Question

PH larutan jenuh Pb(OH)₂ pada T°C adalah 9 - log 2. hitunglah Ksp Pb(OH)₂ pada T°C

Answer 1

PH = 9- log 2

[H⁺] = 2 x 10⁻⁹

10⁻¹⁴ = [OH⁻] [H⁺]
10⁻¹⁴ = [OH⁻] [2 x 10⁻⁹]
[OH⁻] = 10⁻¹⁴/2 x 10⁻⁹
[OH⁻] = 5 x 10⁻⁶

Pb(OH)₂ → Pb²⁺ + 2OH⁻
   s                s          2s

[OH⁻] = 2s
5 x 10⁻⁶ = 2s
s = 2,5 x 10⁻⁶

Ksp = [Pb²⁺] [OH⁻]²
Ksp = [s] [2s]²
Ksp = 4s³
Ksp = 4 (2.5 x 10⁻)²
Ksp = 4 (1,5625 x 10⁻¹⁷)
Ksp = 6,25 x 19⁻¹⁷

Jadi, Ksp Pb(OH)₂pada suhu T⁰C adalah 6,25 x 19⁻¹⁷