20
Ionic Equilibria III:
The Solubility Product
Principle
1
Chapter Goals
• Solubility Product Constants溶解度積常數
• Determination of Solubility Product Constants 決定溶
解度積常數
• Uses of Solubility Product Constants
• Fractional Precipitation分級沉澱
• Simultaneous Equilibria Involving Slightly Soluble
Compounds
• Dissolving Precipitates
2
Aqueous Solutions: An
Introduction
Insoluble
soluble
3
Solubility Product Constants
• Silver chloride, AgCl, is rather insoluble in
water.
• Careful experiments show that if solid AgCl is
placed in pure water and vigorously stirred,
a small amount of the AgCl dissolves in the
water.
Ag+Cl-  Ag+(aq) + Cl-(aq)
• The equilibrium constant expression for this
dissolution is called a solubility product
constant 溶解度積常數.
– Ksp = solubility product constant
Ksp = [Ag+][Cl-] = 1.8x10-10 Appendix H
當溶液解離達平衡後，反應物以及生成物離子的濃度皆不改
變，此離子濃度會達成一個固定的比例即稱溶解度積常數
4
Solubility Product Constants
• Consider the dissolution of silver sulfide in water.
Ag2 S(aq)
H2O
100%
2Ag+ + S2-
• The solubility product expression for Ag2S is:
Ksp = [Ag+]2[S2-] = 1.0x10-49
Ksp的數值愈大，表示溶液中離子的濃度也愈大；也就是水
溶解化合物時，化合物在水中分解成離子的數量也就愈多。
5
Solubility Product Constants
• The dissolution of solid calcium phosphate in
water is represented as:
Ca22+(PO43-)2(s)

H2O
100%
2Ca2+ + 2PO43-
The solubility product constant expression is:
Ksp = [Ca2+]3[PO43-]2 = 1.0x10-25
6
Solubility Product Constants
• In general, the dissolution of a slightly
soluble compound and its solubility product
expression are represented as:
MrYs(s)
H2O
100%
rMs+ + sYr-
Ksp = [Ms+]r[Yr-]s
溶解度的另一表示方法:
•The solubility of a compound molar solubility莫耳溶解度
The number of moles that dissolve to one liter of saturated
solution (M) 每公升飽和水溶液中所含溶質的莫耳數
7
Solubility Product Constants
• The same rules apply for compounds that
have more than two kinds of ions.
• One example of a compound that has more
than two kinds of ions is calcium ammonium
phosphate.
CaNH4PO4(s)
H2O
100%
Ca2+(aq) + NH4+(aq)+PO43-(aq)
Ksp = [Ca2+][NH4+][PO43-]
8
Determination of Solubility
Product Constants
Example 20-1: One liter of saturated silver chloride solution
contains 0.00192 g of dissolved AgCl at 25oC. Calculate the
molar solubility of, and Ksp for, AgCl.
The molar solubility can be easily calculated from the data:
? M AgCl=
0.00192g AgCl
143g AgCl
1.0L
AgCl(s)  Ag+ +
= 1.34x10-5 M
Cl-
1.34x10-5 M
1.34x10-5 M
1.34x10-5 M
Ksp = [Ag+][Cl-] = (1.34x10-5) x (1.34x10-5)
= 1.8x10-10 M
9
Determination of Solubility
Product Constants
Example 20-2: One liter of saturated calcium fluoride solution
contains 0.0167 gram of CaF2 at 25oC. Calculate the molar
solubility of, and Ksp for, CaF2.
? M CaF2=
0.0167g CaF2
78.1g
1.0L
CaF2  Ca2+ +
2.14x10-4 M
2.14x10-4
= 2.14x10-4 M
2F2x(2.14x10-4)
Ksp = [Ca2+][F-]2 = (2.14x10-4) x (4.28x10-4)2
= 3.92x10-11 mol/L
10
Uses of Solubility
Product Constants
Example 20-3: Calculate the molar solubility of barium sulfate,
BaSO4, in pure water and the concentration of barium and
sulfate ions in saturated barium sulfate at 25oC. For barium
sulfate, Ksp= 1.1 x 10-10.
BaSO4  Ba2+ +
xM
xM
SO42-
xM
Ksp = [Ba2+][SO42-] = 1.1x10-10
(x)(x) = 1.1x10-10
x= 1.0x10-5 M
[Ba2+]=[SO42-]= 1.0x10-5 M
?g BaSO4= 1.0x10-5M x 1L = 1.0x10-5 mol
= 1.0x10-5 x 234
= 2.34x10-3 g in 1L
11
Uses of Solubility
Product Constants
Example 20-4: The solubility product constant for magnesium
hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar
solubility of magnesium hydroxide and the pH of a saturated
magnesium hydroxide solution at 25oC.
Mg(OH)2  Mg2+ + 2 OHxM
xM
2x M
Ksp = [Mg2+][OH-]2 = 1.5x10-11
(x)(2x)2 = 1.5x10-11
4x3= 1.5x10-11 M
x3= 3.75x10-12 M
x= 1.6x10-4 M
[OH-]= 2x (1.6x10-4) =3.2x10-4 M
pOH=3.49, pH =10.51
12
The Common Ion Effect in
Solubility Calculations
Example 20-5: Calculate the molar solubility of barium sulfate,
BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC.
Compare this to the solubility of BaSO4 in pure water.
(Example 20-3). (What is the common ion? How was a
common ion problem solved in Chapter 19?)
Na2SO4  2 Na+ + SO42- Soluble ionic salt
0.010 M
2x(0.010)M
BaSO4  Ba2+ +
xM
xM
完全解離
0.010 M
SO42- Slightly Soluble salt
xM
溶解度很低
Ksp = [Ba2+][SO42-] = 1.1x10-10 = (x)(0.01+x)
0.01+x 0.01
(0.01)x = 1.1x10-10
x = 1.1x10-8
Molar solubility of BaSO4
13
The Common Ion Effect in
Solubility Calculations
• The molar solubility of BaSO4 in 0.010 M Na2SO4
solution is 1.1 x 10-8 M.
• The molar solubility of BaSO4 in pure water is 1.0
x 10-5 M.
– BaSO4 is 900 times more soluble in pure water
than in 0.010 M sodium sulfate!
– Adding sodium sulfate to a solution is a fantastic
method to remove Ba2+ ions from solution!
• If your drinking water were suspected to have
lead ions in it, suggest a method to prove or
disprove this suspicion.
14
The Reaction Quotient in
Precipitation Reactions
The reaction quotient, Q, and the Ksp of a compound are
used to calculate the concentration of ions in a
solution and whether or not a precipitate will form.
If Qsp< Ksp Forward process is favored (反應向右)
No precipitation occurs; if solid is present,
more solid can dissolve (不會產生沉澱)
If Qsp= Ksp Solution is just saturated (正好飽和)
Solid and solution are in equilibrium; neither
forward nor reverse process is favored
If Qsp> Ksp Reverse process is favored (反應向左)
precipitation occurs to form more solid (產生
沉澱)
15
The Reaction Quotient in
Precipitation Reactions
Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and
100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate
form?
100mL x 0.10M =0.050 M Pb(NO )
3 2
100mL+100mL
MK2SO4= 100mL x 0.10M =0.050 M K2SO4
100mL+100mL
K2SO4  2 K+ + SO420.05M  0.05x2M 0.05M
Will PbSO4 precipitate??
Pb(NO3)2  Pb2+ + 2 NO30.05M  0.05M 0.05x2M
MPb(NO3)2=
PbSO4  Pb2++ SO42-
Qsp = [Pb2+][SO42-] = (0.050)(0.050) = 2.5x10-4
Ksp= 1.8x10-8 for PbSO4
Qsp> Ksp therefore solid forms
16
The Reaction Quotient in
Precipitation Reactions
Example 20-7: Suppose we wish to remove mercury from an
aqueous solution that contains a soluble mercury compound
such as Hg(NO3)2. We can do this by precipitating mercury (II)
ions as the insoluble compound HgS. What concentration of
sulfide ions, from a soluble compound such as Na2S, is required
to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS,
Ksp=3.0 x 10-53.
HgS  Hg2++ S2Ksp = [Hg2+][S2-] = 3.0x10-53 = [S2-]x(1.0x10-8)
[S2-] = 3.0x10-45
If enough S2-, in th form of Na2S, is added to just
slightly exceed 3.0x10-45 M the mercury will
precipitate
17
The Reaction Quotient in
Precipitation Reactions
Example 20-8: Refer to example 20-7. What volume of the
solution (1.0 x 10-8 M Hg2+ ) contains 1.0 g of mercury?
?L =1.0g
Hg2+
2+
1
mol
Hg
x
201g Hg2+
x
1.0 L
1.0x10-8 mol Hg2+
?L =5.0x105 L
18
Fractional Precipitation
• The method of precipitating some ions from
solution while leaving others in solution is
called fractional precipitation 分級沉澱.
– If a solution contains Cu+, Ag+, and Au+, each ion
can be precipitated as chlorides.
CuCl  Cu++ ClAgCl  Ag++ Cl-
Ksp = [Cu+][Cl-]=1.9x10-7
Ksp = [Ag+][Cl-]=1.8x10-10
AuCl  Au++ Cl-
Ksp = [Au+][Cl-]=2.0x10-13
19
Example 20-9: If solid sodium chloride is slowly added to a
solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which
compound precipitates first? Calculate the concentration of
Cl- required to initiate precipitation of each of these metal
chlorides.
AuCl has the small Ksp, so it precipitates first
By the same reasoning, CuCl will precipitate last
•Calculate the concentration of Cl- required to
precipate AuCl
Ksp = [Au+][Cl-]=2.0x10-13 = (0.010)x[Cl-]
[Cl-]=2.0x10-11
•Repeat the calculation for silver chloride.
Ksp = [Ag+][Cl-]=1.8x10-10 = (0.010)x[Cl-]
[Cl-]=1.8x10-8
•Finally, for copper (I) chloride to precipitate.
Ksp = [Cu+][Cl-]=1.9x10-7 = (0.010)x[Cl-]
[Cl-]=1.9x10-5
20
Fractional Precipitation
• These three calculations give the [Cl-]
required to
– precipitate AuCl ([Cl-] >2.0 x 10-11 M),
– precipitate AgCl ([Cl-] >1.8 x 10-8 M),
– precipitate CuCl ([Cl-] >1.9 x 10-5 M).
• It is also possible to calculate the amount
of Au+ precipitated before the Ag+ begins
to precipitate, as well as the amounts of
Au+ and Ag+ precipitated before the Cu+
begins to precipitate.
21
Example 20-10: Calculate the percentage of Au+ ions that
precipitate before AgCl begins to precipitate.
– Use the [Cl-] from Example 20-9 to determine the [Au+]
remaining in solution just before AgCl begins to
precipitate.
Ksp = [Au+][Cl-]=2.0x10-13
[Au+]x(1.8x10-8)=2.0x10-13
[Au+]=1.1x10-5 unprecipitated
• The percent of Au+ ions unprecipitated just before
AuCl precipitates is:
+]
[Au
unprecipitated
% Au+unprecipitated=
x100%
[Au+]original
1.1x10-5
=
0.01 x100%
=0.1% unprecipitated
• Therefore, 99.9% of the Au+ ions precipitates before
AgCl begins to precipitate.
22
Fractional Precipitation
• A similar calculation for the concentration of Ag+ ions
unprecipitated before CuCl begins to precipitate is:
Ksp = [Ag+][Cl-]=1.8x10-10
[Ag+]x(1.8x10-8)=1.9x10-5
[Ag+]=9.5x10-6 unprecipitated
• The percent of Ag+ ions unprecipitated just before
AgCl precipitates is:
+]
[Ag
unprecipitated
% Ag+unprecipitated=
x100%
+
[Ag ]original
9.5x10-6
=
0.01 x100%
=0.095% unprecipitated
• Therefore, 99.905% of the Ag+ ions precipitates
before CuCl begins to precipitate.
23
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• Many weak acids and bases react with
many metal ions to form insoluble
compounds.許多弱酸和弱鹼可以和金屬反應形
成不可溶的化合物
– The weak acid or weak base equilibrium
as well as the solubility equilibrium
• Equlibria that simultaneously involve two or
more different equilibrium constant
expressions are simultaneous equilibria.
24
Simultaneous Equilibria Involving
Slightly Soluble Compounds
Example 20-12: If 0.10 mole of ammonia and 0.010 mole of
magnesium nitrate, Mg(NO3)2, are added to enough water to make
one liter of solution, will magnesium hydroxide precipitate from the
solution?
– For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5.
Calculate Qsp for Mg(OH)2 and compare it to Ksp.
– Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M.
– Aqueous ammonia is a weak base that we can calculate [OH-].
NH3 + H2O  NH4++ OH-
0.1M
-x
+x
+x
[NH4+] [OH-]
(x)(x)
-5
Kb=
= 1.8x10 =
[NH3]
(0.10-x)
x2=1.8x10-6
x=1.3x10-3
=[OH-]
Mg(NO3) 2  Mg2++ 2NO3- 可溶性鹽類
0.01M
+0.01M +0.01x2M
Qsp = [Mg2+][OH-]2 = (0.010)(1.3x10-3)2 = 1.7x10-8
Qsp> Ksp thus Mg(OH)2 will precipitate
25
Example 20-13: How many moles of solid ammonium chloride,
NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one
liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in
magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this
problem and Example 20-12.)
Calculate the maximum [OH-] that can exist in a solution that is 0.010
M in Mg2+.
Ksp = [Mg2+][OH-]2=1.5x10-11
[OH-]2x(0.010)=1.5x10-11
[OH-]  3.9x10-5 M
[OH-]2=1.5x10-9
Using the maximum [OH-] that can exist in solution, determine the
number of moles of NH4Cl required to buffer 0.10 M aqueous
ammonia so that the [OH-] does not exceed 3.9 x 10-5 M.
NH4Cl → NH4++ Cl-
xM
xM
NH3 + H2O 
xM
NH4++
OH-
Buffer solution
(0.1M)
(3.9x10-5)
[OH-]= Kb x [base] =1.8x10-5 x (x)
[salt]
(0.10)
Because there is 1.0L of solution
26
x= 0.046 =[NH4Cl] There are 0.046 mol NH4Cl
Simultaneous Equilibria Involving
Slightly Soluble Compounds
• Check these values by calculating Qsp for Mg(OH)2.
Qsp = [Mg2+][OH-]2 = (0.010)(3.9x10-5)2 = 1.5x10-11
Qsp = Ksp Thus this system is at equilibrium
• Use the ion product for water to calculate the [H+] and
the pH of the solution.
[H+][OH-] =1.0x10-14
[H+]x(3.9x10-5)=1.0x10-14
[H+] =2.6x10-10
pH=9.59
27
Dissolving Precipitates
• For an insoluble solid, if the ion concentrations (of
either the cation or anion) are decreased, a solid
precipitate can be dissolved.(在不可溶固體中,若任
一離子濃度降低,則此沉澱的固體會被再溶解)
– The trick is to make Qsp < Ksp.
• One method is to convert the ions into weak
electrolytes.採用的方法乃將離子轉成弱電解質
– Make these ions more water soluble. 讓此離子更水
溶
28
Dissolving Precipitates
• If insoluble metal hydroxides are dissolved in
strong acids, they form soluble salts and water. (
不可溶金屬氫氧化合物可溶於強酸而形成可溶性鹽類
及水)
• For example, look at the dissolution of Mg(OH)2
in HCl.
Mg(OH)2(s) + 2 HCl(aq)  MgCl2(aq) + 2H2O(l)
or
Mg(OH)2(s) + 2 H+(aq)  Mg2+(aq) + 2H2O(l)
Notice that the insoluble Mg(OH)2 is
converted to the more soluble MgCl2
29
Dissolving Precipitates
• A second method is to dissolve insoluble
metal carbonates in strong acids.(另一方法是
將不可溶金屬碳酸化合物溶於強酸中)
– The carbonates will form soluble salts, carbon
dioxide, and water.
CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + 2H2O(l)
or
CaCO3(s) + 2 H+(aq)  Ca2+(aq) + CO2(g) + 2H2O(l)
Notice that the formation of carbon dioxide
removes ions and makes the carbonate dissolve
30
Dissolving Precipitates
• A third method is to convert an ion to another
species by an oxidation-reduction reaction.(第三
種方法乃利用氧化還原反應改變離子態)
• For example, the dissolution of insoluble metal
sulfides in hot nitric acid causes the sulfide ions to
be oxidized to elemental sulfur.
3PbS(s) + 8H+ + 2NO3- 3Pb2+ + 3S0(s) +2NO(g) + 4H2O(l)
[Pb2+][S2-]=8.4x10-28=Ksp
The NO3- oxidizes S2- to S0 making the Qsp<Ksp
31
Dissolving Precipitates
• A fourth method is complex ion formation.(第四種
方法是形成錯離子)
• The cations in many slightly soluble compounds
will form complex ions.
– This is the method used to dissolve unreacted AgBr
and AgCl on photographic film.
• Photographic “hypo” is Na2S2O3.
h
2AgBr(s)  2Ag(s) + Br2(l)
In the unexposed portion of the film AgBr(s) is left
AgBr(s) + 2 Na2S3O2(aq)  Na3Ag(S2O3)2 (aq) + NaBr(aq)
AgBr(s) + 2 Na2S32-  [Ag(S2O3)2]3-(aq) + Br-(aq)
32
Dissolving Precipitates
• Copper(II) hydroxide, which is light blue colored,
dissolves in aqueous ammonia to form dark blue,
[Cu(NH3)4]2+.
Cu(OH)2(s) + 4NH3(aq)  [Cu(NH3)4]2+(aq) + 2OHCu(OH)2(s)  Cu2+(aq) + 2OH-(aq)
Cu2+(aq) + 4NH3  [Cu(NH3)4]2+(aq)
33
Complex Ion Equilibria
• A metal ion coordinated to several neutral
molecules or anions forms compounds
called complex ions.
• Familiar examples of complex ions include:
[Ag(NH3)2]+
[Cu(NH3)4]2+
Co(NH3)2Cl2
[Pt(NH3)2]+
34
Complex Ion Equilibria
• The dissociation of complex ions can be
represented similarly to equilibria.
• For example:
[Cu(NH3)4]2+  Cu2+ + 4NH3
[Cu2+] [NH3]4
Kd=
[[Cu(NH3)4]2+]
[Ag(NH3)2]+  Ag+ + 2NH3
[Ag+] [NH3]2
Kd=
[[Ag(NH3)2]+]
Complex ion equilibrium constants are called
dissociation constants Kd.
35
Complex Ion Equilibria
Example 20-14: Calculate the concentration of silver ions in a
solution that is 0.010 M in [Ag(NH3)2]+.
Kd = 6.3 x 10-8
Write the dissociation reaction and equilibrium concentrations.
[Ag(NH3)2]+  Ag+ + 2NH3
(0.01-x)M (x)M
(2x)M
Substitute the algebraic quantities into the dissociation expression.
[Ag+] [NH3]2
Kd=
[[Ag(NH3)2]+]
= 6.3x10-8
4x3= 6.3x10-10
x3= 1.6x10-10
(x)(2x)2
=
(0.010-x)
x= 5.4x10-4 M
36
Example 20-15: How many moles of ammonia must be added to
2.00 L of water so that it will just dissolve 0.010 mole of silver
chloride, AgCl?
? M AgCl = 0.01mol/2.0L = 5x10-3 M
AgCl(s) + 2NH3  [Ag(NH3)2]++ Cl5x10-3 M 2(5x10-3) M
5x10-3 M
5x10-3 M
AgCl  Ag+ + ClKsp=[Ag+][ Cl-] =1.8x10-10
+] [NH ]2
[Ag
3
-10
[Ag(NH3)2]+  Ag+ + 2NH3
Kd=
=6.3x10
[[Ag(NH3)2]+]
The [Ag+] in the solution must satisfy both equilibrium constant
expressions. Because the [Cl-] is known, the equilibrium
concentration of Ag+ can be calculated from Ksp for AgCl.
Ksp=[Ag+][ Cl-] =1.8x10-10 =[Ag+](5x10-3)
Which is the maximum [Ag+] possible
[Ag+] =3.6x10-8
-8)[NH ]2
(3.6x10
[Ag+] [NH3]2
3
-10
=
=6.3x10
Kd=
+
(5x10-3)
[[Ag(NH3)2] ]
[NH3]= 0.094 M
[NH3]2= 8.75x10-3
[NH3]total= 0.094 M + 2x(5.0x10-3) =0.104M
? mol NH3= 2.0L x 0.104M = 0.21
37