PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\) a) Số mol NaOH tham gia: \(n_{NaOH}=0,18.2=0,36\left(mol\right)\) Theo PTHH: \(n_{HCl}=n_{NaOH}=0,36\left(mol\right)\) Nồng độ mol dd HCl: \(C_{M\left(HCl\right)}=\dfrac{0,36}{0,05}=7,2\left(M\right)\) b) dd sau pứ là NaCl. Theo PTHH: \(n_{NaCl}=n_{NaOH}=0,36\left(mol\right)\) Nồng độ mol dd NaCl: \(C_{M\left(NaCl\right)}=\dfrac{0,36}{0,05+0,18}=1,57\left(M\right)\)