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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
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CHE 116 Prof. T.L. Meeks 1Chapter Seventeen: Chapter Seventeen: Additional Aspects of Additional Aspects of Aqueous EquilibriaAqueous Equilibria Water is the most important Water is the most important
solvent on this planet.solvent on this planet. Aqueous solutions Aqueous solutions
encountered in nature encountered in nature contain many solutescontain many solutes
Many equilibria take place in Many equilibria take place in these solutionsthese solutions
CHE 116 Prof. T.L. Meeks 2
The Common Ion EffectThe Common Ion Effect
Common Ion Effect is still important in Common Ion Effect is still important in Acid/Base equilibria. Acid/Base equilibria. Using a salt that contains a Using a salt that contains a
common ion will cause an acid base common ion will cause an acid base equilibria to shift just as we saw in equilibria to shift just as we saw in Chapter 15, using Le Chateliers Chapter 15, using Le Chateliers PrinciplePrinciple
CHE 116 Prof. T.L. Meeks 3
The Common Ion EffectThe Common Ion Effect
Consider:Consider:HCHC22HH33OO22(aq) (aq) H H++(aq) + C(aq) + C22HH33OO22
--(aq)(aq)
- if adding NaC- if adding NaC22HH33OO22, complete , complete dissociation into Na+ and Cdissociation into Na+ and C22HH33OO22
-- ions will occur.ions will occur.- additional C- additional C22HH33OO22
-- ions causes ions causes the the reaction to shift left, using up reaction to shift left, using up HH++ ions, ions, and reducing acidityand reducing acidity
CHE 116 Prof. T.L. Meeks 4
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and ), and 0.10 M potassium nitrite, KNO0.10 M potassium nitrite, KNO22..
CHE 116 Prof. T.L. Meeks 5
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and ), and 0.10 M potassium nitrite, KNO0.10 M potassium nitrite, KNO22..
HNOHNO22 H H++ + NO + NO22--
KKaa = 4.5 x 10 = 4.5 x 10-4-4 = [H = [H++][NO][NO22--]]
[HNO [HNO22]]
CHE 116 Prof. T.L. Meeks 6
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous acid, solution containing 0.085 M nitrous acid, HNOHNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and 0.10 M ), and 0.10 M potassium nitrite, KNOpotassium nitrite, KNO22..
HNOHNO22 H H++ + NO + NO22--
II 0.085 0.085 0 0.10 0 0.10 -x +x +x-x +x +xEE0.085-x0.085-x xx 0.10+x 0.10+x
CHE 116 Prof. T.L. Meeks 7
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and ), and 0.10 M potassium nitrite, KNO0.10 M potassium nitrite, KNO22..
HNOHNO22 H H++ + NO + NO22--
KKaa = 4.5 x 10 = 4.5 x 10-4-4 = [x][0.10+x] = [x][0.10+x] [0.085-x] [0.085-x]
CHE 116 Prof. T.L. Meeks 8
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and 0.10 ), and 0.10 M potassium nitrite, KNOM potassium nitrite, KNO22..
HNOHNO22 H H++ + NO + NO22--
KKaa = 4.5 x 10 = 4.5 x 10-4-4 = [x][0.10+x] = [x][0.10+x] [0.085-x] [0.085-x]*make assumption*make assumption
CHE 116 Prof. T.L. Meeks 9
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and 0.10 ), and 0.10 M potassium nitrite, KNOM potassium nitrite, KNO22..
HNOHNO22 H H++ + NO + NO22--
KKaa = 4.5 x 10 = 4.5 x 10-4-4 = [x][0.10] = [x][0.10] [0.085] [0.085]
CHE 116 Prof. T.L. Meeks 10
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and 0.10 ), and 0.10 M potassium nitrite, KNOM potassium nitrite, KNO22..
HNOHNO22 H H++ + NO + NO22--
KKaa = 4.5 x 10 = 4.5 x 10-4-4 = [x][0.10] = [x][0.10] [0.085] [0.085]x = 3.83 x 10x = 3.83 x 10-4-4
CHE 116 Prof. T.L. Meeks 11
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a solution containing 0.085 M nitrous solution containing 0.085 M nitrous acid, HNOacid, HNO22 (K (Kaa = 4.5 x 10 = 4.5 x 10-4-4), and 0.10 ), and 0.10 M potassium nitrite, KNOM potassium nitrite, KNO22.. [H[H++] = 3.83 x 10] = 3.83 x 10-4-4
pH = -log [HpH = -log [H++] ] pH = 3.42pH = 3.42
CHE 116 Prof. T.L. Meeks 12
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
CHE 116 Prof. T.L. Meeks 13
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
HCHOHCHO22 CHO CHO22-- + H + H++
KKaa = 1.8 x 10 = 1.8 x 10-4-4 = [CHO = [CHO22--][H][H++]]
[HCHO [HCHO22]]
CHE 116 Prof. T.L. Meeks 14
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
HCHOHCHO22 CHO CHO22-- + H + H++
II 0.050 0 0.10 0.050 0 0.10D -x +x +xD -x +x +xEE 0.050-x x 0.10+x 0.050-x x 0.10+x
CHE 116 Prof. T.L. Meeks 15
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
HCHOHCHO22 CHO CHO22-- + H + H++
KKaa = 1.8 x 10 = 1.8 x 10-4-4 = [x][0.10+x] = [x][0.10+x] [0.050-x] [0.050-x]
CHE 116 Prof. T.L. Meeks 16
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
HCHOHCHO22 CHO CHO22-- + H + H++
KKaa = 1.8 x 10 = 1.8 x 10-4-4 = [x][0.10+x] = [x][0.10+x] [0.050-x] [0.050-x]* make assumption* make assumption
CHE 116 Prof. T.L. Meeks 17
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
HCHOHCHO22 CHO CHO22-- + H + H++
KKaa = 1.8 x 10 = 1.8 x 10-4-4 = [x][0.10] = [x][0.10] [0.050] [0.050]
CHE 116 Prof. T.L. Meeks 18
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
HCHOHCHO22 CHO CHO22-- + H + H++
KKaa = 1.8 x 10 = 1.8 x 10-4-4 = [x][0.10] = [x][0.10] [0.050] [0.050]x = 9.0 x 10x = 9.0 x 10-5-5
CHE 116 Prof. T.L. Meeks 19
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
pH will depend on nitric acid, pH will depend on nitric acid, which is a strong acid with 100% which is a strong acid with 100% dissociation.dissociation.
CHE 116 Prof. T.L. Meeks 20
The Common Ion EffectThe Common Ion Effect
Sample exercise: Calculate the formate Sample exercise: Calculate the formate ion concentration and pH of a solution ion concentration and pH of a solution that is 0.050 M in formic acid, HCHOthat is 0.050 M in formic acid, HCHO22 (K(Kaa = 1.8 x 10 = 1.8 x 10-4-4), and 0.10 M in HNO), and 0.10 M in HNO33..
pH = -log[HpH = -log[H++]]pH = -log[0.10]pH = -log[0.10]pH = 1.0pH = 1.0
CHE 116 Prof. T.L. Meeks 21
The Common Ion EffectThe Common Ion Effect
The common ion effect is equally The common ion effect is equally important in the consideration of a important in the consideration of a basic solution!basic solution!
NHNH33(aq) + H(aq) + H22O O NH NH44++(aq) + OH(aq) + OH--(aq)(aq)
- adding NH- adding NH44Cl will cause a shift to the Cl will cause a shift to the left and a decrease in OHleft and a decrease in OH- - ion, ion, increasing acidity.increasing acidity.
CHE 116 Prof. T.L. Meeks 22
Buffered SolutionsBuffered Solutions
Solutions like those just discussed, Solutions like those just discussed, containing weak conjugate acid-base containing weak conjugate acid-base pairs, resist drastic changes in their pairs, resist drastic changes in their pH levels.pH levels.
These solutions are called buffersThese solutions are called buffershuman blood is an extremely human blood is an extremely
important buffer systemimportant buffer system
CHE 116 Prof. T.L. Meeks 23
Buffered SolutionsBuffered Solutions
Buffers resist changes in pH because Buffers resist changes in pH because they contain both an acidic species to they contain both an acidic species to neutralize OHneutralize OH-- and a basic species to and a basic species to neutralize Hneutralize H++..
The species must not actually react with The species must not actually react with each other in a neutralization reaction, each other in a neutralization reaction, and that requirement is only fulfilled and that requirement is only fulfilled by weak conjugate acid/base pairs.by weak conjugate acid/base pairs.
CHE 116 Prof. T.L. Meeks 24
Buffered SolutionsBuffered Solutions
Buffers can be created by dissolving a Buffers can be created by dissolving a salt in its common ion acidic solution.salt in its common ion acidic solution. HCHC22HH33OO22/ NaC/ NaC22HH33OO22
To better understand, consider the To better understand, consider the following:following:HX(aq) HX(aq) H H++(aq) + X(aq) + X--(aq)(aq)KKaa = [H = [H++][X][X--]] [HX] [HX]
CHE 116 Prof. T.L. Meeks 25
Buffered SolutionsBuffered Solutions
Solving for [HSolving for [H++]:]:KKaa[HX] = [H[HX] = [H++]] [X [X--]]
Examining this we can see that the pH Examining this we can see that the pH will be dependent on two factors: the will be dependent on two factors: the value of Kvalue of Kaa and the ratio of the and the ratio of the conjugate acid base pair.conjugate acid base pair.
CHE 116 Prof. T.L. Meeks 26
Buffered SolutionsBuffered Solutions
Adding minute amounts of base will Adding minute amounts of base will alter the proportions of the acid/base alter the proportions of the acid/base concentrations, but not enough to concentrations, but not enough to affect the pH.affect the pH.
Fig. 17.2Fig. 17.2
CHE 116 Prof. T.L. Meeks 27
Buffered SolutionsBuffered Solutions
Two important characteristics of a buffer Two important characteristics of a buffer are its capacity and its pH.are its capacity and its pH.Buffer capacity is the amount of acid Buffer capacity is the amount of acid
or base a buffer can neutralize before or base a buffer can neutralize before the pH begins to change to an the pH begins to change to an appreciable degreeappreciable degree
The pH depends on the KThe pH depends on the Kaa of the of the solution and the relative amounts of solution and the relative amounts of acid/base pair acid/base pair
CHE 116 Prof. T.L. Meeks 28
Buffered SolutionsBuffered Solutions
Two important characteristics of a buffer Two important characteristics of a buffer are its capacity and its pH.are its capacity and its pH.The greater the amounts of conjugate The greater the amounts of conjugate
acid/base pair, the more resistant the acid/base pair, the more resistant the ratio of their concentrations, and ratio of their concentrations, and hence pH, is to change.hence pH, is to change.
Because a common ion is shared, the Because a common ion is shared, the same calculations can be used. same calculations can be used.
CHE 116 Prof. T.L. Meeks 29
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. acid and 0.20 M sodium benzoate.
CHE 116 Prof. T.L. Meeks 30
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic acid buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. and 0.20 M sodium benzoate. Appendix D - KAppendix D - Kaa = 6.3 x 10 = 6.3 x 10-5-5
HCHC77HH55OO22 C C77HH55OO22-- + H + H++
I 0.12 0.20 0I 0.12 0.20 0 -x +x +x-x +x +xE 0.12-x 0.20+x x E 0.12-x 0.20+x x
CHE 116 Prof. T.L. Meeks 31
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. acid and 0.20 M sodium benzoate.
Appendix D - KAppendix D - Kaa = 6.3 x 10 = 6.3 x 10-5-5
HCHC77HH55OO22 C C77HH55OO22-- + H + H++
KKaa = [0.20+x][x] = [0.20+x][x] * make * make [0.12-x] [0.12-x] assumptionassumption
CHE 116 Prof. T.L. Meeks 32
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. acid and 0.20 M sodium benzoate.
Appendix D - KAppendix D - Kaa = 6.3 x 10 = 6.3 x 10-5-5
HCHC77HH55OO22 C C77HH55OO22-- + H + H++
6.3 x 106.3 x 10-5-5 = [0.20][x] = [0.20][x] [0.12] [0.12] x = 3.8 x 10 x = 3.8 x 10-5-5
CHE 116 Prof. T.L. Meeks 33
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the pH of a Sample exercise: Calculate the pH of a buffer composed of 0.12 M benzoic buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. acid and 0.20 M sodium benzoate.
[H[H++] = 3.8 x 10] = 3.8 x 10-5-5
pH = 4.42pH = 4.42
CHE 116 Prof. T.L. Meeks 34
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the Sample exercise: Calculate the concentration of sodium benzoate that concentration of sodium benzoate that must be present in a 0.20 M solution must be present in a 0.20 M solution of benzoic acid to produce a pH of of benzoic acid to produce a pH of 4.00.4.00.
CHE 116 Prof. T.L. Meeks 35
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the concentration of Sample exercise: Calculate the concentration of sodium benzoate that must be present in a 0.20 sodium benzoate that must be present in a 0.20 M solution of benzoic acid to produce a pH of M solution of benzoic acid to produce a pH of 4.00. (K4.00. (Kaa = 6.3 x 10 = 6.3 x 10-5-5))
HCHC77HH55OO22 C C77HH55OO22-- + H + H++
I 0. 20 x 0I 0. 20 x 0 -1 x 10-1 x 10-4-4 +1 x 10 +1 x 10-4-4 +1 x 10 +1 x 10-4-4
E 0. 20 xE 0. 20 x 1 x 10 1 x 10-4-4
CHE 116 Prof. T.L. Meeks 36
Buffered SolutionsBuffered Solutions
Sample exercise: Calculate the concentration Sample exercise: Calculate the concentration of sodium benzoate that must be present in a of sodium benzoate that must be present in a 0.20 M solution of benzoic acid to produce a 0.20 M solution of benzoic acid to produce a pH of 4.00. pH of 4.00. KKaa = 6.3 x 10 = 6.3 x 10-5 -5 = [C= [C77HH55OO22
--][H][H++]] [HC [HC77HH55OO22]] = [x][1 x 10= [x][1 x 10-4-4]] [0. 20] [0. 20] x = x = 0.126 M C0.126 M C77HH55OO22
--
CHE 116 Prof. T.L. Meeks 37
Buffered SolutionsBuffered Solutions
Addition of a strong acid or base to a buffer:Addition of a strong acid or base to a buffer:reactions between strong acids and weak reactions between strong acids and weak
bases proceed essentially to completionbases proceed essentially to completionas long as we do not exceed the buffering as long as we do not exceed the buffering
capacity of the buffer, we can assume capacity of the buffer, we can assume that the strong acid or base will be that the strong acid or base will be completely absorbed by the buffercompletely absorbed by the buffer
CHE 116 Prof. T.L. Meeks 38
Buffered SolutionsBuffered Solutions
Addition of a strong acid or base to a buffer:Addition of a strong acid or base to a buffer:to perform calculationsto perform calculations
1. Consider the acid base 1. Consider the acid base neutralization reaction to determine neutralization reaction to determine stoichiometric proportionsstoichiometric proportions
2. Use K2. Use Kaa and and the new concentrations to calculate [Hthe new concentrations to calculate [H++]]
CHE 116 Prof. T.L. Meeks 39
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
a) before any acid or base is addeda) before any acid or base is addedb) after the addition of 0.015 mol of b) after the addition of 0.015 mol of
HNOHNO33c) after the addition of 0.015 mol of c) after the addition of 0.015 mol of
KOHKOH
CHE 116 Prof. T.L. Meeks 40
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding 0.140 Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer Calculate the pH of the buffer a) before any a) before any acid or base is addedacid or base is added HCNO HCNO CNO CNO-- + H + H++
II 0.140 0.140 0.110 00.110 0 -x +x +x-x +x +xE 0.140-x 0.110+x xE 0.140-x 0.110+x x
CHE 116 Prof. T.L. Meeks 41
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
a) before any acid or base is addeda) before any acid or base is addedHCNO HCNO CNO CNO-- + H + H++
KKaa = 3.5 x 10 = 3.5 x 10-4-4 = [0.110][x] = [0.110][x] [0.140] [0.140]
CHE 116 Prof. T.L. Meeks 42
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer a) a) before any acid or base is addedbefore any acid or base is added HCNO HCNO CNO CNO-- + H + H++
KKaa = 3.5 x 10 = 3.5 x 10-4-4 = [0.110][x] = [0.110][x] [0.140] [0.140]x = 4.5 x 10x = 4.5 x 10-4-4
CHE 116 Prof. T.L. Meeks 43
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
a) pH = - log[H+]a) pH = - log[H+] = -log[4.5 x 10= -log[4.5 x 10-4-4]] = 3.35= 3.35
CHE 116 Prof. T.L. Meeks 44
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding 0.140 Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferCalculate the pH of the buffer b) addition of b) addition of 0.015 mol of HNO0.015 mol of HNO33 HCNO HCNO CNO CNO-- + H + H++
II 0.140+0.015 0.140+0.015 0.110-0.015 x0.110-0.015 x -x +x +x-x +x +xE 0.155-x 0.095+x xE 0.155-x 0.095+x x
CHE 116 Prof. T.L. Meeks 45
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
b) addition of 0.015 mol of HNO b) addition of 0.015 mol of HNO33 HCNO HCNO CNO CNO-- + H + H++
KKaa = 3.5 x 10 = 3.5 x 10-4-4 = [0.095][x] = [0.095][x] [0.155] [0.155]
CHE 116 Prof. T.L. Meeks 46
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding 0.140 Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer Calculate the pH of the buffer b) addition of b) addition of 0.015 mol of HNO0.015 mol of HNO33 HCNO HCNO CNO CNO-- + H + H++
KKaa = 3.5 x 10 = 3.5 x 10-4-4 = [0.095][x] = [0.095][x] [0.155] [0.155]x = 5.7 x 10x = 5.7 x 10-4-4
CHE 116 Prof. T.L. Meeks 47
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding 0.140 Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. potassium cyanate, KCNO, to sufficient water. Calculate the pH of the buffer Calculate the pH of the buffer b) addition of b) addition of 0.015 mol of HNO0.015 mol of HNO33 HCNO HCNO CNO CNO-- + H + H++
KKaa = 3.5 x 10 = 3.5 x 10-4-4 = [0.095][x] = [0.095][x] [0.155] [0.155]x = 5.7 x 10x = 5.7 x 10-4-4
CHE 116 Prof. T.L. Meeks 48
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
b) pH = - log[H+]b) pH = - log[H+] = -log[5.7 x 10= -log[5.7 x 10-4-4]] = 3.24= 3.24
CHE 116 Prof. T.L. Meeks 49
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding 0.140 Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient water. potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferCalculate the pH of the buffer c) addition of c) addition of 0.015 mol of KOH 0.015 mol of KOH HCNO HCNO CNO CNO-- + H + H++
II 0.140-0.015 0.140-0.015 0.110+0.015 00.110+0.015 0 -x +x +x-x +x +xE 0.125-x 0.125+x xE 0.125-x 0.125+x x
CHE 116 Prof. T.L. Meeks 50
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
c) addition of 0.015 mol of KOH c) addition of 0.015 mol of KOH HCNO HCNO CNO CNO-- + H + H++
Ka = 3.5 x 10Ka = 3.5 x 10-4-4 = [0.125][x] = [0.125][x] [0.125] [0.125]
CHE 116 Prof. T.L. Meeks 51
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding 0.140 Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer c) c) addition of 0.015 mol of KOH addition of 0.015 mol of KOH HCNO HCNO CNOCNO-- + H + H++
Ka = 3.5 x 10Ka = 3.5 x 10-4-4 = [0.125][x] = [0.125][x] [0.125] [0.125]x = 3.5 x 10x = 3.5 x 10-4-4
CHE 116 Prof. T.L. Meeks 52
Buffered SolutionsBuffered Solutions
Consider a 1.00 L buffer made by adding Consider a 1.00 L buffer made by adding 0.140 mol cyanic acid, HCNO, and 0.110 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO, to sufficient mol potassium cyanate, KCNO, to sufficient water. Calculate the pH of the bufferwater. Calculate the pH of the buffer
b) pH = - log[H+]b) pH = - log[H+] = -log[3.5 x 10= -log[3.5 x 10-4-4]] = 3.46= 3.46
CHE 116 Prof. T.L. Meeks 53
Acid Base TitrationsAcid Base Titrations
In an acid base titration, a solution containing a In an acid base titration, a solution containing a known concentration of base is slowly added to known concentration of base is slowly added to an acid. an acid.
Indicators can be used to signal the equivalence Indicators can be used to signal the equivalence point of a titrationpoint of a titration
A pH meter can be used to monitor the the A pH meter can be used to monitor the the progress of a reaction producing a pH titration progress of a reaction producing a pH titration curve, a graph of the pH as a function of the curve, a graph of the pH as a function of the volume of base addedvolume of base added
CHE 116 Prof. T.L. Meeks 54
Acid Base TitrationsAcid Base Titrations
The shape of a titration curve makes it The shape of a titration curve makes it possible to determine the equivalence point possible to determine the equivalence point in the titration.in the titration.
The titration curve produced when a strong The titration curve produced when a strong base is added to a strong acid looks like an base is added to a strong acid looks like an elongated S, adding an acid to a base would elongated S, adding an acid to a base would produce an upside down elongated Sproduce an upside down elongated S
CHE 116 Prof. T.L. Meeks 55
Acid Base TitrationsAcid Base Titrations
Strong Base added to Strong AcidStrong Base added to Strong AcidFig 17.6Fig 17.6
CHE 116 Prof. T.L. Meeks 56
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL
CHE 116 Prof. T.L. Meeks 57
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL BaseBase AcidAcid
0.100 M * 0.025 L 0.100 M * 0.02490 L 0.100 M * 0.025 L 0.100 M * 0.02490 L 2.50 x 102.50 x 10-3-3 mol 2.49 x 10 mol 2.49 x 10-3-3 **more moles**more moles
CHE 116 Prof. T.L. Meeks 58
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL BaseBase
0.100 M * 0.025 L 0.100 M * 0.025 L 2.50 x 102.50 x 10-3-3 mol 1.00 x 10 mol 1.00 x 10-5 -5 mol OHmol OH-- **more moles **more moles
CHE 116 Prof. T.L. Meeks 59
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL Get to concentration!!Get to concentration!! 1.00 x 101.00 x 10-5 -5 mol OHmol OH--
0.04990 L0.04990 L
CHE 116 Prof. T.L. Meeks 60
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL Get to concentration!!Get to concentration!! 1.00 x 101.00 x 10-5 -5 mol OHmol OH-- = 2.00 x 10 = 2.00 x 10-4-4 M OH M OH--
0.04990 L0.04990 L
CHE 116 Prof. T.L. Meeks 61
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL Use pOH formula!!Use pOH formula!! pOH = -log[OHpOH = -log[OH--]]
CHE 116 Prof. T.L. Meeks 62
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.a) 24.90 mLa) 24.90 mL
Use pOH formula!!Use pOH formula!! pOH = -log[OHpOH = -log[OH--]] = -log[2.00 x 10= -log[2.00 x 10-4-4]]
CHE 116 Prof. T.L. Meeks 63
Acid Base TitrationsAcid Base TitrationsSample exercise: Calculate the pH when the following Sample exercise: Calculate the pH when the following
quantities of 0.100 M HNOquantities of 0.100 M HNO33 have been added to 25.00 have been added to 25.00 mL of 0.100 M KOH.mL of 0.100 M KOH.a) 24.90 mLa) 24.90 mL
Use pOH formula!!Use pOH formula!! pOH = -log[OHpOH = -log[OH--]] = -log[2.00 x 10= -log[2.00 x 10-4-4]]
= 3.7= 3.7
CHE 116 Prof. T.L. Meeks 64
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mLa) 24.90 mL RememberRemember pOH + pH = 14pOH + pH = 14
CHE 116 Prof. T.L. Meeks 65
Acid Base TitrationsAcid Base TitrationsSample exercise: Calculate the pH when the following Sample exercise: Calculate the pH when the following
quantities of 0.100 M HNOquantities of 0.100 M HNO33 have been added to 25.00 have been added to 25.00 mL of 0.100 M KOH.mL of 0.100 M KOH.a) 24.90 mLa) 24.90 mL
RememberRemember pOH + pH = 14pOH + pH = 14 14 - 3.7 = pH14 - 3.7 = pH
CHE 116 Prof. T.L. Meeks 66
Acid Base TitrationsAcid Base TitrationsSample exercise: Calculate the pH when the following Sample exercise: Calculate the pH when the following
quantities of 0.100 M HNOquantities of 0.100 M HNO33 have been added to 25.00 mL of have been added to 25.00 mL of 0.100 M KOH.0.100 M KOH.a) 24.90 mLa) 24.90 mL
RememberRemember pOH + pH = 14pOH + pH = 14 14 - 3.7 = pH14 - 3.7 = pH
10.3 = pH10.3 = pH
CHE 116 Prof. T.L. Meeks 67
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
b) 25.10 mLb) 25.10 mL
CHE 116 Prof. T.L. Meeks 68
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
c) 25.10 mLc) 25.10 mLBaseBase AcidAcid 0.100 0.100 M * 0.025 L 0.100 M * 0.02510 L 2.50 M * 0.025 L 0.100 M * 0.02510 L 2.50 x 10x 10-3-3 mol 2.51 x 10 mol 2.51 x 10-3-3
**more moles**more moles
CHE 116 Prof. T.L. Meeks 69
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNOfollowing quantities of 0.100 M HNO33 have have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
c) 25.10 mLc) 25.10 mLAcidAcid
1.00 x 101.00 x 10-5-5 mol H mol H++ 0.100 M * 0.02510 L 0.100 M * 0.02510 L 0.0501 L0.0501 L 2.51 x 10 2.51 x 10-3-3
**more moles**more moles
CHE 116 Prof. T.L. Meeks 70
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH when the Sample exercise: Calculate the pH when the following quantities of 0.100 M HNO3 have following quantities of 0.100 M HNO3 have been added to 25.00 mL of 0.100 M KOH.been added to 25.00 mL of 0.100 M KOH.
c) 25.10 mLc) 25.10 mLAcidAcid
2.00 x 102.00 x 10-4-4 M H M H++ 0.100 M * 0.02510 L 0.100 M * 0.02510 L pH = 3.70 2.51 x 10pH = 3.70 2.51 x 10-3-3
**more moles**more moles
CHE 116 Prof. T.L. Meeks 71
Acid Base TitrationsAcid Base Titrations
Optimally an indicator turns colors at the Optimally an indicator turns colors at the equivalence point in a titration, however, in equivalence point in a titration, however, in practice this is not necessary.practice this is not necessary.The pH changes very rapidly near the The pH changes very rapidly near the
equivalence point, and in this region a single equivalence point, and in this region a single drop could change the pH greatly.drop could change the pH greatly.
An indicator beginning and ending its color An indicator beginning and ending its color change anywhere on the rapid rise portion change anywhere on the rapid rise portion of the graph will work.of the graph will work.
CHE 116 Prof. T.L. Meeks 72
Acid Base TitrationsAcid Base Titrations
Weak Acid Strong Base titrations: the curve Weak Acid Strong Base titrations: the curve for a weak acid by a strong base is very for a weak acid by a strong base is very similar in shapesimilar in shape
Fig 17.9Fig 17.9
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 73
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic M NaOH to 40.0 mL of 0.0250 M benzoic acid (HCacid (HC77HH55OO22, K, Kaa = 6.3 x 10 = 6.3 x 10-5-5).).
CHE 116 Prof. T.L. Meeks 74
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid NaOH to 40.0 mL of 0.0250 M benzoic acid (HC(HC77HH55OO22, K, Kaa = 6.3 x 10 = 6.3 x 10-5-5).).
HCHC77HH55OO22 + OH + OH-- C C77HH55OO22-- + H + H22OO
II 1 x 10 1 x 10-3-3 5 x 10 5 x 10-4-4 0 0 -5 x 10-5 x 10-4-4 -5 x 10 -5 x 10-4-4 +5 x 10 +5 x 10-4-4 E 5 x10E 5 x10-4-4 0 5x 10 0 5x 10-4-4
CHE 116 Prof. T.L. Meeks 75
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic M NaOH to 40.0 mL of 0.0250 M benzoic acid (HCacid (HC77HH55OO22, K, Kaa = 6.3 x 10 = 6.3 x 10-5-5).).
HCHC77HH55OO22 + OH + OH-- C C77HH55OO22-- + H + H22OO
E 5 x10E 5 x10-4-4 mol 0 5x 10 mol 0 5x 10-4-4 mol mol 0.050 L 0.050 L 0.050 L 0.050 L
0.01 M0.01 M 0.01 M 0.01 M
CHE 116 Prof. T.L. Meeks 76
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.050 solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic M NaOH to 40.0 mL of 0.0250 M benzoic acid (HCacid (HC77HH55OO22, K, Kaa = 6.3 x 10 = 6.3 x 10-5-5).).
HCHC77HH55OO22 + + C C77HH55OO22-- + H + H++
KKaa = 6.3 x 10 = 6.3 x 10-5-5 = [0.01][x] = [0.01][x] [0.01] [0.01] x = 6.3 x 10x = 6.3 x 10-5-5 pH = 4.20pH = 4.20
CHE 116 Prof. T.L. Meeks 77
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NHM HCl to 20.0 mL of 0.100 M NH33..
CHE 116 Prof. T.L. Meeks 78
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NHM HCl to 20.0 mL of 0.100 M NH33..
HH++ + NH + NH33 NH NH44+ +
II 1 x 10 1 x 10-3-3 2 x 10 2 x 10-3-3 0 0 -1 x 10-1 x 10-3-3 -1 x 10 -1 x 10-3-3 +1 x 10 +1 x 10-3-3 E 0 1 x 10E 0 1 x 10-3-3 1 x 10 1 x 10-3-3
CHE 116 Prof. T.L. Meeks 79
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NHM HCl to 20.0 mL of 0.100 M NH33..
NHNH44++ NH NH33
+ H+ H++
II 0.033 0.033 0 0.033 0.033 0 -x +x +x -x +x +x E 0.033-x 0.033+x xE 0.033-x 0.033+x x
CHE 116 Prof. T.L. Meeks 80
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH in the Sample exercise: Calculate the pH in the solution formed by adding 10.0 mL of 0.100 solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NHM HCl to 20.0 mL of 0.100 M NH33..
NHNH44++ NH NH33
+ H+ H++
Ka = 5.5 x 10Ka = 5.5 x 10-10-10 = [x][0.033] = [x][0.033] [0.033] [0.033] x = 5.5 x 10x = 5.5 x 10-10 -10 MM HH+ +
pH = 9.26pH = 9.26
CHE 116 Prof. T.L. Meeks 81
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when
a) 40.0 mL of 0.025 M benzoic acid is a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOHtitrated with 0.050 M NaOH
CHE 116 Prof. T.L. Meeks 82
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when
a) 40.0 mL of 0.025 M benzoic acid is a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOHtitrated with 0.050 M NaOHHCHC77HH55OO22 + OH + OH-- C C77HH55OO22
-- + H + H22OO
1 x 101 x 10-3-3 mol HC mol HC77HH55OO22 reacts with 1 x 10 reacts with 1 x 10-3-3 mol mol OHOH-- to form 1 x 10 to form 1 x 10-3-3 mol C mol C77HH55OO22
-- in 60 mL in 60 mL of solution, which is a weak base. of solution, which is a weak base.
CHE 116 Prof. T.L. Meeks 83
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when a) 40.0 mL of 0.025 a) 40.0 mL of 0.025 M benzoic acid is M benzoic acid is titrated with 0.050 M titrated with 0.050 M NaOHNaOHCC77HH55OO22
-- + H + H22O O HC HC77HH55OO22 + OH + OH--
I I 1.6 x 10 1.6 x 10-2-2 0 0 0 0 -x-x +x +x +x +xEE 1.6 x 10 1.6 x 10-2-2 -x -x x x x x
CHE 116 Prof. T.L. Meeks 84
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when a) 40.0 mL of 0.025 a) 40.0 mL of 0.025 M benzoic acid is M benzoic acid is titrated with 0.050 M titrated with 0.050 M NaOHNaOHCC77HH55OO22
-- + H + H22O O HC HC77HH55OO22 + OH + OH--
KKbb = 1.6 x 10 = 1.6 x 10-10-10 = [x][x] = [x][x] [1.6 x 10 [1.6 x 10-2-2]]x = 1.6 x 10x = 1.6 x 10-6-6 M OH M OH-- pOH = 5.80pOH = 5.80 pH = 8.20pH = 8.20
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 85
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when b) 40.0 mL 0f b) 40.0 mL 0f 0.100 M NH0.100 M NH33 is titrated is titrated with 0.100 M HClwith 0.100 M HCl
HH++ + NH + NH33 NH NH44++
4 x 104 x 10-3-3 mol NH mol NH33 reacts with 4 x 10 reacts with 4 x 10-3-3 mol H mol H++ to to form 4 x 10form 4 x 10-3-3 mol NH mol NH44 in 80 mL of solution, in 80 mL of solution, which is a weak acid. which is a weak acid.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 86
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when b) 40.0 mL 0f 0.100 b) 40.0 mL 0f 0.100 M NHM NH33 is titrated is titrated with 0.100 M HClwith 0.100 M HCl
NHNH44+ + NH NH3 3 + H+ H++
I I 0.050 0.050 0 0 0 0 -x-x +x +x +x +xEE 0.050 -x 0.050 -x x x x x
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 87
Acid Base TitrationsAcid Base Titrations
Sample exercise: Calculate the pH at the Sample exercise: Calculate the pH at the equivalence point when equivalence point when b) 40.0 mL 0f 0.100 b) 40.0 mL 0f 0.100 M NHM NH33 is titrated is titrated with 0.100 M HClwith 0.100 M HCl
NHNH44+ + NH NH3 3 + H+ H++
KKaa = 5.6 x 10 = 5.6 x 10-10-10 = [x][x] = [x][x] [0.050] [0.050]x = 5.3 x 10x = 5.3 x 10-6-6 M H M H++ pH = 5.28pH = 5.28
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 88
Acid Base TitrationsAcid Base Titrations
The pH titration curves for weak acids and The pH titration curves for weak acids and strong base titrations differ from those of a strong base titrations differ from those of a strong acid strong base titration in 3 strong acid strong base titration in 3 noteworthy ways:noteworthy ways:solutions of weak acids have higher intial solutions of weak acids have higher intial
pH’spH’sthe pH change at the rapid rise portion the pH change at the rapid rise portion
of the curve is much shorterof the curve is much shorter
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 89
Acid Base TitrationsAcid Base Titrations
The titration curve for a weak base and The titration curve for a weak base and strong acid is very similar to the strong base strong acid is very similar to the strong base with strong acid and follows the same 3 with strong acid and follows the same 3 noteworthy differences proportionallynoteworthy differences proportionally
Fig. 17.12Fig. 17.12
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 90
Acid Base TitrationsAcid Base Titrations
Polyprotic Acids: If Polyprotic Acids: If the the acid has more acid has more than than one ionizable one ionizable
proton, then the proton, then the titration curve titration curve
has has more than one more than one equivalence pointequivalence point
CHE 116 Prof. T.L. Meeks 91
Solubility EquilibiaSolubility Equilibia
The equilibria studied so far have involved only The equilibria studied so far have involved only acids and bases. They have also been only acids and bases. They have also been only homogeneous equilibria. homogeneous equilibria.
Another important type of equilibria exists:Another important type of equilibria exists:The dissolution and precipitation of ionic The dissolution and precipitation of ionic compoundscompoundsBy considering solubility equilibria, we can By considering solubility equilibria, we can make quantitative predictions about the amount make quantitative predictions about the amount of a given compound that will dissolveof a given compound that will dissolve
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 92
Solubility EquilibiaSolubility Equilibia
Solubility Product Constant:Solubility Product Constant:Saturated solution - the solution is in contact Saturated solution - the solution is in contact with undissolved solute. A particle of solute with undissolved solute. A particle of solute dissolves and exactly the same rate as a dissolves and exactly the same rate as a dissolved particle precipitates.dissolved particle precipitates.An equilibrium is established between the An equilibrium is established between the dissolved and undissolved particlesdissolved and undissolved particlesBaSOBaSO44(s) (s) Ba Ba2+2+(aq) + SO(aq) + SO44
2-2-(aq)(aq)
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 93
Solubility EquilibiaSolubility Equilibia
BaSOBaSO44(s) (s) Ba Ba2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)
KKeqeq = [Ba = [Ba2+2+][SO][SO442-2-]] [BaSO [BaSO44]]
however, this is a heterogeneous equilibrium and however, this is a heterogeneous equilibrium and solids are not included so…solids are not included so… KKspsp = [Ba = [Ba2+2+][SO][SO44
2-2-]]*the solubility product is equal to the product of *the solubility product is equal to the product of
the concentrations of the ions involved, each the concentrations of the ions involved, each raised to the power of its coefficientsraised to the power of its coefficients
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 94
Solubility EquilibiaSolubility Equilibia
Sample exercise: Give the solubility product Sample exercise: Give the solubility product constant expressions and values of the Kconstant expressions and values of the Kspsp for the following compounds:for the following compounds:
a) barium carbonatea) barium carbonate
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 95
Solubility EquilibiaSolubility Equilibia
Sample exercise: Give the solubility product Sample exercise: Give the solubility product constant expressions and values of the Kconstant expressions and values of the Kspsp for the following compounds:for the following compounds:
a) barium carbonatea) barium carbonateBaBa2+2+ BaCOBaCO33
COCO332-2-
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 96
Solubility EquilibiaSolubility Equilibia
Sample exercise: Give the solubility product Sample exercise: Give the solubility product constant expressions and values of the Kconstant expressions and values of the Ksp sp for the following compounds:for the following compounds:
a) barium carbonatea) barium carbonateBaCOBaCO3 3 Ba Ba2+2+ + CO+ CO33
2-2-
KKspsp = 5.0 x 10 = 5.0 x 10-9-9 = [Ba = [Ba2+2+][CO][CO332-2-]]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 97
Solubility EquilibiaSolubility Equilibia
Sample exercise: Give the solubility product Sample exercise: Give the solubility product constant expressions and values of the Kconstant expressions and values of the Kspsp for the following compounds:for the following compounds:
b) silver sulfateb) silver sulfate
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 98
Solubility EquilibiaSolubility Equilibia
Sample exercise: Give the solubility product Sample exercise: Give the solubility product constant expressions and values of the Kconstant expressions and values of the Kspsp for the following compounds:for the following compounds:
b) silver sulfateb) silver sulfateAgAg++ AgAg22SOSO44
SOSO442-2-
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 99
Solubility EquilibiaSolubility Equilibia
Sample exercise: Give the solubility product Sample exercise: Give the solubility product constant expressions and values of the Kconstant expressions and values of the Kspsp for the following compounds:for the following compounds:
b) silver sulfateb) silver sulfateAgAg22SOSO4 4 2Ag 2Ag+ + + SO+ SO44
2-2-
KKspsp = 1.5 x 10 = 1.5 x 10-5-5 = [Ag = [Ag+ + ]]22[SO[SO442-2-]]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 100
Solubility EquilibiaSolubility Equilibia
Please be careful to distinguish between Please be careful to distinguish between solubility and the solubility product constant.solubility and the solubility product constant.Solubility is the number of grams that will Solubility is the number of grams that will dissolve in a given amount of solvent.dissolve in a given amount of solvent.Solubility product constant is the equilibrium Solubility product constant is the equilibrium constant for the equilibrium between the constant for the equilibrium between the ionic solid and the saturated solutionionic solid and the saturated solution
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 101
Solubility EquilibiaSolubility Equilibia
Sample exercise: A saturated solution of Sample exercise: A saturated solution of AgCl in contact with undissolved solid is AgCl in contact with undissolved solid is prepared at 25°C. The concentration of Agprepared at 25°C. The concentration of Ag++ ions in the solution is found to be ions in the solution is found to be 1.35 x 101.35 x 10-5-5 M. Assuming that AgCl M. Assuming that AgCl dissociates completely in water and that dissociates completely in water and that there are no other simultaneous equilibria there are no other simultaneous equilibria involving Aginvolving Ag++ and Cl and Cl-- ion in the solution, ion in the solution, calculate Kcalculate Kspsp for this compound. for this compound.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Meeks 102
Solubility EquilibiaSolubility Equilibia
Sample exercise: A saturated solution of Sample exercise: A saturated solution of AgCl in contact with undissolved solid is AgCl in contact with undissolved solid is prepared at 25°C. The concentration of Agprepared at 25°C. The concentration of Ag++ ions in the solution is found to be ions in the solution is found to be 1.35 x 101.35 x 10-5-5 M. Calculate K M. Calculate Kspsp for this for this compound.compound.
AgCl AgCl Ag Ag++ + Cl + Cl--
1.35 x 101.35 x 10-5-5 1.35 x 10 1.35 x 10-5-5
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CHE 116 Prof. T.L. Meeks 103
Solubility EquilibiaSolubility Equilibia
Sample exercise: A saturated solution of AgCl Sample exercise: A saturated solution of AgCl in contact with undissolved solid is prepared in contact with undissolved solid is prepared at 25°C. The concentration of Agat 25°C. The concentration of Ag++ ions in the ions in the solution is found to be solution is found to be 1.35 x 101.35 x 10-5-5 M. M. Calculate KCalculate Kspsp for this compound. for this compound.
KKspsp = [Ag = [Ag++][Cl][Cl--] = [1.35 x 10] = [1.35 x 10-5-5][1.35 x 10][1.35 x 10-5-5]]
KKspsp = 1.82 x 10 = 1.82 x 10-10-10
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CHE 116 Prof. T.L. Meeks 104
Solubility EquilibiaSolubility Equilibia
Sample exercise: The KSample exercise: The Kspsp for Cu(N for Cu(N33))22 is is 6.3 x 106.3 x 10-10-10. What is the solubility of Cu(N. What is the solubility of Cu(N33))22 in water in grams per liter?in water in grams per liter?
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CHE 116 Prof. T.L. Meeks 105
Solubility EquilibiaSolubility Equilibia
Sample exercise: The KSample exercise: The Kspsp for Cu(N for Cu(N33))22 is is 6.3 x 10 6.3 x 10-10-10. . What is the solubility of Cu(NWhat is the solubility of Cu(N33))22 in water in grams per in water in grams per liter?liter?
Cu(NCu(N33))22 Cu Cu2-2- + 2N + 2N33--
II 0 0 00 +x +2x+x +2xEE x 2x x 2x
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CHE 116 Prof. T.L. Meeks 106
Solubility EquilibiaSolubility Equilibia
Sample exercise: The KSample exercise: The Kspsp for Cu(N for Cu(N33))22 is is 6.3 x 106.3 x 10-10-10. What is the solubility of Cu(N. What is the solubility of Cu(N33))22 in water in grams per liter?in water in grams per liter?
Cu(NCu(N33))22 Cu Cu2-2- + 2N + 2N33--
Ksp = [CuKsp = [Cu2+2+][N][N33--]]22
6.3 x 106.3 x 10-10-10 = [x][2x] = [x][2x]22
x = 5.4 x 10x = 5.4 x 10-4-4
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CHE 116 Prof. T.L. Meeks 107
Solubility EquilibiaSolubility Equilibia
Sample exercise: The KSample exercise: The Kspsp for Cu(N for Cu(N33))22 is is 6.3 x 106.3 x 10-10-10. What is the solubility of Cu(N. What is the solubility of Cu(N33))22 in water in grams per liter?in water in grams per liter?
Cu(NCu(N33))22 Cu Cu2-2- + 2N + 2N33--
5.4 x 105.4 x 10-4 -4 mol 147.588 g = 0.080 g/Lmol 147.588 g = 0.080 g/L 1L 1 mol 1L 1 mol
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CHE 116 Prof. T.L. Meeks 108
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
The solubility of a substance is affected byThe solubility of a substance is affected bytemperaturetemperaturepresence of other solutespresence of other solutes
presence of common ionpresence of common ion pH of solutionpH of solution presence of complexing agentspresence of complexing agents
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CHE 116 Prof. T.L. Meeks 109
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Common Ion Effect: The presence of a Common Ion Effect: The presence of a common ion reduces the amount the ionic common ion reduces the amount the ionic salt can dissolve shifting the equilibrium to salt can dissolve shifting the equilibrium to the left towards the ionic solidthe left towards the ionic solid
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CHE 116 Prof. T.L. Meeks 110
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: The value for Ksp for Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)manganese (II) hydroxide, Mn(OH)22, is , is 1.6 x 101.6 x 10-13-13. Calculate the molar solubility of . Calculate the molar solubility of Mn(OH)Mn(OH)22 in a solution that contains 0.020 in a solution that contains 0.020 M NaOH.M NaOH.
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CHE 116 Prof. T.L. Meeks 111
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: The value for Ksp for Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)manganese (II) hydroxide, Mn(OH)22, is , is 1.6 x 1.6 x 1010-13-13. Calculate the molar solubility of Mn(OH). Calculate the molar solubility of Mn(OH)22 in a solution that contains 0.020 M NaOH.in a solution that contains 0.020 M NaOH. Mn(OH)Mn(OH)22 Mn Mn2+2+ + 2OH + 2OH--
I 0 0.020I 0 0.020 +x +2x+x +2x E x 2x+0.020E x 2x+0.020
CHE 116 Prof. T.L. Meeks 112
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: The value for Ksp for Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)manganese (II) hydroxide, Mn(OH)22, is , is 1.6 1.6 x 10x 10-13-13. Calculate the molar solubility of . Calculate the molar solubility of Mn(OH)Mn(OH)22 in a solution that contains 0.020 M in a solution that contains 0.020 M NaOH.NaOH. Mn(OH)Mn(OH)22 Mn Mn2+2+ + 2OH + 2OH--
KKspsp = 1.6 x 10 = 1.6 x 10-13-13 = [Mn = [Mn2+2+][OH][OH--]]22
= [x][0.020 + 2x]= [x][0.020 + 2x]22
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CHE 116 Prof. T.L. Meeks 113
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: The value for Ksp for Sample exercise: The value for Ksp for manganese (II) hydroxide, Mn(OH)manganese (II) hydroxide, Mn(OH)22, is , is 1.6 x 1.6 x 1010-13-13. Calculate the molar solubility of Mn(OH). Calculate the molar solubility of Mn(OH)22 in a solution that contains 0.020 M NaOH.in a solution that contains 0.020 M NaOH. KKspsp = 1.6 x 10 = 1.6 x 10-13-13 = [Mn = [Mn2+2+][OH][OH--]]22
= [x][0.020 - 2x]= [x][0.020 - 2x]22
= [x][0.020]= [x][0.020]22
x = 4.0 x 10x = 4.0 x 10-10-10
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CHE 116 Prof. T.L. Meeks 114
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Solubility and pHSolubility and pHThe solubility of any substance whose anion is The solubility of any substance whose anion is basic will be affected to some extent by the pH of basic will be affected to some extent by the pH of the solution.the solution.Anions to be concerned about:Anions to be concerned about:OHOH-- COCO33
2-2-
POPO443-3- CNCN--
SS2-2- SOSO442-2-
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CHE 116 Prof. T.L. Meeks 115
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Write the net ionic equation Sample exercise: Write the net ionic equation for the reaction of the following copper (II) for the reaction of the following copper (II) compounds with acid:compounds with acid:
a) CuSa) CuS
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CHE 116 Prof. T.L. Meeks 116
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Write the net ionic equation Sample exercise: Write the net ionic equation for the reaction of the following copper (II) for the reaction of the following copper (II) compounds with acid:compounds with acid:
a) CuSa) CuSCuS + HCuS + H++ Cu Cu2+ 2+ + HS+ HS--
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CHE 116 Prof. T.L. Meeks 117
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Write the net ionic equation Sample exercise: Write the net ionic equation for the reaction of the following copper (II) for the reaction of the following copper (II) compounds with acid:compounds with acid:
b) Cu(Nb) Cu(N33))22
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CHE 116 Prof. T.L. Meeks 118
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Write the net ionic equation Sample exercise: Write the net ionic equation for the reaction of the following copper (II) for the reaction of the following copper (II) compounds with acid:compounds with acid:
b) Cu(Nb) Cu(N33))22
Cu(NCu(N33))22 + H + H++ Cu Cu2+2+ + 2HN + 2HN33
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CHE 116 Prof. T.L. Meeks 119
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Formation of Complex Ions: a characteristic Formation of Complex Ions: a characteristic property of metal ions is their ability to accept property of metal ions is their ability to accept an electron pair from water molecules.an electron pair from water molecules.
Other molecules than water can also donate Other molecules than water can also donate their electron pair to the metal ions to form their electron pair to the metal ions to form complex ions.complex ions.
The stability of the complex ion depends upon The stability of the complex ion depends upon its size of the Kits size of the Keqeq for its formation for its formation
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CHE 116 Prof. T.L. Meeks 120
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Calculate [CrSample exercise: Calculate [Cr3+3+] in ] in equilibrium with Cr(OH)equilibrium with Cr(OH)44
-- when 0.010 mol when 0.010 mol of Cr(NOof Cr(NO33))33 is dissolved in a liter of solution is dissolved in a liter of solution buffered at pH 10.0.buffered at pH 10.0.
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CHE 116 Prof. T.L. Meeks 121
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Calculate [CrSample exercise: Calculate [Cr3+3+] in ] in equilibrium with Cr(OH)equilibrium with Cr(OH)44
-- when 0.010 mol when 0.010 mol of Cr(NOof Cr(NO33))33 is dissolved in a liter of solution is dissolved in a liter of solution buffered at pH 10.0.buffered at pH 10.0.
KKff is 8 x 10 is 8 x 102929, with a value this large we can , with a value this large we can assume that all the Cr(OH)assume that all the Cr(OH)44
-- dissolves to dissolves to form Crform Cr3+3+ ions ions
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CHE 116 Prof. T.L. Meeks 122
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Calculate [CrSample exercise: Calculate [Cr3+3+] in equilibrium ] in equilibrium with Cr(OH)with Cr(OH)44
-- when 0.010 mol of Cr(NO when 0.010 mol of Cr(NO33))33 is is dissolved in a liter of solution buffered at pH dissolved in a liter of solution buffered at pH 10.0.10.0.CrCr3+3+ + 4OH + 4OH- - Cr(OH) Cr(OH)44
-- x 1x10x 1x10-4-4 0.010 0.010
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CHE 116 Prof. T.L. Meeks 123
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Calculate [CrSample exercise: Calculate [Cr3+3+] in ] in equilibrium with Cr(OH)equilibrium with Cr(OH)44
-- when 0.010 mol when 0.010 mol of Cr(NOof Cr(NO33))33 is dissolved in a liter of solution is dissolved in a liter of solution buffered at pH 10.0.buffered at pH 10.0.
CrCr3+3+ + 4OH + 4OH- - Cr(OH) Cr(OH)44--
KKff = 8 x 10 = 8 x 102929 = [Cr(OH) = [Cr(OH)44--]]
[Cr[Cr3+3+][OH][OH--]]44
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CHE 116 Prof. T.L. Meeks 124
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Sample exercise: Calculate [CrSample exercise: Calculate [Cr3+3+] in equilibrium ] in equilibrium with Cr(OH)with Cr(OH)44
-- when 0.010 mol of Cr(NO when 0.010 mol of Cr(NO33))33 is is dissolved in a liter of solution buffered at pH dissolved in a liter of solution buffered at pH 10.0.10.0.CrCr3+3+ + 4OH + 4OH- - Cr(OH) Cr(OH)44
--
KKff = 8 x 10 = 8 x 102929 = [0.010] = [0.010] [x][1x10[x][1x10-4-4]]44
x = 1.25 x 10x = 1.25 x 10-16-16
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CHE 116 Prof. T.L. Meeks 125
Factors that AffectFactors that AffectSolubility EquilibiaSolubility Equilibia
Amphoterism: Many metal hydroxides and Amphoterism: Many metal hydroxides and oxides that are insoluble in water will oxides that are insoluble in water will dissolve in strong acids and bases. They dissolve in strong acids and bases. They will do this because they themselves are will do this because they themselves are capable of behaving as an acid or a base.capable of behaving as an acid or a base.
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CHE 116 Prof. T.L. Meeks 126
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Equilibrium can be achieved starting with the Equilibrium can be achieved starting with the substances on either side of a chemical substances on either side of a chemical equation.equation.
The use of the reaction quotient, KThe use of the reaction quotient, Keqeq, to , to determine the direction in which a reaction determine the direction in which a reaction must proceed is importantmust proceed is importantIf KIf Keqeq > K > Kspsp, precipitation will occur, precipitation will occur
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CHE 116 Prof. T.L. Meeks 127
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Equilibrium can be achieved starting with the Equilibrium can be achieved starting with the substances on either side of a chemical substances on either side of a chemical equation.equation.
The use of the reaction quotient, KThe use of the reaction quotient, Keqeq, to , to determine the direction in which a reaction determine the direction in which a reaction must proceed is importantmust proceed is importantIf KIf Keqeq = K = Kspsp, equilibrium exists, equilibrium exists
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CHE 116 Prof. T.L. Meeks 128
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Equilibrium can be achieved starting with the Equilibrium can be achieved starting with the substances on either side of a chemical substances on either side of a chemical equation.equation.
The use of the reaction quotient, KThe use of the reaction quotient, Keqeq, to , to determine the direction in which a reaction determine the direction in which a reaction must proceed is importantmust proceed is importantIf KIf Keqeq < K < Kspsp, solid dissolves until K, solid dissolves until Keqeq = K = Kspsp
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CHE 116 Prof. T.L. Meeks 129
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Sample exercise: Will a precipitate form Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 10when 0.050 L of 2.0 x 10-2-2 M NaF is mixed M NaF is mixed with 0.010 L of 1.0 x 10with 0.010 L of 1.0 x 10-2-2 M Ca(NO M Ca(NO33))22??
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CHE 116 Prof. T.L. Meeks 130
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Sample exercise: Will a precipitate form when Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 100.050 L of 2.0 x 10-2-2 M NaF is mixed with 0.010 M NaF is mixed with 0.010 L of 1.0 x 10L of 1.0 x 10-2-2 M Ca(NO M Ca(NO33))22??
Possible reaction:Possible reaction:2NaF + Ca(NO2NaF + Ca(NO33))22 CaF CaF22 + 2NaNO + 2NaNO33
final volume will be 0.060 Lfinal volume will be 0.060 Lsodium salts are very soluble, CaFsodium salts are very soluble, CaF22 has a K has a Kspsp of of
3.9 x 103.9 x 10-11-11
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CHE 116 Prof. T.L. Meeks 131
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Sample exercise: Will a precipitate form when Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 100.050 L of 2.0 x 10-2-2 M NaF is mixed with 0.010 M NaF is mixed with 0.010 L of 1.0 x 10L of 1.0 x 10-2-2 M Ca(NO M Ca(NO33))22??
Molarity of CaMolarity of Ca+2+2:(0.010)(1.0 x 10:(0.010)(1.0 x 10-2-2) = 1.7 x 10) = 1.7 x 10-3-3
(0.060) (0.060)Molarity of FMolarity of F--:(0.050)(2.0 x 10:(0.050)(2.0 x 10-2-2) = 1.7 x 10) = 1.7 x 10-2-2
(0.060)(0.060)KKeqeq = (1.7 x 10 = (1.7 x 10-3-3)(1.7 x 10)(1.7 x 10-2-2) = 2.78 x 10) = 2.78 x 10-5-5
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CHE 116 Prof. T.L. Meeks 132
Precipitation and Precipitation and Separation of IonsSeparation of Ions
Sample exercise: Will a precipitate form Sample exercise: Will a precipitate form when 0.050 L of 2.0 x 10when 0.050 L of 2.0 x 10-2-2 M NaF is mixed M NaF is mixed with 0.010 L of 1.0 x 10with 0.010 L of 1.0 x 10-2-2 M Ca(NO M Ca(NO33))22??
KKeqeq = (1.7 x 10 = (1.7 x 10-3-3)(1.7 x 10)(1.7 x 10-2-2) = 2.78 x 10) = 2.78 x 10-5-5
KKeqeq>K>Kspsp so precipitation will so precipitation will occuroccur
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CHE 116 Prof. T.L. Meeks 133
Qualitative Analysis for Qualitative Analysis for Metallic ElementsMetallic Elements
How can solubility equilibria and complex How can solubility equilibria and complex ion formation be used to detect the presence ion formation be used to detect the presence of particular metal ions in solution?of particular metal ions in solution?
Fig. 17.21Fig. 17.21
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