Bài 4:

`C_{CH_3COOH}^o = 0,1 . (200)/(200 + 300) = 0,04M`

`C_{OH^-}^o = 0,1. (300)/(200+300) = 0,06M`

`CH_3COOH + OH^{-} -> CH_3COO^{-} + H_2O`

`C^o: 0,04 -> 0,06 -> 0`

`TPGH: 0 -> 0,02 -> 0,04`

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-; K_b = \dfrac{K_w}{K_a} = 10^{-9,25}$

`C^o: 0,04 -> -> 0 -> 0,02`

`[]: 0,04 -x -> -> x -> x + 0,02`

`=> (x (x+0,02)/(0,04-x) = 10^{-9,25}`

`=> x = 1,125.10^{-9}`

`=> pH = 14 + log[OH^-] = 14 + log(0,02 + 1,125.10^{-9}) = 12,3`