apabila ksp Cu(OH)2 = 1,08×10^-19. Berapakah ph larutan jenuh Cu(OH)?
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Pertanyaan
apabila ksp Cu(OH)2 = 1,08×10^-19. Berapakah ph larutan jenuh Cu(OH)?
1 Jawaban
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1. Jawaban 93firda
Cu(OH)2 >> Cu^2+ + 2OH^-
Ksp ~ (s) (2s)^2
1.08x10^-19 = 4s^3
s = 1.39x10^-6
pOH = -log [OH]
pOH = -log 1.39x10^-6
pOH = 5.856
pH = 14 - pOH
pH = 14 - 5.856
pH = 8.143