Đáp án:

`a.`

`CH_3COOH+NaHCO_3->CH_3COONa+H_2O+CO_2↑`

`2CH_3COOH+Na_2CO_3->2CH_3COONa+H_2O+CO_2↑`

`b.\ C_(M_(ddCH_3COOH))=0,525(M)`

Giải thích các bước giải:

`a.`

`CH_3COOH+NaHCO_3->CH_3COONa+H_2O+CO_2↑`

`2CH_3COOH+Na_2CO_3->2CH_3COONa+H_2O+CO_2↑`

`b.`

Gọi `a,\ b` là `n_(NaHCO_3),\ n_(Na_2CO_3)`

`=>m_A=84a+106b=1,144(g)\ (1)`

`n_(CO_2)=(0,2464)/(22,4)=0,011(mol)`

`=>n_(NaHCO_3)+n_(Na_2CO_3)=0,011`

`=>a+b=0,011\ (2)`

`(1),\ (2)=>{(a=0text(,)001),(b=0text(,)01):}`

`∑n_(CH_3COOH)=n_(NaHCO_3)+2n_(Na_2CO_3)=a+2b=0,021(mol)`

`C_(M_(ddCH_3COOH))=(0,021)/(0,04)=0,525(M)`

`c.` Đề thiếu dữ kiện để tính được `m_(ddCH_3COOH)` nên không thể tính nồng độ phần trăm các chất trong dung dịch A.

\(\begin{array}{l}
a)\\
2CH_3COOH+Na_2CO_3\to 2CH_3COONa+CO_2+H_2O\\
CH_3COOH+NaHCO_3\to CH_3COONa+CO_2+H_2O\\
b)\\
40ml=0,04l;246,4=0,2464(l)\\
Dat\,n_{Na_2CO_3}=x(mol);n_{NaHCO_3}=y(mol)\\
\to 106x+84y=1,144(1)\\
n_{CO_2}=\frac{0,2464}{22,4}=0,011(mol)\\
Theo\,PT:\,x+y=0,011(2)\\
(1)(2)\to x=0,01(mol);y=0,001(mol)\\
Theo\,PT:\,\sum n_{CH_3COOH}=2x+y=0,021(mol)\\
\to C_{M\,CH_3COOH}=\frac{0,021}{0,04}=0,525M\\
c)\\
\rm Coi\,m_{dd\,CH_3COOH}\,không\,đáng\,kể\\
Theo\,PT:\,\sum n_{CH_3COONa}=\sum n_{CH_3COOH}=0,021(mol)\\
m_{dd\,spu}=0,021.60+25-0,011.44=25,776(g)\\
\to C\%_{CH_3COONa}=\frac{0,021.82}{25,776}.100\%\approx 6,68\%
\end{array}\)