered Solutions Two important characteristics of a buffer are its capacity and its pH. The greater the amounts of conjugate acid/base pair, the more resistant the ratio of their concentrations, and hence pH, is to change. Because a common ion is shared, the same calculations can be used.

Buffered Solutions Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate.

Buffered Solutions Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Appendix D - Ka = 6.3 x 10-5 HC7H5O2  C7H5O2- + H+ I 0.12 0.20 0 D -x +x +x E 0.12-x 0.20+x x

Buffered Solutions Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Appendix D - Ka = 6.3 x 10-5 HC7H5O2  C7H5O2- + H+ Ka = [0.20+x][x] * make [0.12-x] assumption

Buffered Solutions Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Appendix D - Ka = 6.3 x 10-5 HC7H5O2  C7H5O2- + H+ 6.3 x 10-5 = [0.20][x] [0.12] x = 3.8 x 10-5

Buffered Solutions Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. [H+] = 3.8 x 10-5 pH = 4.42

Buffered Solutions Sample exercise: Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid to produce a pH of 4.00.

Buffered Solutions Sample exercise: Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid to produce a pH of 4.00. (Ka = 6.3 x 10-5) HC7H5O2  C7H5O2- + H+ I 0. 20 x 0 D -1 x 10-4 +1 x 10-4 +1 x 10-4 E 0. 20 x 1 x 10-4

Buffered Solutions Sample exercise: Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid to produce a pH of 4.00. Ka = 6.3 x 10-5 = [C7H5O2-][H+] [HC7H5O2] = [x][1 x 10-4] [0. 20] x = 0.126 M C7H5O2-

Buffered Solutions Addition of a strong acid or base to a buffer: reactions between strong acids and weak bases proceed essentially to completion as long as we do not exceed the buffering capacity of the buffer, we can assume that the strong acid or base will be completely absorbed by the buffer

Buffered Solutions Addition of a strong acid or base to a buffer: to perform calculations 1. Consider the acid base neutralization reaction to determine stoichiometric proportions 2. Use Ka and the new concentrations to calculate [H+]

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer a) before any acid or base is added b) after the addition of 0.015 mol of HNO3 c) after the addition of 0.015 mol of KOH

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer a) before any acid or base is added HCNO  CNO- + H+ I 0.140 0.110 0 D -x +x +x E 0.140-x 0.110+x x

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer a) before any acid or base is added HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.110][x] [0.140]

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer a) before any acid or base is added HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.110][x] [0.140] x = 4.5 x 10-4

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer a) pH = - log[H+] = -log[4.5 x 10-4] = 3.35

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer b) addition of 0.015 mol of HNO3 HCNO  CNO- + H+ I 0.140+0.015 0.110-0.015 x D -x +x +x E 0.155-x 0.095+x x

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer b) addition of 0.015 mol of HNO3 HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.095][x] [0.155]

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer b) addition of 0.015 mol of HNO3 HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.095][x] [0.155] x = 5.7 x 10-4

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer b) addition of 0.015 mol of HNO3 HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.095][x] [0.155] x = 5.7 x 10-4

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer b) pH = - log[H+] = -log[5.7 x 10-4] = 3.24

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer c) addition of 0.015 mol of KOH HCNO  CNO- + H+ I 0.140-0.015 0.110+0.015 0 D -x +x +x E 0.125-x 0.125+x x

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer c) addition of 0.015 mol of KOH HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.125][x] [0.125]

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer c) addition of 0.015 mol of KOH HCNO  CNO- + H+ Ka = 3.5 x 10-4 = [0.125][x] [0.125] x = 3.5 x 10-4

Buffered Solutions Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer b) pH = - log[H+] = -log[3.5 x 10-4] = 3.46

Acid Base Titrations In an acid base titration, a solution containing a known concentration of base is slowly added to an acid. Indicators can be used to signal the equivalence point of a titration A pH meter can be used to monitor the the progress of a reaction producing a pH titration curve, a graph of the pH as a function of the volume of base added

Acid Base Titrations The shape of a titration curve makes it possible to determine the equivalence point in the titration. The titration curve produced when a strong base is added to a strong acid looks like an elongated S, adding an acid to a base would produce an upside down elongated S

Acid Base Titrations Strong Base added to Strong Acid Fig 17.6

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Base Acid 0.100 M * 0.025 L 0.100 M * 0.02490 L 2.50 x 10-3 mol 2.49 x 10-3 **more moles

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Base 0.100 M * 0.025 L 2.50 x 10-3 mol 1.00 x 10-5 mol OH- **more moles

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Get to concentration!! 1.00 x 10-5 mol OH- 0.04990 L

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Get to concentration!! 1.00 x 10-5 mol OH- = 2.00 x 10-4 M OH- 0.04990 L

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Use pOH formula!! pOH = -log[OH-]

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Use pOH formula!! pOH = -log[OH-] = -log[2.00 x 10-4]

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Use pOH formula!! pOH = -log[OH-] = -log[2.00 x 10-4] = 3.7

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Remember pOH + pH = 14

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Remember pOH + pH = 14 14 - 3.7 = pH

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. a) 24.90 mL Remember pOH + pH = 14 14 - 3.7 = pH 10.3 = pH

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. b) 25.10 mL

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. c) 25.10 mL Base Acid 0.100 M * 0.025 L 0.100 M * 0.02510 L 2.50 x 10-3 mol 2.51 x 10-3 **more moles

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. c) 25.10 mL Acid 1.00 x 10-5 mol H+ 0.100 M * 0.02510 L 0.0501 L 2.51 x 10-3 **more moles

Acid Base Titrations Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH. c) 25.10 mL Acid 2.00 x 10-4 M H+ 0.100 M * 0.02510 L pH = 3.70 2.51 x 10-3 **more moles

Acid Base Titrations Optimally an indicator turns colors at the equivalence point in a titration, however, in practice this is not necessary. The pH changes very rapidly near the equivalence point, and in this region a single drop could change the pH greatly. An indicator beginning and ending its color change anywhere on the rapid rise portion of the graph will work.

Acid Base Titrations Weak Acid Strong Base titrations: the curve for a weak acid by a strong base is very similar in shape Fig 17.9 Copyright T. L. Heise 2001 - 2002

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 x 10-5).

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 x 10-5). HC7H5O2 + OH-  C7H5O2- + H2O I 1 x 10-3 5 x 10-4 0 D -5 x 10-4 -5 x 10-4 +5 x 10-4 E 5 x10-4 0 5x 10-4

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 x 10-5). HC7H5O2 + OH-  C7H5O2- + H2O E 5 x10-4 mol 0 5x 10-4 mol 0.050 L 0.050 L 0.01 M 0.01 M

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 x 10-5). HC7H5O2 +  C7H5O2- + H+ Ka = 6.3 x 10-5 = [0.01][x] [0.01] x = 6.3 x 10-5 pH = 4.20

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3.

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3. H+ + NH3  NH4+ I 1 x 10-3 2 x 10-3 0 D -1 x 10-3 -1 x 10-3 +1 x 10-3 E 0 1 x 10-3 1 x 10-3

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3. NH4+  NH3 + H+ I 0.033 0.033 0 D -x +x +x E 0.033-x 0.033+x x

Acid Base Titrations Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3. NH4+  NH3 + H+ Ka = 5.5 x 10-10 = [x][0.033] [0.033] x = 5.5 x 10-10 M H+ pH = 9.26

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOH

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOH HC7H5O2 + OH-  C7H5O2- + H2O 1 x 10-3 mol HC7H5O2 reacts with 1 x 10-3 mol OH- to form 1 x 10-3 mol C7H5O2- in 60 mL of solution, which is a weak base.

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOH C7H5O2- + H2O  HC7H5O2 + OH- I 1.6 x 10-2 0 0 D -x +x +x E 1.6 x 10-2 -x x x

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOH C7H5O2- + H2O  HC7H5O2 + OH- Kb = 1.6 x 10-10 = [x][x] [1.6 x 10-2] x = 1.6 x 10-6 M OH- pOH = 5.80 pH = 8.20 Copyright T. L. Heise 2001 - 2002

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when b) 40.0 mL 0f 0.100 M NH3 is titrated with 0.100 M HCl H+ + NH3  NH4+ 4 x 10-3 mol NH3 reacts with 4 x 10-3 mol H+ to form 4 x 10-3 mol NH4 in 80 mL of solution, which is a weak acid. Copyright T. L. Heise 2001 - 2002

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when b) 40.0 mL 0f 0.100 M NH3 is titrated with 0.100 M HCl NH4+  NH3 + H+ I 0.050 0 0 D -x +x +x E 0.050 -x x x Copyright T. L. Heise 2001 - 2002

Acid Base Titrations Sample exercise: Calculate the pH at the equivalence point when b) 40.0 mL 0f 0.100 M NH3 is titrated with 0.100 M HCl NH4+  NH3 + H+ Ka = 5.6 x 10-10 = [x][x] [0.050] x = 5.3 x 10-6 M H+ pH = 5.28 Copyright T. L. Heise 2001 - 2002

Acid Base Titrations The pH titration curves for weak acids and strong base titrations differ from those of a strong acid strong base titration in 3 noteworthy ways: solutions of weak acids have higher intial pH’s the pH change at the rapid rise portion of the curve is much shorter Copyright T. L. Heise 2001 - 2002

Acid Base Titrations The titration curve for a weak base and strong acid is very similar to the strong base with strong acid and follows the same 3 noteworthy differences proportionally Fig. 17.12 Copyright T. L. Heise 2001 - 2002

Acid Base Titrations Polyprotic Acids: If the acid has more than one ionizable proton, then the titration curve has more than one equivalence point

Solubility Equilibia The equilibria studied so far have involved only acids and bases. They have also been only homogeneous equilibria. Another important type of equilibria exists: The dissolution and precipitation of ionic compounds By considering solubility equilibria, we can make quantitative predictions about the amount of a given compound that will dissolve Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Solubility Product Constant: Saturated solution - the solution is in contact with undissolved solute. A particle of solute dissolves and exactly the same rate as a dissolved particle precipitates. An equilibrium is established between the dissolved and undissolved particles BaSO4(s)  Ba2+(aq) + SO42-(aq) Copyright T. L. Heise 2001 - 2002

Solubility Equilibia BaSO4(s)  Ba2+(aq) + SO42-(aq) Keq = [Ba2+][SO42-] [BaSO4] however, this is a heterogeneous equilibrium and solids are not included so… Ksp = [Ba2+][SO42-] *the solubility product is equal to the product of the concentrations of the ions involved, each raised to the power of its coefficients Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: Give the solubility product constant expressions and values of the Ksp for the following compounds: a) barium carbonate Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: Give the solubility product constant expressions and values of the Ksp for the following compounds: a) barium carbonate Ba2+ BaCO3 CO32- Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: Give the solubility product constant expressions and values of the Ksp for the following compounds: a) barium carbonate BaCO3  Ba2+ + CO32- Ksp = 5.0 x 10-9 = [Ba2+][CO32-] Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: Give the solubility product constant expressions and values of the Ksp for the following compounds: b) silver sulfate Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: Give the solubility product constant expressions and values of the Ksp for the following compounds: b) silver sulfate Ag+ Ag2SO4 SO42- Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: Give the solubility product constant expressions and values of the Ksp for the following compounds: b) silver sulfate Ag2SO4  2Ag+ + SO42- Ksp = 1.5 x 10-5 = [Ag+ ]2[SO42-] Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Please be careful to distinguish between solubility and the solubility product constant. Solubility is the number of grams that will dissolve in a given amount of solvent. Solubility product constant is the equilibrium constant for the equilibrium between the ionic solid and the saturated solution Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: A saturated solution of AgCl in contact with undissolved solid is prepared at 25°C. The concentration of Ag+ ions in the solution is found to be 1.35 x 10-5 M. Assuming that AgCl dissociates completely in water and that there are no other simultaneous equilibria involving Ag+ and Cl- ion in the solution, calculate Ksp for this compound. Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: A saturated solution of AgCl in contact with undissolved solid is prepared at 25°C. The concentration of Ag+ ions in the solution is found to be 1.35 x 10-5 M. Calculate Ksp for this compound. AgCl  Ag+ + Cl- 1.35 x 10-5 1.35 x 10-5 Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: A saturated solution of AgCl in contact with undissolved solid is prepared at 25°C. The concentration of Ag+ ions in the solution is found to be 1.35 x 10-5 M. Calculate Ksp for this compound. Ksp = [Ag+][Cl-] = [1.35 x 10-5][1.35 x 10-5] Ksp = 1.82 x 10-10 Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: The Ksp for Cu(N3)2 is 6.3 x 10-10. What is the solubility of Cu(N3)2 in water in grams per liter? Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: The Ksp for Cu(N3)2 is 6.3 x 10-10. What is the solubility of Cu(N3)2 in water in grams per liter? Cu(N3)2  Cu2- + 2N3- I 0 0 D +x +2x E x 2x Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: The Ksp for Cu(N3)2 is 6.3 x 10-10. What is the solubility of Cu(N3)2 in water in grams per liter? Cu(N3)2  Cu2- + 2N3- Ksp = [Cu2+][N3-]2 6.3 x 10-10 = [x][2x]2 x = 5.4 x 10-4 Copyright T. L. Heise 2001 - 2002

Solubility Equilibia Sample exercise: The Ksp for Cu(N3)2 is 6.3 x 10-10. What is the solubility of Cu(N3)2 in water in grams per liter? Cu(N3)2  Cu2- + 2N3- 5.4 x 10-4 mol 147.588 g = 0.080 g/L 1L 1 mol Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia The solubility of a substance is affected by temperature presence of other solutes presence of common ion pH of solution presence of complexing agents Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Common Ion Effect: The presence of a common ion reduces the amount the ionic salt can dissolve shifting the equilibrium to the left towards the ionic solid Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)2, is 1.6 x 10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH. Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)2, is 1.6 x 10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH. Mn(OH)2  Mn2+ + 2OH- I 0 0.020 D +x +2x E x 2x+0.020

Factors that Affect Solubility Equilibia Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)2, is 1.6 x 10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH. Mn(OH)2  Mn2+ + 2OH- Ksp = 1.6 x 10-13 = [Mn2+][OH-]2 = [x][0.020 + 2x]2 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)2, is 1.6 x 10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH. Ksp = 1.6 x 10-13 = [Mn2+][OH-]2 = [x][0.020 - 2x]2 = [x][0.020]2 x = 4.0 x 10-10 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Solubility and pH The solubility of any substance whose anion is basic will be affected to some extent by the pH of the solution. Anions to be concerned about: OH- CO32- PO43- CN- S2- SO42- Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Write the net ionic equation for the reaction of the following copper (II) compounds with acid: a) CuS Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Write the net ionic equation for the reaction of the following copper (II) compounds with acid: a) CuS CuS + H+  Cu2+ + HS- Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Write the net ionic equation for the reaction of the following copper (II) compounds with acid: b) Cu(N3)2 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Write the net ionic equation for the reaction of the following copper (II) compounds with acid: b) Cu(N3)2 Cu(N3)2 + H+  Cu2+ + 2HN3 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Formation of Complex Ions: a characteristic property of metal ions is their ability to accept an electron pair from water molecules. Other molecules than water can also donate their electron pair to the metal ions to form complex ions. The stability of the complex ion depends upon its size of the Keq for its formation Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Kf is 8 x 1029, with a value this large we can assume that all the Cr(OH)4- dissolves to form Cr3+ ions Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Cr3+ + 4OH-  Cr(OH)4- x 1x10-4 0.010 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Cr3+ + 4OH-  Cr(OH)4- Kf = 8 x 1029 = [Cr(OH)4-] [Cr3+][OH-]4 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Sample exercise: Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Cr3+ + 4OH-  Cr(OH)4- Kf = 8 x 1029 = [0.010] [x][1x10-4]4 x = 1.25 x 10-16 Copyright T. L. Heise 2001 - 2002

Factors that Affect Solubility Equilibia Amphoterism: Many metal hydroxides and oxides that are insoluble in water will dissolve in strong acids and bases. They will do this because they themselves are capable of behaving as an acid or a base. Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Equilibrium can be achieved starting with the substances on either side of a chemical equation. The use of the reaction quotient, Keq, to determine the direction in which a reaction must proceed is important If Keq > Ksp, precipitation will occur Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Equilibrium can be achieved starting with the substances on either side of a chemical equation. The use of the reaction quotient, Keq, to determine the direction in which a reaction must proceed is important If Keq = Ksp, equilibrium exists Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Equilibrium can be achieved starting with the substances on either side of a chemical equation. The use of the reaction quotient, Keq, to determine the direction in which a reaction must proceed is important If Keq < Ksp, solid dissolves until Keq = Ksp Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 10-2 M NaF is mixed with 0.010 L of 1.0 x 10-2 M Ca(NO3)2? Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 10-2 M NaF is mixed with 0.010 L of 1.0 x 10-2 M Ca(NO3)2? Possible reaction: 2NaF + Ca(NO3)2  CaF2 + 2NaNO3 final volume will be 0.060 L sodium salts are very soluble, CaF2 has a Ksp of 3.9 x 10-11 Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 10-2 M NaF is mixed with 0.010 L of 1.0 x 10-2 M Ca(NO3)2? Molarity of Ca+2:(0.010)(1.0 x 10-2) = 1.7 x 10-3 (0.060) Molarity of F-:(0.050)(2.0 x 10-2) = 1.7 x 10-2 (0.060) Keq = (1.7 x 10-3)(1.7 x 10-2) = 2.78 x 10-5 Copyright T. L. Heise 2001 - 2002

Precipitation and Separation of Ions Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 10-2 M NaF is mixed with 0.010 L of 1.0 x 10-2 M Ca(NO3)2? Keq = (1.7 x 10-3)(1.7 x 10-2) = 2.78 x 10-5 Keq>Ksp so precipitation will occur Copyright T. L. Heise 2001 - 2002

Qualitative Analysis for Metallic Elements How can solubility equilibria and complex ion formation be used to detect the presence of particular metal ions in solution? Fig. 17.21 Copyright T. L. Heise 2001 - 2002