The ksp of lead(ii) hydroxide, pb(oh)2, is 1.43 × 10-20. calculate the molar solubility of this compound.

When this compound is dissolved in water, the dissociation reaction would be:

Pb(OH)₂ --> Pb²⁺ + 2 OH⁻

Then, using the ICE approach, let s be the molar solubility.

                Pb(OH)₂ --> Pb²⁺ + 2 OH⁻
I                    s              0            0
C                 -s              +s         +2s
E                  0               s            2s

The expression for Kp is:

Kp = [Pb²⁺][OH⁻]² = 1.43×10⁻²⁰
1.43×10⁻²⁰ = (s)(2s)²
Solving for s,
s = 1.53×10⁻⁷ M

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