Ta có: \(n_{CH_3COOH}=\dfrac{18}{60}=0,3\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
Theo PT: \(n_{NaOH}=n_{CH_3COOH}=0,3\left(mol\right)\)
Ta có: \(C_{M_{NaOH\:}}=\dfrac{0,3}{V_{dd_{NaOH}}}=0,25M\)
=> \(V_{dd_{NaOH}}=1,2\left(lít\right)\)