PTHH: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có: \(n_{NaOH}=\dfrac{30\cdot20\%}{40}=0,15\left(mol\right)=n_{CH_3COONa}=n_{CH_3COOH}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COONa}=0,15\cdot82=12,3\left(g\right)\\C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)