Solubility Product Constant
A special case of equilibrium
involving dissolving.
Solid  Positive Ion + Negative Ion
Mg(NO3)2  Mg2+ + 2NO3Keq = [Mg2+ ] [NO3-]2
Because the constant is a product of
solubility, we call it the solubility
product constant Ksp
Solubility Product Problems
•Given Ksp, find solubility
•Given solubility, find Ksp
•Find solubility in a solution with a
common ion
•Predicting precipitation
Solubility Product Constant (Ksp)
BaF2(S)  Ba+2(aq) + 2F-(aq)
Write out the equilibrium law expression…
Ksp = [Ba+2][F-]2
Solubility Generalizations
•All nitrates are soluble
•All compounds of the alkali metals
are soluble (Li, Na, K, etc.)
•All compounds of the ammonium
(NH4+) are soluble
Given Ksp, Find Solubility
What is the solubility of Silver Bromide
(Ksp = 5.2 x 10-13)
AgBr  Ag+ + BrKsp = [Ag+][Br-] = 5.2 x 10-13
Let x = the solubility
(x)(x) = 5.2 x 10-13
X2
= 5.2 x
10-13
X = 7.2 x 10-7
Another Example
What is the solubility of PbI2 (Ksp =
7.1 x 10-9)
PbI2(s)  Pb+2 + 2IKsp = [Pb+2][I-]2 = 7.1 x 10-9
Let x = the solubility
(x)(2x)2 = 7.1 x 10-9
4x3 = 7.1 x 10-9
(x)(4x2) = 7.1 x 10-9
X = 1.21x10-3
Find Ksp Given Solubility
What is the Ksp of Boric Acid, given its
solubility of 2.15 x 10-3 Moles/liter?
H3BO3  3H+ + BO3-3
Ksp = [H+]3[BO3-3]
[3(2.15x10-3)]3 [2.15x10-3]
= 5.8 x 10-10
Solubility with a Common Ion
What is the solubility of lead iodide
(PbI2) in a .15M solution of KI ?
PbI2  Pb+2 + 2IKI  K+ + IKsp = [Pb+2][I-]2 = 7.1 x 10-9
Solubility with a Common Ion
What is the solubility of lead iodide
(PbI2) in a .15M solution of KI ?
PbI2  Pb+2 + 2IKsp = [Pb+2][I-]2
KI  K+ + I-
Let x = solubility
Then: [Pb+] = x
[I-] = .15+2x
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Let x = solubility of PbI2 in the
solution
Then, [Pb+2] = x
[I-] = .15 + 2x
Ksp = (x)(.15+2x)2 = 7.1 x 10-9
Ksp = [Pb+2][I-]2 = 7.1 • 10-9
Ksp = (x)(.15+2x)2 = 7.1 • 10-9
(x)(4x2 + .6x +.152) = 7.1 • 10-9
4x3 + .6x2 + .152x = 7.1 • 10-9
4x3 + .6x2 + .0225 – 7.1 • 10-9 = 0
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Ksp = (x)(.15+2x)2 = 7.1 x 10-9
Assume .15>>2x
then, .15+2x  .15
Bill Gates is the
Richest Man in the
World.
Woopie! It’s my lucky day!
He is worth about…
$34 Billion Dollars
Now, suppose one day
he finds a penny.
$34,000,000,000.01 $34 Billion
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Ksp = (x)(.15+2x)2 = 7.1 x 10-9
Assume .15>>2x
then, .15+2x  .15
Ksp = (x)(.15)2 = 7.1 x 10-9
X = 3.16 x 10-7 moles/liters
Predicting Precipitation
A student mixes 0.010 mole Ca(NO3)2 in
2 liters of 0.10M Na2C03 solution. Will a
precipitate form?
Step 1: Write out the dissolving equations
Ca(NO3)2  Ca2+ + 2NO3Na2CO3  2Na+ + CO32-
Step 2: Determine the most likely
precipitate & write out it’s equation.
Ca(NO3)2  Ca2+ + 2NO3-
Na2CO3  2Na+ + CO32-
Recall the solubility generalizations…
•All nitrates are soluble
•All compounds of the alkali metals
are soluble (Li, Na, K, etc.)
•All compounds of the ammonium
(NH4+) are soluble
Step 2: Determine the most likely
precipitate & write out it’s equation.
Ca(NO3)2  Ca2+ + 2NO3-
Na2CO3  2Na+ + CO32CaCO3  Ca2+ + CO32-
CaCO3  Ca2+ + CO32-
Ksp = [Ca2+ ][CO32-] = 4.7 x 10-9
Step 3: Determine the molar
concentrations & calculate the reaction
quotient (Q).
[Ca2+] = .01 mole/2 liters = .005M
[CO32-] = 0.10 M (given)
Reaction Quotient (Q)
The product of the Ksp equation
using the ion concentration before any
reaction interaction.
If Q > Ksp Then a precipitate will form.
Q = [Ca2+ ][CO32-]
[Ca2+] = .01mole/2 liters = .005M
[CO32-] = 0.10 M (given)
Q = (.005)(0.10) = .0005
Q > Ksp  a precipitate will
form.
You try one….
.015 moles of AgNO3 is
mixed with 5 liters of .02M NaCl
solution. What is the most likely
precipitate and will it form?
You try one….
.015 moles of AgNO3 is mixed with 5 liters of
.02M NaCl solution. What is the most likely
precipitate and will it form?
AgNO3  Ag+ + NO3-
NaCl
 Na+ + Cl-
AgCl  Ag+ + Cl-
AgCl  Ag+ + Cl-
Q = [Ag+][Cl-]
Q = (.015/5)(.02) = .00006
Look up the Ksp
(1.0 x 10-10)
Q > Ksp So a precipitate will form!