Question

Jika ph larutan jenuh L(OH)2 dalam air adalah 8,maka ksp garam tersebut adalah...

Answer 1

Ph = 8
Ph = 14- POH
POH = 14-8
POH = 6
POH = -log[OH⁻]
6    = -log[OH⁻]
[OH⁻] = 10⁻⁶
[OH-] = n.M
10⁻⁶ = 2.M
M = 5 x 10⁻⁷

L(OH)₂      ->       L²⁺    +      2 OH
     s                     s               s
5 x 10⁻⁷              5 x 10⁻⁷      2(5 x 10⁻⁷) = 10⁻⁶

Ksp L(OH)₂ = [L²⁺]  [OH⁻]²
                  = (5 x 10⁻⁷) (10⁻⁶)²
                  = 5 x 10⁻⁷ (10⁻¹²)
                  = 5 x 10⁻¹⁹

Answer 2

PH = 8
pOH = 6
OH- = 10^-6

L(OH)2 ⇒ L2+ + 2 OH-
 5 x 10^-7  5x10^-7   10^-6

Ksp = (L2+)(OH-)² = (5x10^-7)(10^-12) = 5x10^-19