Ksp – Solubility
Product
Chapter 16
Ksp - background
The equilibrium between solids and ions is different from the
equilibrium between gases
The equilibrium between solids and ions is a “phase” equilibrium
(e.g. NaCl(aq))
Na+(aq) + Cl-(aq)
NaCl(s)
Ksp deals with a phase equilibrium: (s)  (aq)
Why use Ksp instead of Kc?
Consider: NaCl  Na+(aq) + Cl-(aq)
[NaCl(s)] doesn’t
[Na+(aq)] [Cl-(aq)]
K=
change (it’s a konstant)
[NaCl(s)]
K • [NaCl(s)] = [Na+(aq)] [Cl-(aq)]
Ksp = [Na+(aq)] [Cl-(aq)]
We use Ksp because it
indicates that the
+
Na (aq) + Cl (aq)
equilibrium is a phase
equilibrium (between a
solid and it’s ions)
NaCl(s)
Ksp - molar solubility
 A solid always dissolves until no more can dissolve - called
molar solubility , s, (mol/L or M)
 An equilibrium is established when the amount dissolving
equals the amount precipitating
 This can only be true if there is some solid (thus, we can
usually see if there is an equilibrium)
 A solution with solid remaining is called “saturated”…this
means that the solution is at equilibrium
Relative solubilities
• Ksp will only allow us to compare the solubility of solids that
fall apart into the same number of ions.
• The bigger the Ksp of those the more soluble.
Relative Solubilities
AgI
Ksp = 1.5 x 10-16
CuI
Ksp = 5.0 x 10-12
CaSO4 Ksp = 6.1 x 10-5
All 3 dissociate into two ions, therefore:
Ksp = [cation][anion] = x2
X = √Ksp = solubility
CaSO4 > CuI > AgI
Most soluble
Least soluble
Relative Solubilities
If they fall apart into different number of pieces you have to do
the math.
CuS
Ag2S
Bi2S3
Ksp = 8.5 x 10-45
Ksp = 1.6 x 10-49
Ksp = 1.1 x 10-73
Dissociate each to write a Ksp expression for each and solve for
solubility, x.
Relative Solubilities
CuS
Cu2+ + S2x = 9.2 x 10-23
Ksp = 8.5 x 10-45
Ag2S
2Ag+ + S2x = 3.4 x 10-17
Ksp = 1.6 x 10-49 = (2x)2x
Bi2S3
2Bi3+ + 3S2x = 1.0 x 10-15
Ksp = 1.1 x 10-73 = (2x)2(3x)3
Bi2S3 > Ag2S > CuS
Most Soluble
Least Soluble
= x2
Example #1- If 1.0L of water dissolves 7.1 x 10
-7
mol of AgBr, what is the Ksp?
Step 1: write the balanced chemical equation based on your
knowledge of valences:
AgBr(s)  Ag+(aq) + Br–(aq)
Step 2: solve problem using RICE chart
R
I
C
E
AgBr(s)
Ag+(aq)
0
+7.1x10–7
7.1x10–7
Ksp = [Ag+][Br –]
Ksp = [7.1x10–7][7.1x10–7] = 5.0 x 10–13
Br –(aq)
0
+7.1x10–7
7.1x10–7
Example #2 – The molar solubility of PbCl is 1.7 x 10 M in
2
-3
a 0.10M NaCl solution. What is the Ksp?
1: write balanced chemical equations:
NaCl(s)  Na+(aq) + Cl–(aq) ([Cl–] = 0.10 M)
PbCl2(s)  Pb2+(aq) + 2Cl–(aq)
2: RICE Chart
R
I
C
E
PbCl2(s)
Pb2+(aq)
0
+1.7x10-3
1.7x10-3
2Cl–(aq)
0.10
+2(1.7x10-3)
0.1034
Ksp = [Pb2+(aq)][Cl–(aq)]2
Ksp = [1.7x10-3][0.1034]2 = 1.7 x 10–5
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