Transcript Document

Solubility
Lesson 3
Calculating
Ksp
The Molar Solubility is the molarity required to saturate or fill the solution at any
given temperature.
1.
The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the
solubility product or Ksp.
The Molar Solubility is the molarity required to saturate of fill the solution at any
given temperature.
1.
The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the
solubility product or Ksp.
BaCO3(s) ⇌ Ba2+ + CO32s
s
s
Ksp = [Ba2+][CO32-]
Ba2+
CO32-
BaCO3(s)
Ksp = [s][s]
Ksp =
s2
Ksp =
(5.1 x 10-5)2
Ksp =
2.6 x 10-9
Ksp
Solubility Product
Saturated solutions- at equilibrium
No Units
Increasing Temperature increases the Ksp
2.
The solubility of PbBr2 is 0.012 M @ 25 0C. Calculate the Ksp.
⇌
dissociation equation
PbBr2(s)
solubility
s
equilibrium expression
Ksp = [Pb2+][Br-]2
substitute & solve
Ksp = [s][2s]2
Pb2+
s
Ksp = 4s3
Ksp = 4(0.012)3
Ksp = 6.9 x 10-6
Note that the Br- is doubled and then squared!
+
2Br2s
3.
If 0.00243 g of Fe2(CO3)3 is required to saturate 100.0 mL of solution.
What is the solubility product?
Fe2(CO3)3
⇌
s
s
=
2Fe3+
2s
0.00243 g x 1 mole
291.6 g
0.100 L
+
3CO323s
Ksp = [Fe3+]2[CO32-]3
Ksp = [2s]2[3s]3
Ksp = 108s5
=
8.333 x 10-5 M
Ksp = 108(8.333 x 10-5)5
Ksp = 4.34 x 10-19
4.
A 200.0 mL sample of a saturated solution of Mg(OH)2 weighs 222.1210 g.
When the beaker containing the solution is evaporated to dryness it weighs
22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp.
Mass of Beaker + Mg(OH)
-Mass of Beaker
Mass of Mg(OH)2
22.1213 g
-22.1200 g
0.0013 g
note sig figs- 4th decimal
Mg(OH)2⇌ Mg2+ +
s
s
=
0.0013 g x 1 mole
58.3 g
0.2000 L
=
1.1149 x 10-4 M
Ksp
s
2OH2s
= [Mg2+][OH-]2
= [s][2s]2 = 4s3
=
4(1.1149 x 10-4)3
=
5.5 x 10-12
5.
40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL
of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.
Titration
2HCl
+
0.02910 L
0.300 M
[Ba(OH)2]
=
1Ba(OH)2
0.0400 L
?M
0.02910 L HCl x 0.300 moles
1L
0.0400 L
s
=
0.1091 M
x
1 mole Ba(OH)2
2 moles HCl
Ksp
⇌
Ba(OH)2
s
Ba2+
s
Ksp
= [Ba2+][OH-]2
= [s][2s]2
= 4s3
= 4(0.1091)3
= 5.20 x 10-3
+
2OH2s