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jika [L²⁺] = s
maka [OH⁻] = 2s
Ksp = [L²⁺][OH-]²
= 4s²
5x10⁻⁴ = 4s²
s = 10⁻² (hasil pembulatan)
[OH⁻] = 2x10⁻²
pOH = 2-log2
pH = 12-log2
L(OH)2 (s) --> L^2+ (aq) + 2OH^- (aq)
s s 2s
OH^- = 2s
= 2 x 5x10^-4
= 10 x 10^-4
= 1 x10^-3
pOH = - log 1 x 10^-3
= 3 - log 1
= 3
pH = 14 - 3 = 11 (B)