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