Ionic EquilibriumHard

Question

What is the pH of 4 × 10–3 M–Y(OH)2 solution assuming the first dissociation to be 100% and second dissociation to be 50%, where Y represents a metal cation? (log 2 = 0.3, log 3 = 0.48)

Options

A.11.78
B.11.22
C.2.22
D.2.78

Solution

Y(OH)2 $\rightleftharpoons$ Y(OH)+ + OH–

4 × 10–3 M 0 0

Final 0 4 × 10–3 M 4 × 10–3 M

Y(OH)+ $\rightleftharpoons$ Y2+ + OH–

4 × 10–3 –2 × 10–3 2 × 10–3 M 4 × 10–3 +2 × 10–3

= 2 × 10–3 M = 6 × 10–3 M

∴ POH = –log (6 × 10–3) = 2.22 ⇒ PH = 11.78

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