a) HCl+NaOH-->NaCl +H2O
b) n\(_{HCl}=0,25.1=0,25\left(mol\right)\)
Theo pthh
n\(_{NaOH}=n_{HCl}=0,25\left(mol\right)\)
V\(_{NaOH}=\frac{0,25}{0,25}=1\left(M\right)\)
\(a,PTHH:NaOH+HCl\rightarrow NaCl2+H2O\)
\(\text{b) Ta có: nHCl=0,25.1=0,25 mol}\)
Theo phản ứng: nNaOH=nHCl=0,25 mol
\(\Rightarrow\text{V dung dịch NaOH= nNaOH / CM NaOH}\)
\(\Rightarrow V_{NaOH}=\frac{0,25}{0.25}=1M\)