jika ksp X(OH)2 adalah 4×10pangkat 12 maka pH larutan jenuh X(OH)2 adalah...

X(OH)2 —> X+ + 2OH-
Misalkan kelarutan X(OH)2 = a M
Mk [X+] = a M dan [OH-] = 2a M
Ksp = [X+] [OH-]²
4×10⁻¹² = a (2a)²
4×10⁻¹² = 4a³
a = 10⁻⁴ M
[OH-] = 2a = 2×10⁻⁴ M
pOH = - log 2×10⁻⁴ = 4 - log 2
pH = 14 - (4-log 2) = 10 + log 2 = 10,3