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maka pOH = 14 -9 = 5
berarti konsentrasi OH- = 10⁻⁵
X(OH)₂ <-----> 1 X²⁺ + 2OH⁻
s s 2s
Ksp X(OH)₂ = [X²⁺ ]¹.[OH⁻]²
Ksp X(OH)₂ = s¹. (2s)² = 4s³
[OH⁻] = 2 s = 10⁻⁵
maka s = (1/2). 10⁻⁵ = 5 .10⁻⁶
berarti Ksp X(OH)₂ = 4 (5 .10⁻⁶)³ = 4 . 125 . 10⁻¹⁸ = 500. 10⁻¹⁸ = 5 .10⁻¹⁶