Chapter 15: Solubility, Precipitation, and Complexation
A: MULTIPLE CHOICE QUESTIONS
1. Which one of the following is the correct equation for the dissolution of calcium hydroxide,
Ca(OH)2(s) in water?
a) Ca(OH)2(aq) + H2O(l) ⇋ Ca2+(aq) + 2OH-(aq) + H2O(l)
b) Ca(OH)2(aq) ⇋ Ca2+(aq) + 2OH-(aq)
c) Ca(OH)2(s) ⇋ Ca2+(aq) + 2OH-(aq)
d) Ca(OH)2(s) ⇋ Ca(OH)2(aq)
ANS: c
REF: Pages, 554, 555
2. Which of the following is the correct form of the expression for Q during the dissolution of calcium
hydroxide?
[𝐶𝑎2+][𝑂𝐻−]
a) [𝐶𝑎(𝑂𝐻)]^2
b) [Ca2+][OH-]
c) [Ca2+][OH-]2
d) [Ca2+][2OH-]2
ANS: c
REF: Page 555
3. The group II fluorides including that of barium, magnesium, leads, and calcium, have Ksp’s,
respectively, of 2.4 x 10-5, 6.4 x 10-9, 4.0 x 10-8, and 4.0 x 10-11. The order of increasing solubility is
which one of the following?
a) CaF2 >MgF2>PbF2>BaF2
b) BaF2>PbF2>MgF2>CaF2
c) BaF2>MgF2>PbF2>CaF2
d) CaF2>PbF2>MgF2>BaF2
ANS: b
REF: Pages 556, 557
4. Which one of the following is likely to have a higher solubility than predicted by a simple dissolution
reaction?
a) KNO3
b) NaCl
c) LiF
d) Rb2SO4
ANS: d
REF: Pages, 558, 559
5. Which one of the following is not predicted to have an effect on the solubility of nickel II sulfide?
a) A decrease in the pH using nitric acid
b) Addition of H2S
c) Addition of nickel nitrate
d) Addition of more solid nickel sulfide
ANS: d
REF: Section 15.2
6. For the imaginary ionic compound XY, which one of the following scenarios will lead to formation of
a precipitate?
a) Q<Ksp
b) Q<Ksp but the concentration of the cation exceeds the normal saturated concentration
c) Q = Ksp
d) Only when Q>Ksp
ANS: d
REF: Page564
7. A saturated solution of silver chloride will precipitate if Q is greater than Ksp but addition of
potassium chloride results in disappearance of the precipitate. This counterintuitive event occurs
because of which one of the following?
a) Potassium chloride is acidic and the chloride ions are basic
b) Silver chloride can acquire another chloride ion and become soluble
c) Potassium forms a solid with chloride ions in solution and shifts the equilibrium
d) Dissolution of potassium chloride is exothermic and the rise in temperature results in a higher
solubility for silver chloride
ANS: b
REF: Page 571
8. In the case of competing equilibria such as seen with AgCl and Ag(NH3)2, the overall equilibrium
constant called Knet, is calculated by which of the following methods?
a) Summing the equilibrium constants Ksp and Kf
b) Multiplying the equilibrium constanst Ksp and Kf
c) Subtracting the equilibrium constants Ksp and Kf
d) Dividing the Kf into Ksp
ANS: b
REF: Pages 569, 570
9. The competitive solubilization of silver chloride using ammonia is best explained by which one of the
following?
a) The ammonia ions complex with silver ions and pull them from the solid
b) The ammonia ions pick up protons from water to form ammonium and hydroxide, the hydroxide
ions complex with aquated silver ions and shift the equilibrium
c) The ammonia molecules form a complex with the AgCl crystals that is soluble
d) The ammonia molecules complex with aquated silver ions and shift the solubility equilibrium of
AgCl
ANS: d
REF: Page 569
10. What is the pH of a saturated solution of silver hydroxide if Ksp = 2.0 x 10-8?
a) 7.70
b) 10.15
c) 3.85
d) 8.00
ANS: b
REF: Entire chapter 15
B: SHORT ANSWER QUESTIONS
1. The Ksp of strontium chloride, SrCl2, is 7.9 x 10 -10 at 250C. What is its’ solubility?
ANS:
Ksp is the equilibrium constant for a saturated solution of strontium chloride. It dissociates
according to the following equation.
SrCl2(s) ⇋ Sr2+(aq) + 2Cl-(aq)
At equilibrium, the concentrations of strontium ions and chloride ions are x and 2x, respectively.
Ksp = 7.9 x 10-10 = x(2x)2 = 4x3 solving for x we have; x = 5.82 x 10 -4 = [Sr2+] in a saturated
solution at this temperature. I have chosen strontium because one strontium ion forms for every
one formula unit of strontium chloride that goes into solution. Therefore the strontium chloride
solubility is the same as the strontium ion concentration, i.e., 5.82 x 10-4 M
REF: Section 15.3
2. A student adds 0.2500 grams of Ni(OH)2 to 25.00 mL of water at 25 0C and stirs until no more
Ni(OH)2 will dissolve.
Ksp for Ni(OH)2 is 1.6 x 10 -16
What mass of Ni(OH)2 remains as solid?
ANS:
The solubility of Ni(OH)2 depends upon its’ Ksp at this temperature. The dissociation reaction is;
Ni(OH)2(s) ⇋ Ni2+(aq) + 2OH-(aq)
At equilibrium (saturation) there will be x M nickel ions and 2x M hydroxide.
Ksp = 1.6 x 10-16 = x(2x)2 = 4x3 solving for x we have; x = 3.42 x 10 -6
x equals the concentration of nickel ion in the saturated solution. It also equals the solubility of
the Ni(OH)2 in moles per litre.
We have 0.02500 L of solution. 0.02500 L x (3.42 x 10-6 moles/L) = 8.498797 x 10 -8 moles of
Ni(OH)2 in solution.
8.498797 moles of Ni(OH)2 x (58.71 + 32.00 + 2.02)g/mole = 7.928 x 10 -6 g in solution.
We started with 0.2500 grams of Ni(OH)2 and only the above mass dissolved. This leaves
(0.2500 – 7.928 x 10-6) g = 0.249992072 g = 0.2500 g The difference in mass is inconsequential.
REF: Page’s 566, 567
3.
Due to the common ion effect, barium fluoride is less soluble in a solution of potassium fluoride
than it is in pure water. Ksp of BaF2 is 2.4 x 10-5. What is the solubility of BaF2 in a solution that is
1.000 mM for KF?
ANS:
Potassium fluoride as a salt of a group 1metal with a halide is highly soluble (refer back to earlier
chapters). Therefore a 1.000 millimolar solution of KF would be 1.000 mM for fluoride ions.
That is, it is assumed to be 100% soluble.
The dissolution reaction for barium fluoride is as follows;
BaF2(s) ⇋
Ba2+(aq) +
2F-(aq)
I
C
E
0
+x
x
0.001
+2x
0.001 + 2x
Ksp = x(0.001 + 2x)2 = x(1.0 x 10 -6 + 0.002x + 0.002x + 4x2) = 10-6x + 0.004x2 + 4x3 = 2.4 x
10-5
And solving for x (using an equation solver) we have; x = 0.0178
x = [Ba2+] and x = solubility of BaF2
REF: Pages 560, 561
4. Dissolving of lead arsenate in an acidic solution results in a higher concentration of lead ions in
solution than would occur in pure water. This is due to the reaction of hydronium ions with the
arsenate ions of the salt. This shifts the equilibrium toward dissolution as the arsenate ions are
converted to hydrogen arsenate. Hydrogen arsenate has a Ka of 8.0 x 10-8
The overall equilibrium depends upon the following two equations;
PbAsO4(s) ⇋ Pb2+(aq) + AsO42-(aq)
Ksp = 4.0 x 10-36
Their sum is;
AsO42-(aq) + H3O+(aq) ⇋ HAsO4-(aq) + H2O(l)
PbAsO4(s) + H3O+(aq) ⇋ HAsO4-(aq) + H2O(l) + Pb2+(aq)
Karsenate = ?
What is the value for the equilibrium constant for the net reaction?
ANS:
The equilibrium constant for the second reaction is the inverse of the acid dissociation constant
since it is simply the reverse equation of the acid reacting with water. Thus; Karsenate = 1/Ka =
1/8.0 x 10 -8 = 1.25 x 107
The overall Knet = Ksp x Karsenate = (4.0 x 10-36)(1.25 x 107) = 5.0 x 10 -29
REF: Pages, 562, 563
5. If a student mixes 25 mL of 0.00015 M sodium hydroxide with 40.0 mL of 0.0011 M silver nitrate,
will a precipitate form at 250C in water?
ANS:
We are concerned with the following solubility equilibrium;
AgOH(s) ⇋ Ag+(aq) + OH-(aq)
In a saturated solution, K = [Ag+][OH-] = Q
In this instance; the
[Ag+] = (0.040L x 0.0011 moles/L) / (0.025L + 0.040L) = 6.769 x 10-4
M
[OH-] = (0.025L x 0.00015 moles/L) / (0.025L + 0.040L) = 5.7692 x 10-5
Q = [Ag+][OH-] = (6.769 x 10-4 M) x (5.7692 x 10-5) = 3.905 x 10-8
Referring to a table of solubility product constants we find that silver hydroxide has a solubility
product constant of 2.0 x 10-8. Q is greater than Ksp therefore a precipitate can form.
REF: Pages 564, 565
6. A photographer wishes to reclaim the remaining silver from his dipping tank containing silver nitrate.
From past experience he knows that the residual concentration is about 0.250 moles/L.
Ksp for silver bromide is 5.0 x 10-13. Assuming his estimate is accurate, what is the concentration of
bromide ion still in solution once he has precipitated 99% of the silver from a 5.00 L tank?
ANS:
If 99% is precipitated, then 1% of the silver remains in solution as silver ions. The amount
remaining in solution is dependent upon the concentration of the counter ion, Br-. Ksp is a
constant, [Ag+] can be calculated and [Br-] is unknown.
Ksp =[Ag+][Br-] becomes;
[Br-] = Ksp/[Ag+]
+
[Ag ] = 1% of 0.250 M = 0.01 x 0.250M = 0.0025M
Then;
[Br-] = 5.0 x10-13/0.0025 = 2.0 x 10-10 M
REF: Pages 566,567
7. A company doing electroplating has a residual mix of plating solution that contains both silver nitrate
and nickel nitrate ions in equal concentration. What differential solubility criteria could be used to
separate the two cations selectively but still do so using a common counter ion?
ANS:
Either a sulfide or a hydroxide salt could be used to selectively precipitate these two ions. Silver
hydroxide has a Ksp of 2.0 x 10-8 and nickel hydroxide has a Ksp of 1.6 x 10-16. If we solve for
hydroxide concentration at saturation we see that a nickel solution is saturated at a concentration
of 6.84 x10-6M and the silver solution becomes saturated at a much higher concentration of 1.414
x 10-4 M in solutions made from the hydroxides. By carefully controlling addition of sodium or
potassium hydroxide to just below the saturation point of the silver, we could effectively
precipitate all of the nickel from solution while leaving the silver solvated.
After filtering, more hydroxide could be added to precipitate the still aquated silver ions as silver
hydroxide. Hydroxide is preferable to sulfide for this protocol due to its less significant
environmental risk.
If we have an inkling of the starting concentration of the silver and nickel in solution, the amount
of sodium or potassium hydroxide to be added can be reasonably predicted prior to
commencement of the isolation protocol.
REF: Pages 567, 568
8.
A student has a solution of copper nitrate and wishes to precipitate the copper II ions as copper
sulfide. However, instead of adding sodium sulfide, she inadvertently adds sodium amide (NaNH2).
The amide ion forms ammonia and hydroxide ions when added to water. The solution changes color
from a medium blue to a dark blue and no precipitate forms. Why did the solution change color?
ANS:
Prior to adding the sodium amide, the copper solution contained aquated Cu2+ ions and aquated
nitrate ions. Upon adding the sodium amide, the amide reacted with water to form ammonia and
hydroxide.
At this point, either the remaining ammonia or the hydroxide could react with the copper ions.
Copper is a Lewis acid. It will react with the strongest Lewis base in solution. The hydroxide is
the strongest Lewis base when comparing hydroxide, ammonia, and water as potential electron
donors. Thus, complexes form with the hydroxide ions and the copper begins to precipitate as
copper hydroxide. However, the ammonia molecules are able to complex with the Cu(OH)2 to
form [Cu(NH3)4(H2O)2](OH)2 complexes which are soluble. These complexes have a large
formation constant and lend the solution a dark blue color.
REF: http://en.wikipedia.org/wiki/Schweizer's_reagent and Pages 569 to 573 of text
9.
What would be the expected effect of adding nitric acid to a solution containing a metal hydroxide
precipitate?
ANS:
All metal hydroxides react in acid since the hydroxide ions of the solid, accept protons to form
water. Given enough acid, the metal ions become fully hydrated. An observer would see the
solid seem to dissolve. In reality, new complexes are forming.
REF: Pages 561 to 563
10. A student mixes silver nitrate with sodium chloride and obtains a white precipitate. They then add
ammonia in water and the precipitate disappears. Finally, they add nitric acid dropwise and
eventually a precipitate reappears. Explain the observations above.
ANS:
Silver nitrate is soluble so the solution contains aquated silver ions and aquated nitrate ions.
Upon addition of sodium chloride, the majority of the silver ions precipitate as AgCl(s). Addition
of ammonia permits formation of Ag(NH3)2+ from silver ions still in solution. This removal of
silver ions in solution drives dissolution of the silver chloride (which is slightly soluble) until all
of the silver chloride is converted to the silver diamine complex and aquated chloride ions.
Addition of acid converts the ammonia in the Ag(NH3)2+ complexes to ammonium, NH4+. The
ammonium is not a Lewis base and no longer complexes with the Lewis acid, Ag+. Instead, the
chloride ions, which are still in solution and not basic to protons, form complexes with the silver
ions and re-precipitate as AgCl(s).
REF: Page 570 of text
C: LONG-ANSWER QUESTIONS
1. Cd(OH)2 has a Ksp of 5.9 x 10-15 at 25 0C . Answer the following questions.
a) The solubility of this hydroxide in micrograms/mL
b) The pH of a saturated solution
c) The expected behavior in acid
ANS:
a) The dissolution equation for cadmium hydroxide is as follows;
Cd(OH)2(s) ⇋ Cd2+(aq) + 2OH-(aq)
In a saturated equilibrium the concentrations of Cd2+ and hydroxide are, respectively, x and
2x.
Ksp = 5.9 x 10-15 = [Cd2+][OH-]2 = x(2x)2 = 4x3
solved for x = 1.138 x 10-5 = solubility
in moles/L
In grams/mL we have;
(1.138 x 10-5 moles/L) x (1litre/1000 mL) x (146.43 g/mole) = 2.02 x 10-6 g/mL = 2.02
ug/mL
b) The pH is dependent upon the hydroxide ion concentration in the saturated solution. It was
calculated in the prior step to be 2x where x = 1.138 x 10-5 thus; [OH-] = 2.276 x 10-5
pH = 14 – pOH = 14 –(-log 2.276 x 10-5) = 9.36
c) The Cd(OH)2 will accept protons forming hydrated cadmium ions.
REF: Entire chapter
2. Potassium oxalate at a concentration of 0.210M and a volume of 0.100 L is added to 0.170 L of 0.250
M barium bromide. A precipitate forms. Barium oxalate has a Ksp of 2.3 x 10-8. What are the residual
concentrations of all ionic species in solution after the precipitation reaction has gone to completion?
ANS:
The potassium oxalate is soluble, as is the barium bromide. The barium oxalate, on the other
hand, has a low solubility. We can make the assumption that formation of the barium oxalate
goes to completion due to its low solubility relative to the starting concentrations. Because of this
we can say the following;
 Moles of potassium oxalate = 0.210 moles/L x 0.100L = 0.0210 moles
 Moles of barium bromide = 0.170L x 0.250 moles/L = 0.0425 moles
I
C
Final
c=
K2C2O4(aq) +
0.0210 moles
-0.0210
0 moles
0M
BaBr2(aq) 
0.0425 moles
- 0.0210
0.0215 moles
0.0215 moles
/0.27L = 0.0796M
2KBr(aq) + BaC2O4(s)
0 moles
0 moles
+(2 x 0.0210) +(0.0210)
0.0420 moles 0.0210 moles
= 0.156 M
0.0778 M
Then we can use the dissociation reaction of barium carbonate to calculate the concentration of
oxalate and barium in solution through dissolution. The amount of each ion that aquates will be
dependent upon the Ksp and the amount of the common ion, Ba2+, that is in solution.
BaC2O4(s) ⇋
I
C
E
Ba2+(aq) +
C2O42-(aq)
0.0796M
+x
0.0796 + x
0
+x
x
Ksp = 2.3 x 10-8 = (0.0796 + x)(x) = 0.0796x + x2
[Ba2+] = 0.0796 + 2.89 x 10-7 = 0.0796M
[K+] = [KBr] = 0.156M
[Br-] = [KBr] = 0.156M
REF: Exercise 15.2
x = 2.89 x 10-7 = [C2O42-]
3. Magnesium hydroxide has a low solubility with a Ksp of 1.8 x 10-11. In the presence of ammonia, a
solution of magnesium chloride will be susceptible to precipitation since ammonia reacts with water
to form the hydroxide ion and ammonium. As a weak base with a Kb of 1.8 x 10-5, the solution is
tolerant of ammonia to a finite concentration. A given solution is 0.010M for MgCl2.
a) What is the maximum hydroxide concentration permitted before precipitation occurs and at what
pH does this occur?
b) If the solution is 0.100 M for NH3 and the source of the hydroxide and the ammonium is through
construction of a buffer with ammonium chloride and not through reaction of ammonia with water,
what is the NH4+ concentration at this concentration of hydroxide?
c) Would a higher concentration of ammonium, favour formation of the hydroxide or inhibit it?
d) What could you add to increase the ammonium ion concentration without increasing the
ammonia concentration and, given this is a buffer, what would be an appropriate concentration of
the ammonium?
ANS:
a) The maximum hydroxide concentration depends upon Ksp for Mg(OH)2 and the concentration
of the magnesium ion.
Ksp = 1.8 x 10-11 = [Mg2+][OH-] 2
solving for [OH-] = (1.8 x 10-11/[Mg2+])1/2 = (1.8
-11
1/2
1/2
x 10 ) / (0.010M) )= (1.8 x 10-9) = 4.24 x 10-5 M
b) In a buffer solution of ammonia and ammonium, the following equilibrium applies;
NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH-(aq) Kb = 1.8 x 10-5
Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5
and; [NH4+] = Kb[NH3]/[OH-] = (1.8
-5
-5
x10 ) (0.100M) / (4.24 x10 M) = 0.0423M
c) A higher concentration of ammonium would inhibit the formation of precipitate because the
buffer equilibrium shown above would shift to the left as ammonium concentration rises.
This would consume hydroxide ion and thereby prevent a hydroxide concentration high
enough to cause precipitation of the magnesium ions.
d) Add more ammonium chloride until the concentration of ammonium equals the concentration
of ammonia. That is; ideally 0.10 M in ammonium chloride and 0.10M in ammonia.
REF: Section 15.3 and chapter 14
4. Inorganic mercury, as Hg22+ is in equilibrium with elemental mercury and the +2 ion, according to
the following equilibrium;
Hg2+(aq) + Hg(l) ⇋ Hg22+(aq)
The product, Hg22+ is mercury in the +1 oxidation state. The equilibrium lies far to the right.
a) Suggest a strategy to remove all of the mercury from a solution containing these species once
prepared from digestion, solubilization, and oxidation of a biological sample. Assume there are no
cross reactions with unidentified organic species.
b) why will a saturating solution of your chosen ion not be adequate for precipitating all of the
mercury in solution and what must be done, instead?
ANS:
a) Since the equilibrium lies far to the right, there is likely to be very little of the liquid mercury
and mercury II present in solution. Nonetheless, driving the equilibrium forward will convert
all of the mercury II and elemental mercury to mercury I. This can be accomplished by
removing the mercury II from solution as a precipitate. Mercury I precipitates well with
iodide ion as Hg2I2 (Ksp = 2.9 x 10-29).
Iodide ion could be supplied from sodium iodide. The sodium will not react with any species
in solution and sodium iodide is highly soluble. Either solid sodium iodide can be dissolved
into the solution, or a concentrated solution of the sodium iodide can be added.
b) A saturating solution means that there is just enough of the counter ion to induce
precipitation. As soon as some of the mercury I precipitates from solution as Hg2I2, the
concentration of the iodide drops below a saturating level. So does the level of Hg22+ in
solution. You can think of this as a decrease in the ion concentrations without a decreasing
change in volume. Since concentration is moles/L and moles have decreased, but volume
hasn’t, then concentration has decreased. To precipitate all of the Hg22+, a large excess of the
fluoride ion must be used; not just saturating levels.
REF: Pages 570, 571
5.
The Ksp of AgCl is 1.8 x10-10 and that of CaF2 is 5.3 x 10-11. Clearly, that for silver chloride is a
larger number than that of CaF2.
a) Explain how silver chloride solubility is lower than calcium fluoride even though the Ksp of
silver chloride is a larger number. Use the saturated concentrations of the solutions to illustrate
your explanation.
b) What characteristic of silver chloride makes it important to gradually precipitate the salt as solid
silver chloride rather than adding a large excess of chloride in one single step?
ANS:
a)
AgCl(s) ⇋ Ag+(aq) + Cl-(aq)
At saturation;
x
x
Ksp = x2 = 1.8 x10-10
solved for x = solubility = 1.34 x 10-5
CaF2(s) ⇋ Ca2+(aq) + 2F-(aq)
At saturation:
x
2x
Ksp = 5.3 x 10-11 = x(2x)2 = 4x3 solved for x = solubility = 2.36 x 10-4
The solubility of the calcium fluoride is higher
b) Silver chloride will form AgCl2- as the chloride ion concentration rises. This complex is
soluble. The formation constant for this product is large and thus strongly favoured. By
adding the chloride ion slowly, just enough chloride is present to form the AgCl(s) and not
enough to favour formation of the dichloride complex.
REF: Page 571