suatu larutan jenuh Mg(OH)2 mempunyai Ksp = 6x10-12 tentukan harga PH saat mulai terbentuk endapan Mg(OH)2

Mg(OH)2 —> Mg2+ + 2OH-
Mis kelarutan Mg(OH)2 = a mol/liter mk
[Mg2+] = a M
[OH-] = 2a M
Ksp = [Mg2+] [OH-]²
6.10^-12 = a (2a)²
6.10^-12 = 4a³
a³ = 1,5.10^-12
a = 1,14.10^-4 M
[OH-] = 2a = 2,28.10^-4 M
pOH = - log (2,28.10^-4) = 3,64
pH = 14 - 3,64 = 10,36