CHEM 1412
Zumdahl & Zumdahl
Practice Exam II (Ch. 14, 15 & 16)
Multiple Choices: Please select one best answer. Answer shown in bold.
1. Consider the equilibrium: PO4-3 (aq) + H2O (l)  HPO42- (aq) + OH- (aq).
Which of the following pair is a conjugate acid-base pair for the above equilibrium?
(A) HPO42- (aq), PO43-(aq)
(B) HPO42-(aq), OH- (aq)
(C) OH- (aq), HPO42- (aq)
(D) HPO4- (aq), H2O(l)
-3
(E) PO4 (aq), H2O (l)
Hint: For 9th ed., see Section 14.1: p. 654. Also see End-of-Chapter Exercises 37 & 38.
What is the conjugate acid for HN3-?
(A) N3(B) HN2(C) H2N(D) H3N
(E) H2N2Hint: For 9th ed., see Section 14.1: p. 654. Also see End-of-Chapter Exercises 37 & 38.
2. Which of the following lies predominately to the right upon equilibrium?
(A) SO42- (aq) + HCO3- (aq)  HSO4- (aq) + CO32-(aq)
(B) NO2- (aq) + HCO3- (aq)  HNO2 (aq) + CO32-(aq)
(C) CH3COOH(aq) + HCOO- (aq)  CH3COO- (aq) + HCOOH(aq)
(D) HCOOH (aq) + CN-(aq)  HCOO- (aq) + HCN (aq)
(E) HCN (aq) + F-(aq)  CN- (aq) + HF (aq)
Hint: For 9th ed., see Section 14.2: p.p. 656-661. The reaction direction for acid and
base is always from “stronger acid + stronger base  weaker conjugate base + weaker
conjugate acid”. You may or may not be given Ka from Tables 14.2, 14.4 and A5.1 in p.
A22 as reference.
For (A), HSO4- (Ka= 1.2x10-2); HCO3- (Ka= 5.6x10-11);
For (B), HNO2 (Ka= 4.0x10-4); HCO3- (Ka= 5.6x10-11);
For (C), CH3COOH (Ka= 1.8x10-5); HCOOH (Ka= 1.8x10-4);
For (D), HCOOH (Ka= 1.8x10-4); HCN (Ka= 6.2x10-10);
For (E), HCN (Ka= 6.2x10-10); HF (Ka= 7.2x10-4);
3. Which of the following is the weakest acid?
(A) CCl3COOH
(B) CBr3COOH
(C) CI3COOH
(D) CClBrICOOH
(E) CClF2COOH
Hint: For 9th ed., see Section 14.2: p. 657: carboxylic acid and its derivatives.
Inductive effect for organic acids derived from acetic acid, CH3COOH:
(i) More halogens, same kind, attached to C of CH3, the stronger the acid.
(ii) More electronegative halogens attached to C of CH3, the stronger the acid.
Electronegativity: F (4.0), O (3.5), N (3.0), Cl (3.0), Br (2.8), C (2.5), H(2.1)
Which of the following is the strongest acid?
(A) HNO3
(B) HNO2
(C) H2O
(D) HIO
1
(E) HClO2
Hint: For 9th ed., see Sections 14.2 & 14.9. Comparing the oxoacids (or oxyacids) and
remembering the periodic trend.
4. What is the pH of 0.011 M of Ca(OH)2? What is the pH of 1.0x10-10 M HCl?
(A) 1.66; 10 (B) 1.96; 4
(C) 7; 7
(D) 12.04; 7 (E) 12.34; 7
th
Hint: For 9 ed., see Section 14.3: Interactive Examples 14.5 and 14.7. There is a very
important concept in Interactive Example 14.7(b). See Section 14.6: Bases: 2nd
paragraph of p. 675. Also see End-of-Chapter Exercise: Solution of Bases: 90 (a).
Since Ca(OH)2 is a strong base and thus the [OH-] = 2 x 0.011 M = 0.022 M 
pOH = - log 0.022 = 1.66  pH = 14 – pOH = 12.34
What is the pH of an aqueous solution of 1.110-10 M NaOH?
(A) 9.96
(B) 7.00
(C) 4.04
(D) 2.66
(E) None of the above
Hint: The same concept as Interactive Example 14.7(b).
5. The Kb for NH3 at 25oC is 1.8×10-5. What’s the pH for 0.4 M NH4NO3 (the conjugate
acid of ammonia) aqueous solution?
(A) 2.57
(B) 4.83
(C) 5.44
(D) 11.43
(E) 13.95
Hint: For 9th ed., see Section 14.8: Interactive Example 14.19 (calculate the pH of
conjugate acid) or 14.18 (calculate the pH of conjugate base). For (1) any weak base and
its conjugate acid or (2) any weak acid and its conjugate base, the relationship of K a x Kb
= Kw is always valid.
Here, Ka, NH4NO3 = Kw / Kb = 1×10-14/1.8×10-5 = 5.56×10-10
Using equilibrium table (textbook method) to calculation the concentration of proton:
NH4+ 
H+
+
NH3
[initial]
0.4
0
0
[change]
-x
+x
+x
0.4-x
+x
+x
Ka = [H+][NH3]/[ NH4+] = (+x)(+x)/0.4-x = 5.56×10-10
Assuming that x << 0, then Ka = [H+][NH3]/[ NH4+] = (+x)(+x)/0.4 = 5.56×10-10
Thus, x = square root of (0.4 x 5.56×10-10) = 1.49×10-5 = [H+]
pH = -log 1.49×10-5 = 4.826
For mono-protic acid: Use equilibrium table once to calculate [H+].
Short-cut: [H+] = {Ka x [HA]}1/2 = square root of {Ka x [HA]}; this is only valid if
[HA]initial ≥ Ka x 400
6. Which of the following substances will be more soluble in basic solution than in
acidic solution?
(A) KAl(SO4)2
(B) NaCl
(C) K2CO3
(D) CH3COONH4
(E) KNO3
th
Hint: For 9 ed., see Section 14.8. Interactive example 14.21. Also see lab manual.
Consider the acidity or basicity of solutions. To be more soluble in basic solution, the
acidic substance will be more soluble than neutral substance, which in turns is more
2
soluble than the basic substance. That is, the basic substance is the least one to be
soluble in basic solution.
For (A): IA & IIA metal ions are neutral; anions from strong acids are neutral. K +: neutral;
Al3+: acidic; SO42-: neutral. Thus, overall, KAl(SO4)2, is acidic.
For (B): IA & IIA metal ions are neutral; anions from strong acids are neutral. Na+:
neutral; Cl-: neutral. Thus, overall, NaCl, is neutral.
For (C): Anions from weak acids are weak bases; K+: neutral; CO32-: weak base. Thus,
overall, K2CO3, is basic.
For (D): Anion from a weak acid is basic; conjugate acid of a weak base is a weak acid. So
we need to see the Kb of CH3COO- (Kb = Kw/ Ka, CH3COOH = 1×10-14/1.8×10-5 = 5.56×10-10)
and the Ka of NH4+ (Ka = Kw/ Kb, NH3 = 1×10-14/1.8×10-5 = 5.56×10-10). Since the Kb of
CH3COO- is same as the Ka of NH4+, and thus CH3COONH4 is neutral.
For (E): K+: neutral; NO3-: neutral. Thus, KNO3 is neutral.
7. Which of the following is a not a Lewis base?
(A) BF3
(B) AlCl4(C) BF4(D) O2(E) PF3
th
Hint: For 9 ed., see Section 14.11: Definition. Lewis acid is an electron-pair acceptor;
while the Lewis base is an electron-pair donor. This question is asking for Lewis acid.
Q8-10. A buffer contains 0.30 mole HC2H3O2 and 0.20 mole NaC2H3O2 in 1 L. Ka of
HC2H3O2 is 1.8×10-5.
8. What is the pH of this buffer?
(A) 4.57
(B) 4.64
(C) 4.74
(D) 4.88
(E) 4.49
Hint: For 9th ed., see Section 15.2: Interactive Example 15.2. Use equilibrium table or
Short-cut: Henderson-Hasselbalch equation:
[HC2H3O2 ] = 0.30 mol/1 L = 0.30 M; [NaC2H3O2] = 0.20 mol/1 L = 0.20 M
pH = pKa + log [weak base] /[weak acid] = pKa + log [salt] /[weak acid]
= -log 1.8×10-5 + log [0.2] /[0.3] = 4.7447 + (-0.1761) = 4.5686
9. What is the pH of this buffer after the addition of 0.02 mole of KOH?
(A) 3.66
(B) 4.64
(C) 4.74
(D) 4.88
(E) 4.49
Hint: For 9th ed., see Section 15.2: Interactive Example 15.3. First, use stoichiometric
table, and then use the equilibrium table (TEXTBOOK METHOD) or HendersonHasselbalch equation to solve it. Or use short-cut (if you want to use it, you need to
memorize it): the modified Henderson-Hasselbalch equation.
Use Modified Henderson-Hassbach Equations:
a. Add strong acid into a buffer
pH = pKa + log { [weak base] – [strong acid] }/{ [weak acid] + [strong acid] }
b. Add strong base into a buffer
pH = pKa + log { [weak base] + [strong base] }/{ [weak acid] - [strong base]
3
Since this is adding the strong base (limiting reagent: used up) to the buffer,
pH = pKa + log { [weak base] + [strong base] }/{ [weak acid] - [strong base]
= -log 1.8×10-5 + log {(0.2+0.02)/(0.3-0.02)} = 4.7447 + (-0.1047) = 4.6370
10. What is the pH of this buffer after the addition of 0.02 mole of HCl?
(A) 3.66
(B) 3.77
(C) 4.74
(D) 4.88
(E) 4.49
Hint: For 9th ed., see Interactive Example 15.6, and Example 15.7. First, use
stoichiometric table, and then use the equilibrium table (TEXTBOOK METHOD) or
Henderson-Hasselbalch equation to solve it. Or use short-cut (if you want to use it,
you need to memorize it): the modified Henderson-Hasselbalch equation.
Since this is adding the strong acid (limiting reagent: used up) to the buffer,
pH = pKa + log { [weak base] - [strong acid] }/{ [weak acid] + [strong acid]
= -log 1.8×10-5 + log {(0.2-0.02)/(0.3+0.02)} = 4.7447 + (-0.2499) = 4.4948
11. Which of the following can make a buffer solution? (Multiple Answers)
(A) C4H9COOH and CH3COH
(B) H2SO4 and NaOH
(C) HS and NaF
(D) H2SO4 and CH3COOH
(E) NaOH and NH3
(F) CH3COH and CH3OH
(G) NH4Cl (in excess) and NaOH
(H) NH3 (in excess) and HCl
Hint: For 9th ed., see Section 15.2, p.p. 715-724. There are FOUR ways to make a
buffer solution (memorize them): (1) weak acid and its conjugated base; (2) weak base
and its conjugate acid; (3) strong acid and excessive weak base; and (4) strong base
and excessive weak acid.
CH3COH is aldehyde; C4H9COOH is carboxylic acid; CH3OH is alcohol.
More practice on pH of buffer calculation:
What is the pH for a buffer that contains 0.46 M HNO2 and 0.40 M KNO2? The Ka for
HNO2 is 4.5×10-4.
(A) 3.14
(B) 3.29
(C) 3.95
(D) 4.04
(E) 4.44
Hint: Similar to Q 8. For 9th ed., see Section 15.2: Interactive Examples 15.2, 15.3, 15.4,
and 15.5. Two methods to solve it: (1) use the equilibrium table or (2) HendersonHasselbalch equation (memorize it).
pH = pKa + log [weak base] /[weak acid] = pKa + log [salt] /[weak acid]
= -log 4.5×10-4 + log [0.4] /[0.46] = 3.3468 + (-0.0607) = 3.2861
What is the pH for a buffer that is prepared containing 0.25 M HCOOH (formic acid) and
0.10 M HCOONa (sodium formate)? The Ka for HCOOH is 1.8×10-4.
(A) 3.62
(B) 3.57
(C) 3.35
(D) 3.29
(E) 3.10
th
Hint: Similar to Q 8. For 9 ed., see Section 15.2: Interactive Examples 15.2, 15.3, 15.4,
and 15.5. Two methods to solve it: (1) use the equilibrium table or (2) HendersonHasselbalch equation (memorize it).
pH = pKa + log [weak base] /[weak acid] = pKa + log [salt] /[weak acid]
4
= -log 1.8×10-4 + log [0.1] /[0.25] = 3.7447 + (-0.3979) = 3.3468
Q 12-14. Strong Acid - Strong Base Titration: Basic type of titration without buffer and
hydrolysis regions to consider. For 9th ed., see Section 15.4: p.p. 728-730.
Calculate the pH when the following quantities of 0.1 M HCl have been added to 25.0
mL of 0.1 M NaOH solution: (a) 0.0 mL, (b) 24.9 mL, (c) 25.0 mL, (d) 25.1 mL
Answer:
(a) Before the titration: NaOH is in the beaker and HCl is in the buret; as no HCl has
been added into the beaker, and the pH meter is placed in the beaker, the pH of
the solution is determined by the substance in the beaker.
As NaOH is a strong base, so pOH = - log[OH-] = -log[NaOH] = -log 0.1 = 1.
Thus, pH = 14 – pOH = 13.
12. (b) Before the equivalence point: At the equivalence point, the mole of hydrogen
ions must be equal to the mole of hydroxide ions. Therefore, to reach the equivalence
point, the volume of HCl should be 25.0 mL. Thus, this is the region before the
equivalence point, which the NaOH is in excess. The pH of the solution is determined by
the substance in excess in the beaker.
[OH-]new = [OH-]left over = (0.1x25.0-0.1x24.9)mmol/(25.0+24.9)mL
={0.1x(25.0-24.9)/(25.0+24.9)} = 0.00020M,
so pOH = -log 0.00020 = 3.70; pH = 14 – pOH = 10.30
13. (c) At the equivalence point: For a strong acid-base (vise versa) titration, the pH at
the equivalence point is 7 as it is the hydrolysis of a neutral salt. No calculation is
needed.
14. (d) After the equivalence point: This is a region after the equivalence point. Thus,
HCl is in excess.
[H+]new = [H+]left over = (0.1x25.1-0.1x25.0)mmol/(25.1+25.0)mL
={0.1x(25.1-25.0)/(25.1+25.0)} = 0.00020M,
so pH = -log 0.00020 = 3.70.
Q15-17. A 50.0 mL of 0.025 M HC7H5O2 (Ka = 6.5×10-5) is titrated with 0.050 M KOH
solution.
15. What is the pH after adding 10 mL of base is added?
(A) 3.32
(B) 4.01
(C) 6.43
(D) 8.20
(E) 13.10
th
Hint: For 9 ed., see Section 15.4: p.p. 730-740: Titration of Weak Acids with Strong
Bases. This is before the equivalence point. So the solution is a buffer. First, use
5
stoichiometric table, and then use the equilibrium table or Henderson-Hasselbalch
equation to solve it.
Moles of HC7H5O2 = 0.025 x (50.0/1000) = 1.25x10-3 = 0.00125
mole of KOH = 0.050 x (10/1000) = 0.0005
Before the equivalence point: Note that this is a buffer region as shown below: To
reach an equivqlence point, the volume of KaOH should be 1.25/0.05 = 25.0 mL. Since
now KOH only has 10 mL < 25.0 mL, and thus, this is the region before the equivalence
point, which acetic acid is in excess.
HC7H5O2
+
KOH 
KC7H5O2
+ H2O
Before reaction 0.00125 mol
0.0005mol
0 mol
Change
-0.0005 mol
-0.0005mol
+0.0005mol
After reaction
0.00075 mol
0 mol
0.0005 mol
Weak acid
Conjugate base
Total Volume = 50 mL + 10 mL = 60 mL
So, apply pH = pKa + log {[weak base]/[weak acid]}; since the volume is the same, we
can use mole to represent the concentration; it is because the ratio is the same.
pH = pKa + log [weak base] /[weak acid] = pKa + log [salt] /[weak acid]
= -log 6.5×10-5 + log [0.0005] /[0.00075] = 4.1871 + (-0.1761) = 4.0110
16. What is the pH after adding 25 mL of base is added?
(A) 3.32
(B) 4.01
(C) 6.43
(D) 8.20
(E) 13.10
Hint: This is at the equivalence point. So the pH of the solution is determined by the
hydrolysis of the salt (product).
Mole KOH = 0.05 x (25/1000) = 0.00125
HC7H5O2
+
KOH 
Before reaction 0.00125 mol
0.00125mol
Change
-0.00125 mol
-0.00125mol
After reaction
0.00000 mol
0.00000 mol
KC7H5O2
+ H2O
0 mol
+0.00125mol
0.00125 mol
Conjugate base
Total Volume = 50 mL + 25 mL = 75 mL = 0.075 L
[NaC7H5O2] = 0.00125/0.075 = 0.0167 M
The salt will hydrolyze:
[initial]
[change]
C7H5O2- + H2O  HC7H5O2
0.0167
0
-x
+x
0.0167-x
+x
+
OH0
+x
+x
Kb = Kw/Ka = 1x10-14/6.5×10-5 = 1.538 x 10-10 = (x)(x)/0.0167-x
Assuming x << 0.0167; then 1.538 x 10-10 = (x)(x)/0.0167;
6
Thus x = square root of (1.538 x 10-10)(0.0167) = 1.6029 x 10-6
pOH = -log x = - log (1.6029 x 10-6) = 5.795
So pH = 14 – pOH = 14 - 5.795 = 8.205
Or use the short-cut, that is the formula (memorize):
pOH = square root of (Kb x C) = square root of {(Kw/Ka) x C}
= square root of (1.538 x 10-10)(0.0167) = 1.6029 x 10-6
pOH = -log x = - log (1.6029 x 10-6) = 5.795
So pH = 14 – pOH = 14 - 5.795 = 8.205
17. What is the pH after adding 50 mL of base is added?
(A) 3.32
(B) 4.74
(C) 6.43
(D) 8.81
(E) 12.10
Hint: This is after the equivalence point. So the pH is determined by the excess
amount of the strong base. First, use stoichiometric table, and then use the
pOH = –log [OH-] and the pH = 14 – pOH.
Mole of KOH = 0.050 x (50/1000) = 0.0025
Before reaction
Change
After reaction
HC7H5O2
+
0.00125 mol
-0.00125 mol
0.00000 mol
KOH 
0.0025mol
-0.00125mol
0.00125 mol
Strong base
KC7H5O2
+ H2O
0 mol
+0.00125mol
0.00125 mol
Weak base
Total Volume = 50 mL + 50 mL = 100 mL = 0.1 L
The pH is determined by the excess amount of the KOH.
[KOH] = 0.00125/0.1 = 0.0125 M
 pOH = -log 0.0125 = 1.903  pH = 14 – pOH = 12.097
Q 18-21. Calculate the pH when the following quantities of 0.1 M HCl have been added
to 25.0 mL of 0.1 M NH3 solution: (a) 0.0 mL, (b) 24.9 mL, (c) 25.0 mL, (d) 25.1 mL
Hint: For 9th ed., see Section 15.4, p.p. 740-742: Titration of Weak Bases with Strong
Acids.
Answer:
18. (a) Before the titration: NH3 is in the beaker and HCl is in the buret; as no HCl has
been added into the beaker, and the pH meter is placed in the beaker, the pH of the
solution is determined by the substance in the beaker, that is, ammonia. As ammonia is
a weak base, the equilibrium problem at this point is the familiar one of calculating the
pOH of a solution of a weak base. Thus, the [OH-] can be obtained from either of the two
methods: (1) equilibrium table or (2) short-cut formula [OH-] = square root of (Kb x C) =
square root of {(1.8x10-5) x (0.1)} = 1.34x10-3.
pOH = -log 1.34x10-3 = 2.87. Thus, pH = 14 – pOH = 11.13.
7
19. (b) Before the equivalence point: Note that this is a buffer region as shown below:
To reach an equivqlence point, the volume of HCl should be 25.0 mL. Thus, this is the
region before the equivqlence point, which acetic acid is in excess.
NH3
+
HCl

NH4+
+ ClBefore reaction 0.00250 mol
0.00249mol
0 mol
Change
-0.00249 mol
-0.00249mol
+0.00249mol
After reaction
0.00001 mol
0 mol
0.00249 mol
Weak base Conjugate acid
[NH3] = 0.00001 mol/0.0499 L = 0.0002 M
[NH4+] = 0.00249/0.0499 = 0.0499 M
So, apply pH = pKa + log {[weak base]/[weak acid]}
= (14 – pKb) + log {[weak base]/[weak acid]}
= {14 – (-log 1.8x10-5 )}+ log 0.00001/0.00249
= (14 – 4.7447) + (-2.3962) = 6.8591
20. (c) At the equivalence point: This is the region of hydrolysis of salt.
NH3
+
HCl

NH4+
+ ClBefore reaction 0.00250 mol
0.00250mol
0 mol
Change
-0.00250 mol
-0.00250mol
+0.00250mol
After reaction
0.00000 mol
0.00000 mol
0.00250 mol
8
[NH4+] = 0.00250/0.0500 = 0.050 M
NH4+
+
H2O ↔
NH3 +
H3O+
[Initial]
0.050
0
0
[Change]
-x
+x
+x
[Equilibrium]
0.050-x
+x
+x
+
+
-14
-5
-10
So, Ka = [H3O ][NH3]/[NH4 ] = Kw/Kb = 1.0x10 /1.8x10 = 5.56 x 10
Thus, (x)(x)/(0.050 – x) = 5.56 x 10-10
Assume 0 < x << 0.050 and thus the above quadric equation can be simplified as
(x)(x)/(0.050) = 5.56 x 10-10
Therefore, x = square root of 0.050 x (5.56 x 10-10)= 5.27 x 10-6
Thus, pH = -log 5.27 x 10-6 = 5.28.
21. (d) After the equivalence point: Thus, HCl is in excess.
[H+]left over = (0.1x25.1-0.1x25.0)/(25.1+25.0)
={0.1x(25.1-25.0)/(25.1+25.0)} = 0.00020M,
so pH = -log 0.00020 = 3.70.
Additional Practice:
A 50.0 mL of 0.1 M NH3 (Kb = 1.8x10-5) solution is titrated with 0.25 M HCl solution. (See
9th ed., p.p. 740-742: same weak base, same strong acid)
*What is the pH after adding 0.0 mL of acid is added?
(A) 11.13
(B) 9.26
(C) 5.28
(D) 4.74
(E) 2.87
Hint: Before adding any strong acid, the pH of the solution is solely determined by the
WEAK base, NH3. Use equilibrium table to calculate the hydroxide ion concentration and
then calculate the pOH and thus pH = 14 – pOH.
*What is the pH after adding 15 mL of acid is added?
(A) 8.78
(B) 4.74
(C) 7.0
(D) 9.26
(E) 11.13
Hint: This is before the equivalence point and thus the solution is a buffer solution. To
check whether the titration reaches the equivalence point, you need to calculate the
moles or milli-moles of both acid and base. First, use stoichiometric table to calculate
how much ammonia is left (1.25 milli-moles) and how much ammonium ion (3.75
milli-moles) is produced. Then use the equilibrium table or modified HendersonHasselbalch equation to solve it. See 9th ed. lecture notes. If you use equilibrium table,
you need to remember that Kw = Ka x Kb = 1.00 x 10-14 to calculate Ka for ammonium ion
who will undergo hydrolysis. Hydrogen ion concentration calculated is 1.67 x 10-9. Thus
pH = 8.78.
*What is the pH after adding 20 mL of acid is added?
(A) 1.51
(B) 2.87
(C) 5.19
(D) 7.0
(E) 11.13
Hint: This is at the equivalence point. The pH is determined by the hydrolysis of the
salt, ammonium chloride. Chloride ion is neutral and thus it does not hydrolyze but
ammonium does as it’s a weak acid. Total volume of solution now is 70 mL. The
9
hydrogen ion concentration calculated is 6.3 x 10-6. Since this number, 6.3 x 10-6, is
much smaller than 7.14.0 x 10-2 M of hydrogen ion concentration calculated by
stoichiometry. Thus we can drop the x. However, when calculating the total hydrogen
ion concentration = 6.3 x 10-6 + 1.0 x 10-7 = 6.4 x 10-6M. Thus, pH = – log 6.4 x 10-6 =
5.19. Therefore, pOH = 14 – pH = 8.81. See 9th ed. lecture notes.
*What is the pH after 30 mL of acid is added?
(A) 1.51
(B) 2.87
(C) 8.81
(D) 7.0
(E) 9.26
Hint: This is after the equivalence point. Thus, the pH of the solution is solely
determined by the excess strong acid, HCl. Use stoichiometric table to calculate how
much of the HCl left. Then calculate the new concentration of HCl. The total volume of
the solution now is 30 + 50 = 80 mL. The new HCl concentration = 3.125 x 10-2 M = new
hydrogen ion concentration. Thus, the pH = 1.51.
22. The Ksp for Ag2CrO4 in water at 25oC is 1.1×10-12. What is the molar concentration for
CrO4 2- and Ag+, respectively?
(A) 1.3×10-4 and 6.5×10-5
(B) 6.5×10-5 and 1.3×10-4
(C) 3.7×10-11 and 4.8×10-14
-11
-14
-18
-18
(D) 3.7×10 and 4.8×10
(E) 1.9×10 and 1.9×10
Hint: For 9th ed., see Section 16.1: Interactive Exercise 16.3.
[Equilibrium]
Ag2CrO4 (s)  2Ag+ (aq) + CrO4 2- (aq)
--2x
x
Ksp = [Ag+]2[CrO42-] = (2x)2(x) = 4x3 = 1.1×10-12  x3 = (1.1×10-12)/4 = 2.75×10-13
 x = cubic root of 2.75×10-13 = (2.75×10-13)1/3 = 6.50×10-5 M = [CrO42-]
 2x = [Ag+] = 2x(6.50×10-5)= 1.3×10-4 M
23. What is the molar solubility for Cu(OH)2 at pH 10.5? Ksp for Cu(OH)2 is 4.8×10-20.
(A) 1.3×10-4 (B) 6.3×10-10 (C) 4.8×10-13 (D) 4.8×10-14 (E) 4.9×10-7
Hint: For 9th ed., see Section 16.1: p.p. 765-768. This is the common-ion effect
concerning pH and solubility. The question indirectly gives you the concentration of
hydroxide ion via pH.
Given pH 10.5  pOH = 14-10.5 = 3.5  [OH-] = 10-3.5 = 3.16×10-4 M
[Equilibrium]
Cu(OH)2(s)  Cu+2 + 2OH--x
2x+3.16×10-4
Ksp = [Cu+2][OH-]2 = (x)(2x+3.16×10-4)2= 4.8×10-20
 Assuming that 2x << 3.16×10-4 and thus 2x+3.16×10-4 = 3.16×10-4
 Ksp = [Cu+2][OH-]2 = (x)(3.16×10-4)2= 4.8×10-20
 x = (4.8×10-20)/(3.16×10-4)2 = (4.8×10-20)/(9.986×10-8) = 4.8069×10-13 M
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Additional Practice:
*The molar solubility for CaF2 in water at 25oC is 1.24×10-3 M. What is its Ksp at this
temperature?
(A) 1.3×10-4 (B) 8.22×10-7 (C) 7.6×10-9 (D) 6.93×10-13 (E) 8.77×10-5
Hint: For 9th ed., see Section 16.1: Interactive Examples 16.1 and 16.2.
*What is the molar solubility for cerium (III) hydroxide, Ce(OH)3, in 0.2 M Ba(OH)2
solution? Ksp for Ce(OH)3 is 1.5×10-20.
(A) 2.34×10-19 (B) 1.3×10-4 (C) 6.3×10-6 (D) 3.7×10-11 (E) 4.8×10-14
Hint: For 9th ed., see Section 16.1: p.p. 765-768. see Q. 23; This is the common-ion
effect. The question indirectly gives you the concentration of hydroxide ion via
concentration of the strong base, di-basic base. Thus, [OH-] = 0.2 x 2 = 0.4 M.
Ce(OH)3 (s)  Ce+3 (aq) + 3 OH - (aq)
[Equilibrium]
--x
3x+0.4
+3
- 3
3
-20
Thus, Ksp = [Ce ][OH ] = (x)(3x+0.4) = 1.5×10
Assuming 3x << 0.4, then Ksp = [Ce+3][OH-]3 = (x)(0.4)3 = 1.5×10-20
 x = 1.5×10-20 /(0.4)3 = 2.34375×10-19 M
*Will Mn(OH)2 to precipitate when the pH of a 0.050 M solution of MnCl2 is adjusted to
10.00? Ksp for Mn(OH)2 is 1.6×10-13.
(A) Yes
(B) No
(C) Not sure
Hint: For 9th ed., see section 16.2. The question is indirectly given the [OH-]. Calculate
Q and compare it with Ksp. If Q is greater than Ksp, then there will be a solid formed.
24. Which of the following action may/will NOT decrease the solubility of a precipitate?
(A) Add the common ion into the solution.
(B) Add a complexing agent into the solution.
(C) Decrease the temperature of the solution.
(D) Add a precipitating agent into the solution.
(E) Increase the concentration of solution.
Hint: For 9th ed., see Section 16.3.
25. The solubility for SrF2 in water at 45oC is found 2 g/100 mL. What is its Ksp?
The equilibrium is
SrF2 (s)  Sr2+ (aq) + 2F - (aq)
(A) 1.6×10-2 (B) 2.7×10-9 (C) 7.6×10-9 (D) 3.7×10-11 (E) 2.7×10-12
Hint: For 9th ed., see Section 16.1: The definition of Ksp is made up by the molar
concentration product of each ion involved and raised the coefficient as its exponent.
11
The molar concentration is molarity (mole/Liter). So the solubility in g/mL must be
converted to mole/L. Thus, Ksp= [Sr2+][F -]2 = (0.159)(2x0.159)2 = 1.612×10-2.
NOTE: Ksp and other equilibrium constants do not have units.
Additional information:
[H+] = [H+]solvent + [H+]solute
[OH-] = [OH-]solvent + [OH-]solute
pH = -log[H+]
pOH = -log[OH-]
ax2 + bx + c = 0 where x = {-b  (b2-4ac)1/2}/2a
Kp = Kc(RT)n(g)
12