Solubility Product Constant
6-5
Ksp
• is a variation on the equilibrium constant
for a solute-solution equilibrium.
• remember that the solubility equilibrium is
based on a saturated solution.
• Solids are not written in the equilibrium
expression.
CaF2 (s)
Ca 2+ (aq) + 2F - (aq)
Ksp = [Ca 2+] [F -]2
Where Ksp is the solubility product
constant, or simply the solubility product.
CaCl2 (s)
Ca 2+ (aq) + 2 Cl - (aq)
Ksp = [Ca 2+] [Cl -]2
• Given the solubility, calculate the
Ksp of the substance.
1).Write the chemical equation for the
substance dissolving and
dissociating.
2) Write the Ksp expression.
3) Insert the concentration of each ion
and multiply out.
Calculating the solubility product:
CuBr has a measured solubility of 2.0 10-4 M
at 25 C. Calculate the Ksp value.
Cu + (aq) + Br - (aq)
CuBr (s)
Ksp
= [Cu +] [Br -]
= (2.0 10-4 )(2.0 10-4 )
= 4.0 10-8
EX: Calculate the Ksp for Bi2 S3 , which
has a solubility of 1.36 x 10-15 mol/L at
25°C.
x
Bi2S3 (s)
x
2 Bi3+ (aq)
+ 3 S2- (aq)
2x
2(1.36 x 10-15 )
2.72 x 10-15
3(1.36 x 10-15)
4.08 x 10-15
Ksp =
[Bi 3+]2 [S 2-]3
=
(2.72 x 10-15)2 (4.08 x 10-15)3
=
5.02 x 10-73
• Determine the Ksp of calcium
fluoride (CaF2 ), given that its molar
solubility is 2.14 x 10-2 M.
• CaF2 (s) → Ca2+ (aq) + 2 F¯ (aq)
2x
x
•
• The Ksp expression is:
Ksp = [Ca2+ ] [F¯]2
Ksp = [Ca2+ ] [F¯]2
• Ksp = (2.14 x 10-2 ) (4.28 x 10-2 )2
• Ksp = 3.92 x 10-5
Convert g/l to mol/l
The solubility of Fe(OH)2 is found to be 1.4 x 10-3
g/l. What is the Ksp value?
Change the g/l to mol/l.
Molar mass of Fe(OH)2 is 89.87g/mol
Solubility is
= 1.56 x 10-5 mol/l
Fe(OH)2(s) → Fe2+(aq) + 2OH-(aq)
x
x
2x
Ksp = [Fe2+ ][OH- ]2
=(1.56 x 10-5 ) (2[1.56 x 10-5 ]) 2
=1.5 x 10-14
• Calculate the Ksp of
silver bromide: ( AgBr ), if the
solubility is 1.3 x 10 -4 g / L .
•
•
•
•
We can also use a known value of Ksp to
find out the solubility of a substance.
Write out the chemical equation for the
substance
Write the Ksp expression
Solve for x
• If the Ksp is 1.8 x 10 -8 of silver
chloride, find the solubility.
• write Ksp expression for silver
chloride.
• AgCl (s) ------> Ag+ (aq) + Cl- (aq)
•
x
x
• Ksp = [Ag + (aq) ] [Cl - (aq)]
• 1.8 x 10 -8
•
1.8x10
= x2
8
• 1.3 x 10-4 = x
x
• The Ksp value for Cu(IO3)2 is
1.4 x 10-7 at 25°C. Calculate its
solubility at this temperature.
Cu(IO3)2 (s)
x
Cu2+ (aq) + 2 IO31- (aq)
x
2x
•
3
Ksp
=
[Cu 2+] [IO3 1-]2
• 1.47 x 10-7 =
(x) (2x)2
• 1.47 x 10-7 =
4x3
• 1.47 x 10-7 =
4
4x3
4
3.67 x10
8
x
3.3 x 10-3 = x
Solubility of Cu(IO3)2
• If Ksp of calcium hydroxide is 5.5 x 10-6,
find the solubility.
Ca(OH)2 (s)→ Ca+2 (aq) + 2OH-(aq)
x=0.011 M