Applications of Aqueous Equilibria II: Solubility and Ksp
Reading: Moore chapter 17, sections 17.4-17.6
Questions for Review and Thought: 54, 56, 58, 60, 62, 66, 72, 74, 86.
Key Concepts and Skills:
•
definition of solubility product constant, complex ion, solubility, sparingly soluble
(insoluble) salt, freely soluble salt, common ion effect, selective precipitation
•
Review calculations of Ksp from solubility of ionic salts, using Ksp, calculate the
solubility of a sparingly soluble salt, how to use Q to predict whether a solution is
unsaturated, saturated, or supersaturated (a precipitate forms), how to use Ksp and
ionic equilibria for selective precipitation of ions, show how complex ion
formation and pH can alter the solubility of a sparingly soluble salt.
Lecture Topics:
I.
Solubility and Ksp.
Solubility (S) is defined as the amount of solute per unit volume of solution that
dissolves to form a saturated solution (on the verge of precipitation).
For freely soluble salts, at least several grams of salt dissolves per 100g of water.
For sparingly soluble (or insoluble) salts, a mass of salt << 1 gram dissolves in
100g of water.
Review the solubility rules presented in the text on page 164 (Chapter 5)
Example:
AgNO3(s)  AgNO3(aq)  Ag +(aq) + NO3-(aq)
S
S
S
S can be moles/liter or grams/liter or mg/liter. Concentrations [] are always moles/L
The solubility product constant, K sp is the equilibrium constant for the balanced
chemical equation representing a solubility equilibrium. The magnitude of Ksp indicates
the extent to which the solid solute dissolves to give ions in solution:
AgNO3(s)  Ag +(aq) + NO3-(aq)
Ksp = [Ag+][NO3-]=(S)(S)=S2
n+
mIn general, for
AxBy(s)  xA + yB ,
Ksp = [An+]x[Bm-]y
Thus, Ag2SO4(s)  2 Ag+ (aq) + SO42- (aq)
Ksp = [Ag+]2[SO42-]=(2S)2(S)=4S3
2+
–
And Mn(OH)2 (s)  Mn (aq) + 2 OH (aq)
Ksp = [Mn2+][HO-]2=(S)(2S)2=4S3
Values of Ksp for numerous salts are tabulated in appendix H
Example: calculate the solubility of CuCl: CuCl(s)  Cu+ (aq) + Cl-(aq) Ksp=1.9x10-7
Ksp = [Cu+][Cl-] = S2; S=√1.9x10-7 = 4.3 x 10-4 moles/liter
Ksp, the equilibrium constant for the equilibrium between an ionic solute and its ions in
solution, can very accurately be used to describe experimentally observed equilibria
when: 1.) salts contain +1 and –1 ions, and 2.) the ions do not react with water
Example of potential complications:
Suppose CaCO3  Ca2+ (aq) + CO32-(aq)
And: CO32- (aq) + H2O(l)  HCO3- (aq) + HO-(aq) ; if the second reaction occurs to
any extent, the solubility of CaCO3 will be increased! (Le Chatelier’s principle)
II.
Factors Affecting the Aqueous Solubility of Ionic Compounds
1. Temperature: endothermic dissolutions lead to greater solubility at
elevated temperature; exothermic dissolutions lead to less solubility at
elevated temperature.
2. Competing equilibria (see above and below)
3. pH
A. Effect of pH on solubility of sparingly soluble salts
• An insoluble salt can be dissolved by acid if one or both of its ions are basic
For CaCO3, the acidity of water determines the extent of dissolution; less acidic aqueous
media leads to precipitation.
CaCO3  Ca2+ (aq) + CO32-(aq)
CO32- (aq) + H3O+(aq)  HCO3- (aq) + H2O(l)
(H3O+ is a strong acid)
HCO3- (aq) + H3O+(aq)  H2CO3 (aq) + H2O(l);
H2CO3 (aq)  CO2(g) + H2O(l)
The escape of CO2(g) shifts
equilibrium to the right
In general, insoluble salts containing ions that are Bronsted bases (CO32-, S2-, PO43-)
dissolve in solutions of low pH (exception: Cl- containing salts, eg., AgCl)
B.
Common Ion Effect
• A process of removing ions by forming an insoluble precipitate
• A “common ion” displaces equilibrium to the left, resulting in precipitation
•The presence of a second solute that provides a common ion lowers the solubility of an
ionic compound.
Example: What is the solubility of AgCl in 0.55 M NaCl?
Note that NaCl dissolves completely in water: NaCl(s) Na+(aq) + Cl-(aq), thus initial
[Cl-] = 0.55M. Thus,
AgCl(s)  Ag+ (aq) + Cl-(aq)
At equilibrium:
S
S+ 0.55M
Thus, Ksp = (S)(S + 0.55)= 1.8 x 10-10; S= 3.3 x 10-10.
Without NaCl, Ksp = S2 ; S=1.3 x10-5
•The extra Cl- suppresses dissolution of AgCl, as we’ d expect from LeChatelier’s
principle.
C.
Complex Ion Formation
Complex ion formation occurs when a metal in solution reacts with a Lewis basic
substance:
AgCl (s) + 2 CN-(aq)  [Ag (CN)2]+(aq) + Cl-(aq)
Complex ion formation can dissolve an insoluble metal salt:
AgCl(s)  Ag+(aq) + Cl-(aq)
Ksp= 1.8 x 10-10
+
+
Ag (aq) + 2CN  [Ag (CN)2] (aq)
Kf = 5.6 x 1018
Net: AgCl (s) + 2 CN-(aq)  [Ag (CN)2]+(aq) + Cl-(aq)
K=Ksp*Kf = 1 x109
Kf is the formation constant which gives quantitative information about the complexforming equilibrium step; note the net reaction is now favorable for dissolution of AgCl.
•Complex ions are non-neutral metal complexes
•metal salts insoluble in water can be brought into solution by reaction with a Lewis base
(NH3, OH-, CN-, S2O32-, etc), forming a complex ion.
•Le Chatelier’s principle at work: equilibrium is shifted so that more insoluble metal salt
dissolves.
D. Amphoterism
Metal hydroxides have both acidic and basic character: Al(OH)3.
E. Precipitation – Prediction using Q.
Comparing Q and Ksp allows one to predict whether a solution is unsaturated, saturated,
or supersaturated.
• if Q<Ksp, the solution is unsaturated and no precipitate forms. More solute needs to be
added in order for precipitation to occur.
•if Q>Ksp, the solution contains a higher concentration of ions than it can hold at
equilibrium, and the solution is supersaturated. Precipitation occurs till Q=Ksp.
•if Q=Ksp, the solution is saturated, and is at equilibrium at the point of precipitation.
Example: if you mix 0.2L of 4.5 x 10-3 M AgNO3 with 0.1 L of 7.5 x 10-2M NaBrO3, will
a precipitate form? (Ksp (AgBrO3) = 6.7 x 10-5) How much AgBrO3 is precipitated?
[Ag+] = 4.5x10-3M*(0.2L/0.3L) = .003M; [BrO3-] = 7.5x10-2M*(0.1L/0/3L)= .025M
AgBrO3(s)  Ag+ (aq) + BrO3-(aq) Q=[Ag+][BrO3-]=7.5x10-5;
Q>Ksp, thus a precipitate will form. How much?
AgBrO3(s)  Ag+ (aq) + BrO3-(aq)
x
.003-x
.025-x
Ksp=6.7x10-5 =[Ag+][BrO3-]=(.003-x)(.025-x)
-4
using quadratic to solve, x=2.8x10 moles/L; since 0.3L, 8.4 x 10-5 moles AgBrO3
precipitate.
F. Selective Precipitation
• Principle: more soluble compounds remain in solution; less soluble compounds
precipitate.
•Two ions of the same charge can be separated from each other by using an ion of
opposite charge that forms a sparingly soluble salt of each.
Example: A solution is 0.1M in both Ca2+ and Ba2+ ions. Ksp (CaF2)=1.7 x 10-10; Ksp
(BaF2) = 1.7 x 10-6) NaF is added slowly. How high can F- rise before BaF2 begins to
precipitate? What fraction of the Ca2+ remains in solution?
Ksp (BaF2) = 1.7 x 10-6 = [Ba2+][F-]2 = (0.1M)[F-]2;
[F-] = .0041M (F- can rise until this concentration is reached);
How much Ca2+ is left in solution?
Ksp (CaF2)=1.7 x 10-10= [Ca2+][F-]2 = [Ca2+](.0041M)2
[Ca2+] = 1 x 10-5M; fraction remaining unprecipitated: 1 x10-5M/0.1M = 1x10-4 or .01%
Problems
1. A 0.1L sample of a saturated solution of copper (II) periodate [Cu(IO4)2]
contains 0.146 g of the dissolved salt. What is the Ksp of copper (II)
periodate?
2. What is the solubility of Hg2Cl2 in moles/liter and g/liter if Ksp = 1.1x10-18
and the cation Hg22+ is the cation in solution?
3. The Ksp for lead iodide (PbI2) is 1.4x10-8. Calculate the solubility of lead
iodide in
a. water
b. 0.1M Pb(NO3)2
c. 0.01M NaI
4. What are the concentrations of all ions in solution after the following solutions
are mixed and have reached equilibrium? Will a precipitate form? If so, how
many grams of solid are formed?
a. 100 mL of 0.02M Pb(NO3)2 and 100.0 mL of 0.02M NaCl (assume
both NaCl and Pb(NO3)2 are freely soluble in water and Ksp
PbCl2=1.6x10-5)
b. 75mL of 0.02M BaCl2 and 125ml of 0.04M Na2SO4
(assume BaCl2 and Na2SO4 are freely soluble in water and that
Ksp (BaSO4)=1.1 x 10-10)
Note: if a precipitate forms in either of the above cases, simplify the math by
assuming that as much solid forms as possible.
5. For each of the following ionic compounds, state whether the solubility will
increase, decrease, or remain unchanged as a solution at pH 7 is made acidic.
a. PbI2
b. AgOH
c. Ca3(PO4)2
6. The cations in an aqueous solution that contains 0.1M Hg2(NO3)2 and 0.05M
Pb (NO3)2 are to be separated by taking advantage of the difference in the
solubilities of their iodides Ksp (PbI2) = 1.4 x 10-8; Ksp (Hg2I2) = 1.2 x 10-28.
What should be the concentration of iodide ion for the best separation (one in
which one of the cations should remain entirely in solution and the other
should precipitate as fully as possible)? Assume Hg22+ is the cation in solution
from Hg2(NO3)2.
7. When 50 mL of 0.1M AgNO3 and 30 mL of 0.06M Na2CrO4 are mixed, a
precipitate of of silver chromate (Ag2CrO4) is formed. The Ksp of silver
chromate in water is 1.9 x 10-12. Calculate the [Ag+] and [CrO42-] remaining in
solution at equilibrium.
8. The enamel of your teeth contains hydroxyapatite (Ca5(PO4)3OH,
Ksp= 6.8x10-37) write out the chemical equation & solubility product
expression for the dissolution of this substance in water. It is known that tooth
decay starts when bacteria in the mouth convert sugar into organic acids. Why
might this be so? Flouridation of drinking water (0.7 to 1.2 ppm) and in dental
products converts hydroxyapatite to Ca5(PO4)3F (Ksp=1 x10-60). How does this
prevent tooth decay?