suatu larutan garam ca(ch3coo)2 dengan konsentrasi 0,1m ka ch3cooh=1,8×10^-5 hitung ph larutannya

Ca(CH3COO)2 → Ca2+ + 2CH3COO-

[OH-] = √Kw / Ka . n . M
[OH-] = √10^-14 / 18 x 10^-6 . 2 . 10^-1
[OH-] = √10^-8 / 18 (2 . 10^-1)
[OH-] = √1/9 x 10^-9
[OH-] = √1/9 x 1/10 x 10^-8
[OH-] = 1/3√0,1 x 10^-4

pOH = -log[OH-]
pOH = -log1/3√0,1 x 10^-4
pOH = 4 - log1/3√0,1

pH = 14 - (4 - log1/3√0,1)
pH = 10 + log1/3√0,1
pH = 10 + 0,1
pH = 10,1