Test Bank – Solubility and Complex-Ion Equilibria: Complete Practice Questions, Detailed Solutions, and Verified Answers for College Chemistry Exam Preparation (Updated 2025)

CHAPTER 17
Solubility and Complex-Ion Equilibria


■ SOLUTIONS TO EXERCISES
Note on significant figures If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.

17.1. a. BaSO4(s)  Ba2+(aq) + SO42−(aq) Ksp = [Ba2+][SO42−]

b. Fe(OH)3(s)  Fe3+(aq) + 3OH−(aq) Ksp = [Fe3+][OH−]3

c. Ca3(PO4)2(s)  3Ca2+(aq) + 2PO43−(aq) Ksp = [Ca2+]3[PO43−]2

17.2. Calculate the molar solubility. Then assemble the usual concentration table, and substitute the
equilibrium concentrations from it into the equilibrium-constant expression. (Because no
concentrations can be given for solid AgCl, dashes are written. In later problems, similar spaces
will be left blank.)
1.9  103 g 1 mol
 = 1.33  10−5 M
1L 143 g

Set up the table as usual.
Conc. (M) AgCl(s)  Ag+ + Cl−
Starting — 0 0
Change — +1.33  10−5 +1.33  10−5
Equilibrium — 1.33  10−5 1.33  10−5
Ksp = [Ag+][Cl−] = (1.33  10−5)(1.33  10−5) = 1.768  10−10 = 1.8  10−10

17.3. Calculate the molar solubility. Then assemble the usual concentration table, and substitute from it
into the equilibrium-constant expression. (Because no concentrations can be given for solid
Pb3(AsO4)2, spaces are left blank.)
3.0  105 g 1 mol
 = 3.34  10−8 M
1L 899 g




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, 658 Chapter 17: Solubility and Complex-Ion Equilibria



Conc. (M) Pb3(AsO4)2(s)  3Pb2+ + 2AsO43−
Starting 0 0
Change +3(3.34  10−8) +2(3.34  10−8)
Equilibrium 3(3.34  10−8) 2(3.34  10−8)
Ksp = [Pb2+]3[AsO43−]2 = [3  (3.34  10−8)]3(2  3.34  10−8)2 = 4.489  10−36 = 4.5  10−36

17.4. Assemble the usual concentration table. Let x equal the molar solubility of CaSO4. When x mol
CaSO4 dissolves in 1 L of solution, x mol Ca2+ and x mol SO42− form.
Conc. (M) CaSO4(s)  Ca2+ + SO42−
Starting 0 0
Change +x +x
Equilibrium x x
Substitute the equilibrium concentrations into the equilibrium-constant expression, and solve for
x. Then convert to g CaSO4 per liter.
[Ca2+][SO42−] = Ksp
(x)(x) = x2 = 2.4  10−5

x= (2.4  105 ) = 4.89  10−3 M

4.89  103 mol 136 g
 = 0.6664 = 0.67 g/L
L 1 mol

17.5. a. Let x equal the molar solubility of BaF2. Assemble the usual concentration table, and
substitute from the table into the equilibrium-constant expression.
Conc. (M) BaF2(s)  Ba2+ + 2F−
Starting 0 0
Change +x +2x
Equilibrium x 2x
[Ba2+][F−]2 = Ksp
(x)(2x)2 = 4x3 = 1.0  10−6

1.0  106
x= 3 = 6.299  10−3 = 6.3  10−3 M
4
b. At the start, before any BaF2 dissolves, the solution contains 0.15 M F−. At equilibrium, x
mol of solid BaF2 dissolves to yield x mol Ba2+ and 2x mol F−. Assemble the usual
concentration table, and substitute the equilibrium concentrations into the equilibrium-
constant expression. As an approximation, assume x is negligible compared to 0.15 M F−.
Conc. (M) BaF2(s)  Ba2+ + 2F−
Starting 0 0.15
Change +x +2x
Equilibrium x 0.15 + 2x




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, Chapter 17: Solubility and Complex-Ion Equilibria 659

[Ba2+][F−]2 = Ksp
(x)(0.15 + 2x)2  (x)(0.15)2  1.0  10−6
1.0  106
x = 4.444  10−5 = 4.4  10−5 M
(0.15) 2

Note that adding 2x to 0.15 M will not change it (to two significant figures), so 2x is
negligible compared to 0.15 M. The solubility of 4.4  10−5 M in 0.15 M NaF is lower than
the solubility of 6.3  10−3 M in pure water.

17.6. Calculate the ion product, Qc, after evaporation, assuming no precipitation has occurred. Compare
it with the Ksp.
Qc = [Ca2+][SO42−] = (2  0.0052)(2  0.0041) = 8.528  10−5
Since Qc > Ksp (2.4  10−5), precipitation occurs.

17.7. Calculate the concentrations of Pb2+ and SO42−, assuming no precipitation. Use a total volume of
0.456 L + 0.255 L, or 0.711 L.
0.00016 mol
 0.255 L
[Pb ] =
2+ L = 5.74  10−5 M
0.711 L
0.00023 mol
 0.456 L
[SO42−] = L = 1.48  10−4 M
0.711 L
Calculate the ion product, and compare it to Ksp.
Qc = [Pb2+][SO42−] = (5.74  10−5)(1.48  10−4) = 8.49  10−9
Since Qc is less than the Ksp of 1.7  10−8, no precipitation occurs, and the solution is unsaturated.

17.8. The solubility of AgCN would increase as the pH decreases, because the increasing concentration
of H3O+ would react with the CN− to form the weakly ionized acid HCN. As CN− is removed,
more AgCN dissolves to replace the cyanide:
AgCN(s)  Ag+(aq) + CN−(aq) [+ H3O+  HCN + H2O]

In the case of AgCl, the chloride ion is the conjugate base of a strong acid and would, therefore,
not be affected by any amount of hydrogen ion.

17.9. Because Kf = 4.8  1012 and because the starting concentration of NH3 is much larger than that of
the Cu2+ ion, you can make a rough assumption that most of the copper(II) is converted to
Cu(NH3)42+ ion. This ion then dissociates slightly to give a small concentration of Cu2+ and
additional NH3. The amount of NH3 remaining at the start after reacting with 0.015 M Cu2+ is
[0.100 M − (4  0.015 M)] = 0.040 M starting NH3




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, 660 Chapter 17: Solubility and Complex-Ion Equilibria


Assemble the usual concentration table using this starting concentration for NH3 and assuming
the starting concentration of Cu2+ is zero.
Conc. (M) Cu(NH3)42+  Cu2+ + 4NH3
Starting 0.015 0 0.040
Change −x +x +4x
Equilibrium 0.015 − x +x 0.040 + 4x
Even though this reaction is the opposite of the equation for the formation constant, the
formation-constant expression can be used. Simply substitute all exact equilibrium concentrations
into the formation-constant expression; then simplify the exact equation by assuming x is
negligible compared to 0.015 and 4x is negligible compared to 0.040.
[Cu(NH3 ) 4 2+ ] (0.015  x) (0.015)
Kf = 2+ 4
= 4
  4.8  1012
[Cu ][NH3 ] (x)(0.040 + 4x) (x)(0.040) 4

Rearrange and solve for x:
x = [Cu2+]  (0.015)  [(4.8  1012)(0.040)4]  1.22  10−9 = 1.2  10−9 M

17.10. Start by calculating the [Ag+] in equilibrium with the Ag(CN)2− formed from Ag+ and CN−. Then
use the [Ag+] to decide whether or not AgI will precipitate by calculating the ion product and
comparing it with the Ksp of 8.3  10−17 for AgI. Assume all the 0.0045 M Ag+ reacts with CN− to
form 0.0045 M Ag(CN)2−, and calculate the remaining CN−. Use these as starting concentrations
for the usual concentration table.
[0.20 M KCN − (2  0.0045 M)] = 0.191 M starting CN−
Conc. (M) Ag(CN)2−  Ag+ + 2CN−
Starting 0.0045 0 0.191
Change −x +x +2x
Equilibrium 0.0045 − x x 0.191 + 2x
Even though this reaction is the opposite of the equation for the formation constant, the
formation-constant expression can be used. Simply substitute all exact equilibrium concentrations
into the formation-constant expression; then simplify the exact equation by assuming  is
negligible compared to 0.0045 and 2x is negligible compared to 0.191.
[Ag(CN) 2  ] (0.0045  x) (0.0045)
Kf =  2
= 2
  5.6  1018

[Ag ][CN ] (x)(0.191 + 2x) (x)(0.191) 2

Rearrange and solve for x:
x = [Ag+]  (0.0045)  [(5.6  1018)(0.191)2]  2.202  10−20 M
Now, calculate the ion product for AgI:
Qc = [Ag+][I−] = (2.20  10−20)(0.15) = 3.30  10−21 = 3.3  10−21
Because Qc is less than the Ksp of 8.3  10−17, no precipitate will form, and the solution is
unsaturated.




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