Larutan Ch3cook 0.1 M(ka=1×10-5) sebanyak 200ml memiliki ph sebesar

≈ Menghitung pH Garam ≈ Hidrolisis Garam ≈ Kelas 11 ≈

Ka CH3COOH = 1 × 10^-5
[CH3COOK] = 0,1 M

CH3COOK => CH3COO^{-} + K^{+}
..... 0,1 M ................ 0,1 M ......... 0,1 M

Garam berasal dari Asam Lemah CH3COOH dan Basa Kuat KOH maka Garam bersifat Basa

Kh = Kw / Ka
Kh = 10^-14 / 10^-5 = 10^-9

[OH-] = √ Kh × M
[OH-] = √ 10^-9 × 0,1
[OH-] = 10^-5

pOH = - log [OH-] = - log (10^-5) = 5
pH = pKw - pOH = 14 - 5 = 9