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= 0,0005 M
Hidrolisis:
[OH⁻]² = Kw.M/Ka
= 10⁻¹⁴ x 5 x 10⁻⁴/1,7x10⁻⁵
= 29,4 x10⁻¹⁴
[OH⁻] = 5,42 x 10⁻⁷
pOH = -log 5,42 x 10⁻⁷
= 7 - log 5,42
pH = 14 - ( 7 - log 5,42)
= 7 + log 5,42