Question

Harga pH larutan jenuh Pb(OH)2 sekitar .... (Ksp Pb(OH)2 = 2,57.10-11)


A. 3
B. 5
C. 10
D. 11
E. 13

Answer 1

Pb(OH)2 => Pb²⁺     +  2 OH-
   s               s                2s
Ksp Pb(OH)2 = [pB²⁺] [OH-]²
2,57 x10⁻¹¹ = [s] [2s]²
4s³ = 2,57 x10⁻¹¹
s³ = 6,425 x10⁻¹²
s = 1,859 x10⁻⁴ mol/liter
[OH-]=2s = 2x1,859 x10⁻⁴ = 3,718 x10⁻⁴
pOH = - log 3,718 x10⁻⁴
pOH = 4 - log 3,718

pH = 14 - pOH
pH = 14 - (4 - log 3,718)
pH = 10 + log 3,718
pH = 10 + 0,57 = 10,57