hitung Ph dari A.CH3COOH 0,001 M (ka= 4 x 10^-5) B.NH4OH 0,01 M (Kb= 9x10^4)

A. CH3COOH
[OH-] = √kb . M
√4 x 10^-5 . 10^-3
√4 x 10^-8
2 x 10^-4
pOH = - log [OH-]
= 4 - log 2
ph = 14 - (4 - log 2)
= 10 + log 2 = 10,3

B. NH4OH
[OH-] = √kb. M
√9x10^4 . 10^-2
√9x10^-6
3 x 10^-3
pOH = - log [OH-]
3 - log 3
Ph = 14 - (3 - log 3)
11 + log 3 = 11,4