Đặt số mol $CH_3COONa$ là $x$ 

$n_{CH_3COOH}=0,2.2=0,4(mol)$

$n_{NaOH}=0,05(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

$\to CH_3COOH$ dư 

$\to 0,2+0,05=0,25l$ $Y$ có: $\begin{cases} CH_3COO^-: x+0,05(mol) \\ CH_3COOH: 0,4-0,05=0,35(mol)\end{cases}$

$\to$ sau khi trộn: $C_{CH_3COO^-}=\dfrac{x+0,05}{0,25}=4x+0,2(M); C_{CH_3COOH}=\dfrac{0,35}{0,25}=1,4M$

$[H^+]=10^{-4,3098}$

$CH_3COOH\rightleftharpoons CH_3COO^-+H^+\quad K_a=1,75.10^{-5}$

$C_{CH_3COOH\text{phân li}}=C_{CH_3COO^-\text{tạo ra}}=[H^+]=10^{-4,3098}(M)$

Tại thời điểm cân bằng:

$Y\begin{cases} [CH_3COOH]=1,4-10^{-4,3098}(M) \\ [CH_3COO^-]=4x+0,2+10^{-4,3098}(M) \\ [H^+]=10^{-4,3098}(M) \end{cases}$

$K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}=1,75.10^{-5}$

$\to 10^{-4,3098}(4x+0,2+10^{-4,3098})=(1,4-10^{-4,3098}).1,75.10^{-5}$

$\to x=0,075$

Vậy $m=82x=6,15g$