larutan jenuh m (oh)2 mempunyai ph : 10+log2 hasil kali kelarutan m (oh)2 adalah

M(oh)2 --> M2+ + 2OH-
PH=10+log2
POH = 4-log2
[OH-] = 2.10^-4
2 s = 2.10^-4
S = 10^-4
Ksp = [M2+] [OH-]^2
= S (2S)^2
= 4 S^3
= 4 ( 10^-4)^3
= 4.10^-12

pH = 10 + log 2
pOH = 4 - log 2
OH⁻ = 2 . 10⁻⁴

Ksp M(OH)₂ = [M²⁺] [2 OH⁻]²
Ksp M(OH)₂ = [10⁻⁴] [2 . 10⁻⁴]²
Ksp M(OH)₂ = 2 . 10⁻¹² ⇒ Jawab