CHE 116 Prof. T.L. Heise 1 Chapter Seventeen: Additional Aspects of Aqueous Equilibria v Water is...

CHE 116 Prof. T.L. Heise 1Chapter Seventeen: Chapter Seventeen: Additional Aspects of Additional Aspects of Aqueous EquilibriaAqueous Equilibria

Chapter 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

Copyright T. L. Heise 2001 - 2002

CHE 116 Prof. T.L. Heise 2

The Common Ion EffectThe Common Ion EffectThe 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




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 reaction to shift left, using up the reaction to shift left, using up HH++ ions, and reducing acidity ions, and reducing acidity




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]]




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..


II 0.085 0.085 0 0.10 0 0.10

-x +x +x-x +x +x

EE 0.085-x0.085-x xx 0.10+x 0.10+x






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]






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






KKaa = 4.5 x 10 = 4.5 x 10-4-4 = [x][0.10] = [x][0.10] [0.085] [0.085]






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





[H[H++] = 3.83 x 10] = 3.83 x 10-4-4

pH = -log [HpH = -log [H++] ]

pH = 3.42pH = 3.42




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]]






II 0.050 0 0.10 0.050 0 0.10

D -x +x +xD -x +x +x

EE 0.050-x x 0.10+x 0.050-x x 0.10+x






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]






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






KKaa = 1.8 x 10 = 1.8 x 10-4-4 = [x][0.10] = [x][0.10] [0.050] [0.050]






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





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.





pH = -log[HpH = -log[H++]]

pH = -log[0.10]pH = -log[0.10]

pH = 1.0pH = 1.0




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.



Buffered SolutionsBuffered SolutionsBuffered 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




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 each other in a neutralization reaction, and that requirement is only reaction, and that requirement is only fulfilled by weak conjugate acid/base fulfilled by weak conjugate acid/base pairs.pairs.




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]




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.




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




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




Two important characteristics of a Two important characteristics of a buffer are its capacity and its pH.buffer are its capacity and its pH.The greater the amounts of The greater the amounts of

conjugate acid/base pair, the more conjugate acid/base pair, the more resistant the ratio of their resistant the ratio of their concentrations, and hence pH, is to concentrations, and hence pH, is to change.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.




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.




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 +x

E 0.12-x 0.20+x x E 0.12-x 0.20+x x





Appendix D - KAppendix D - Kaa = 6.3 x 10 = 6.3 x 10-5-5


KKaa = [0.20+x][x] = [0.20+x][x] * make * make [0.12-x] [0.12-x] assumptionassumption





Appendix D - KAppendix D - Kaa = 6.3 x 10 = 6.3 x 10-5-5


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





[H[H++] = 3.8 x 10] = 3.8 x 10-5-5

pH = 4.42pH = 4.42




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.




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. (KpH of 4.00. (Kaa = 6.3 x 10 = 6.3 x 10-5-5))


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




Sample exercise: Calculate the concentration of Sample exercise: Calculate the concentration of sodium benzoate that must be present in a 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 = 0.126 M Cx = 0.126 M C77HH55OO22--




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




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++]]




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

HNOHNO33

c) after the addition of 0.015 mol of c) after the addition of 0.015 mol of KOHKOH




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 a) before a) before any acid or base is addedany acid 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 +x

E 0.140-x 0.110+x xE 0.140-x 0.110+x x





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]




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

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]

x = 4.5 x 10x = 4.5 x 10-4-4





a) pH = - log[H+]a) pH = - log[H+]

= -log[4.5 x 10= -log[4.5 x 10-4-4]]

= 3.35= 3.35




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 b) addition of 0.015 mol of HNOof 0.015 mol of HNO33 HCNO HCNO CNOCNO-- + 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 +x

E 0.155-x 0.095+x xE 0.155-x 0.095+x x





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]




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) b) addition of 0.015 mol of HNOaddition 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]

x = 5.7 x 10x = 5.7 x 10-4-4




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) b) addition of 0.015 mol of HNOaddition 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]

x = 5.7 x 10x = 5.7 x 10-4-4





b) pH = - log[H+]b) pH = - log[H+]

= -log[5.7 x 10= -log[5.7 x 10-4-4]]

= 3.24= 3.24




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) c) addition of 0.015 mol of KOH addition of 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 +x

E 0.125-x 0.125+x xE 0.125-x 0.125+x x





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]




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) 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]

x = 3.5 x 10x = 3.5 x 10-4-4





b) pH = - log[H+]b) pH = - log[H+]

= -log[3.5 x 10= -log[3.5 x 10-4-4]]

= 3.46= 3.46



Acid Base TitrationsAcid Base TitrationsAcid Base TitrationsAcid Base Titrations

In an acid base titration, a solution containing In an acid base titration, a solution containing a known concentration of base is slowly a known concentration of base is slowly added to an acid. added to an acid.

Indicators can be used to signal the Indicators can be used to signal the equivalence point of a titrationequivalence point 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 progress of a reaction producing a pH titration curve, a graph of the pH as a titration curve, a graph of the pH as a function of the volume of base addedfunction of the volume of base added




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




Strong Base added to Strong AcidStrong Base added to Strong Acid

Fig 17.6Fig 17.6




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





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





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





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





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





a) 24.90 mLa) 24.90 mL

Use pOH formula!!Use pOH formula!!

pOH = -log[OHpOH = -log[OH--]]





a) 24.90 mLa) 24.90 mL



= -log[2.00 x 10= -log[2.00 x 10-4-4]]




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 been have been added to 25.00 mL of 0.100 M KOH.added to 25.00 mL of 0.100 M KOH.

a) 24.90 mLa) 24.90 mL



= -log[2.00 x 10= -log[2.00 x 10-4-4]]

= 3.7= 3.7





a) 24.90 mLa) 24.90 mL

RememberRemember

pOH + pH = 14pOH + pH = 14




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 been have been added to 25.00 mL of 0.100 M KOH.added to 25.00 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




Sample 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 have been added to 25.00 mL of 0.100 M KOH.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

10.3 = pH10.3 = pH





b) 25.10 mLb) 25.10 mL





c) 25.10 mLc) 25.10 mL

BaseBase 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





c) 25.10 mLc) 25.10 mL

AcidAcid 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




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 mL

AcidAcid 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




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 equivalence point, and in this region a single drop could change the pH greatly.single drop could change the pH greatly.

An indicator beginning and ending its An indicator beginning and ending its color change anywhere on the rapid rise color change anywhere on the rapid rise portion of the graph will work.portion of the graph will work.




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




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

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






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





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




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





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





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




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







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.






CC77HH55OO22-- + 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 +x

EE 1.6 x 10 1.6 x 10-2-2 -x -x x x x x




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 NaOHNaOH

CC77HH55OO22-- + 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





b) 40.0 mL 0f 0.100 M NHb) 40.0 mL 0f 0.100 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 form 4 x 10to form 4 x 10-3-3 mol NH mol NH44 in 80 mL of in 80 mL of solution, which is a weak acid. solution, which is a weak acid.





b) 40.0 mL 0f 0.100 M NHb) 40.0 mL 0f 0.100 M 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 +x

EE 0.050 -x 0.050 -x x x x x




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 M NHb) 40.0 mL 0f 0.100 M NH33 is titrated is titrated with with 0.100 M HCl0.100 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




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




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




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


Solubility EquilibiaSolubility EquilibiaSolubility 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 compoundscompounds

By 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




Solubility Product Constant:Solubility Product Constant:

Saturated solution - the solution is in Saturated solution - the solution is in contact with undissolved solute. A particle contact with undissolved solute. A particle of solute dissolves and exactly the same rate of solute dissolves and exactly the same rate as a dissolved particle precipitates.as a dissolved particle precipitates.

An equilibrium is established between the An equilibrium is established between the dissolved and undissolved particlesdissolved and undissolved particles

BaSOBaSO44(s) (s) Ba Ba2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)




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][SO442-2-]]

*the solubility product is equal to the product of the *the solubility product is equal to the product of the concentrations of the ions involved, each raised to concentrations of the ions involved, each raised to the power of its coefficientsthe power of its coefficients




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





a) barium carbonatea) barium carbonate

BaBa2+2+ BaCOBaCO33

COCO332-2-




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 carbonate

BaCOBaCO3 3 Ba Ba2+2+ + CO+ CO332-2-

KKspsp = 5.0 x 10 = 5.0 x 10-9-9 = [Ba = [Ba2+2+][CO][CO332-2-]]





b) silver sulfateb) silver sulfate






AgAg++ AgAg22SOSO44

SOSO442-2-






AgAg22SOSO4 4 2Ag 2Ag+ + + SO+ SO442-2-

KKspsp = 1.5 x 10 = 1.5 x 10-5-5 = [Ag = [Ag+ + ]]22[SO[SO442-2-]]




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 Solubility is the number of grams that will dissolve in a given amount of solvent.will dissolve in a given amount of solvent.

Solubility product constant is the Solubility product constant is the equilibrium constant for the equilibrium equilibrium constant for the equilibrium between the ionic solid and the saturated between the ionic solid and the saturated solutionsolution




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.




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




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




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?




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 +2x

EE x 2x x 2x





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





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



Factors that AffectFactors that AffectSolubility EquilibiaSolubility EquilibiaFactors 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




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




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.




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



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 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




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




Solubility and pHSolubility and pH

The 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 basic will be affected to some extent by the pH of the solution.of the solution.

Anions to be concerned about:Anions to be concerned about:

OHOH-- COCO332-2-

POPO443-3- CNCN--

SS2-2- SOSO442-2-




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





a) CuSa) CuS

CuS + HCuS + H++ Cu Cu2+ 2+ + HS+ HS--





b) Cu(Nb) Cu(N33))22





b) Cu(Nb) Cu(N33))22

Cu(NCu(N33))22 + H + H++ Cu Cu2+2+ + 2HN + 2HN33




Formation of Complex Ions: a characteristic Formation of Complex Ions: a characteristic property of metal ions is their ability to property of metal ions is their ability to accept an electron pair from water molecules.accept 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




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





-- when 0.010 mol of when 0.010 mol of Cr(NOCr(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--

x 1x10x 1x10-4-4 0.010 0.010







KKff = 8 x 10 = 8 x 102929 = [Cr(OH) = [Cr(OH)44--]]

[Cr[Cr3+3+][OH][OH--]]44





-- when 0.010 mol of when 0.010 mol of Cr(NOCr(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 = 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




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.



Precipitation and Precipitation and Separation of IonsSeparation of IonsPrecipitation 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





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





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




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??




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 M NaF is mixed with 0.010 L of 1.0 x 100.010 L 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 L

sodium 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




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 M NaF is mixed with 0.010 L of 1.0 x 100.010 L 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




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



Qualitative Analysis for Qualitative Analysis for Metallic ElementsMetallic ElementsQualitative 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


Date post:	16-Jan-2016
Category:	Documents
Upload:	rosalind-miles
View:	221 times
Download:	0 times

CHE 116 Prof. T.L. Heise 1 Chapter Seventeen: Additional Aspects of Aqueous Equilibria v Water is...

Documents