$n_{CH_3COOH}=0.1(mol)$

$n_{NaOH}=0.06(mol)$

`->`$n_{CH_3COONa}=0.06(mol);n_{CH_3COOH,dư}=0.04(mol)$

`=>`$C_{M,CH_3COONa}=0.3(mol);C_{M,CH_3COOH,dư}=0.2(mol)$

$K_{a,CH_3COOH}=\dfrac{10^{-14}}{K_{b,CH_3COO^-}}=10^{-4.74}$

$[H^+]=\dfrac{C_a}{C_m}K_a=\dfrac{0.2}{0.3} \times 10^{-4.74}=10^{-4.916}(M)$

`=>`$pH=-log[H^+]=4.916$

`pKb = -log(5,5.10^{-10}) = 9,26`

`=> pKa = 14 - pKb = 4,74`

Trộn hai dd với nhau:

`C_{CH_3COOH}^o = (1.0,1)/(0,1 + 0,1) = 0,5M`

`C_{NaOH}^o = (0,6.0,1)/(0,1 + 0,1) = 0,3M`

`PTHH:CH_3COOH + NaOH -> CH_3COONa + H_2O`$\\$ `C^o: \text{          0,5                    0,3 
                0}` $\\$ `[C]: \text{          0,2                     0                  0,3}`

`CH_3COONa -> CH_3COO^{-} + Na^{+}`

`0,3 ------> 0,3`

Ta có các CB:

$CH_3COOH \rightleftharpoons CH_3COO^- + H^+ (1); K_a = 10^{-4,74}$

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-(2); K_b = 10^{-9,26}$

$H_2O \rightleftharpoons H^+ + OH^-; K_w = 10^{-14}$

Nhận xét:

`C_{CH_3COOH}^o . K_a >> K_w =>` Bỏ qua `H^+` do `H_2O` phân ly

`=> (1)` là CB chính tạo ra `H^+`

`C_{CH_3COO^-}^o . K_b >> K_w =>` Bỏ qua `OH^-` do `H_2O` phân ly

`=> (2)` là CB chính tạo ra `OH^-`

Mà `CH_3COO^-// CH_3COOH` là cặp acid-base liên hợp

`=> pH = pKa + log( (C_{CH_3COO^-})/(C_{CH_3COOH}) ) = 4,74 + log( (0,3)/(0,2) ) = 4,916`