Chapter 16
SOLUBILITY EQUILIBRIA
Saturated solutions of “insoluble” salts are another type of chemical equilibria. Ionic compounds that are
termed “insoluble” actually have a low, but definite solubility.
For a saturated solution of AgCl, the equation would be: AgCl(s) ⇄ Ag+(aq) + Cl−(aq)
The solubility product expression would be: Ksp = [Ag+][Cl−]
The AgCl(s) is left out since solids are left out of equilibrium expressions (constant concentrations).
For Ag2CO3, Ag2CO3 ⇄ 2Ag+ + CO32− Ksp = [Ag+]2[CO32−]
Ksp = solubility product constant
The Ksp of a substance is a measure of its solubility.
Ksp values cannot be directly compared if the number of ions in the formula differs. Smaller Ksp values
indicate lower solubility.
The Ksp of AgCl is 1.6×10−10. This means that if the product of [Ag+][Cl−] is < 1.6×10−10, the solution is
unsaturated and no solid would be present. If the product = 1.6×10−10, the product is exactly saturated and
no solid would be present. If the product is > 1.6×10−10, the solution is saturated and a solid (precipitate)
would form. The product of the ions (raised to the power of their coefficients) is called the ion product
quotient or Q.
If Q > Ksp, ppt forms.
If Q < Ksp, no ppt forms.
4 Main Types of Ksp Problems
1. Given molar solubility, find Ksp.
Ex. The molar solubility of silver sulfate is 1.5×10−2 mol/L. Calculate the solubility product of the
salt.
R
Ag2SO4(s)  2Ag+ + SO42−
Remember that molar
I
---0
0
−2
solubility is “x”!
C
−x
+2x
+x
x = 1.5×10
E
---3.0×10−2 1.5×10−2
Since 1.5×10−2 mol/L of Ag2SO4 dissolve, 1.5×10−2 mol/L of SO42- form and 2(1.5×10−2 mol/L) of Ag+
form.
Ksp = [Ag+]2[SO42−] = (3.0×10−2)2(1.5×10−2)= 1.4×10−5
2. Given Ksp, find molar solubility.
Ex. Calculate the molar solubility of calcium phosphate. The Ksp of calcium phosphate is 1.2×10−26.
R
Ca3(PO4)2  3Ca2+ + 2PO43−
Remember that molar
I
--0
0
C
−x
+3x
+2x
solubility is “x”!
E
--3x
2x
Ksp = [Ca2+]3[PO43−]2
1.2×10−26 = (3x)3(2x)2 = 108x5
x5 = 1.1×10−28
−6
x = 2.6×10 M = molar solubility
3. Given Ksp, find molar solubility in the presence of a common ion.
Ex. What is the molar solubility of lead(II) iodide in a 0.050 M solution of sodium iodide?
(Common ion effect problem)
R PbI2  Pb2+ + 2I−
I
---0
0.050M
Don’t forget to put in the initial
C
−x
+x
+2x
concentration of the common
E
---x
0.050 + 2x
ion!
2+
− 2
Ksp = [Pb ][I ]
Ksp = 1.4×10−8 = (x)(0.050+2x)2  x(0.050)2
x = 5.6×10−6M
The molar solubility of PbI2 in pure water is 1.5×10−3M. This shows the decreased solubility of a salt in
the presence of a common ion.
4. Will a precipitate form when these two solutions are mixed?
Ex. Exactly 200 mL of 0.040 M BaCl2 are added to exactly 600 mL of 0.080 M K2SO4. Will a
precipitate form? The Ksp of BaSO4 is 1.1×10−10.
BaCl2 + K2SO4  BaSO4 + 2KCl
Barium sulfate is the likely precipitate.
0.200 L × 0.040 M BaCl2 = 8.0×10−3 mol Ba2+
8.0×10−3 mol/0.800L total volume = 1.0×10−2 M Ba2+
0.600 L × 0.080 M K2SO4 = 4.8×10−2 mol SO42−
4.8×10−2 mol/0.800L = 6.0×10−2M SO42−
Ksp = [Ba2+][SO42−] = 1.1×10−10
Q = (1.0×10−2)(6.0×10−2) = 6.0×10−4
Q > Ksp 6.0×10−4 > 1.1×10−10
A precipitate of BaSO4 forms.
Shortcut Alert!
In a problem involving dilution (like the previous problem), the volumes are often equal. When
the volumes are equal and then combined, concentrations are halved.
For example, if 400 mL of each solution in the previous problem were combined, the resulting
concentrations would be:
[Ba2+]= 0.0400/2 = 0.0200 M and
[SO42] = 0.0800/2 = 0.0400 M