$\rm PTHH: CH_3COOH +NaOH \rightarrow{} CH_3COONa +H_2O$ 

$\rm n_{CH_3COOH}=0,5.0,4=0,2(mol)$ 

theo pt 

$\rm n_{NaOH}=n_{CH_3COOH}=0,2(mol)$ 

$\rm → V_{NaOH}=\dfrac{0,2}{0,5}=0,4(l)$ 

Đáp án:

 400 ml

Giải thích các bước giải:

PTHH: $CH_{3}$$COOH_{}$ + $NaOH_{}$ → $CH_{3}$$COONa_{}$ +$H_{2}$$O_{}$ 

-Ta có: $n_{CH_{3}COOH_{} }$ =$\frac{400}{1000}$. 0,5 = $0,2_{}$ $(mol)_{}$ 

-Theo PTHH:$n_{NaOH}$ =$n_{CH_{3}COOH_{} }$ = 0,2 (mol)

→ $V_{NaOH}$ =   = $\frac{n_{NaOH} }{C_{M_{NaOH}}}$ =$\frac{0,2}{0,5}$=0,4(l) = 400 (ml)

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