Question

1) Calculate the metal cation concentration of the following sparingly soluble salts in a 0.0167 M...

1) Calculate the metal cation concentration of the following sparingly soluble salts in a 0.0167 M solution of Mg(ClO4)2 using activities:

(a) SnI2 (Ksp = 8.3 x 10¯6)

(b) PbSO4 (Ksp = 6.3 x 10¯7)

Homework Answers

Answer #1

From the given information, reaction occur in 1st case

Mg(ClO4)2 + SnI2 ------------> Sn(ClO4)2 + MgI2

Mg(ClO4)2 = 0.0167 M

Ksp = 8.3 X 10-6

0.0167 M Mg(ClO4)2

Ksp of Mg(ClO4)2= 49.6 from solublity table

(ClO4)2 = 0.0167 / 49.6 = 3.37 X 10-4 M

Mg2+ = 0.0167 / 49.6 = 3.37 X 10-4 M
Ionic product of Sn(ClO4)2 = [Sn2+ ][(ClO4)2]

Ksp = [Sn2+ ][(ClO4)2]
8.3 x 10-6 = [Sn2+ ] X 3.37 X 10-4

[Sn2+ ] = 8.3 x 10-6/3.37 X 10-4

[Sn2+ ] = 2.46 X 10-2 M

For Second solution PbSO4 (Ksp = 6.3 x 10-7)

Mg(ClO4)2 + PbSO4 ------------> Pb(ClO4)2 + MgSO4

Ionic product of Sn(ClO4)2 = [Pb2+ ][(ClO4)2]

Ksp = [Pb2+][(ClO4)2]
6.3 x 10-7 = [Pb2+] X 3.37 X 10-4

[Pb2+] = 6.3 x 10-7/3.37 X 10-4

[Pb2+] = 0.0187 X 10-3 M

[Pb2+] = 1.87 X 10-5 M

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