Gọi \(V_{HCl}=a\left(l\right)\)
\(\rightarrow V_{NaOH}=3a\left(l\right)\)
\(\rightarrow\left\{{}\begin{matrix}n_{HCl}=2a\left(mol\right)\\n_{NaOH}=0,3.3a=0,9a\left(mol\right)\end{matrix}\right.\)
PTHH: NaOH + HCl ---> NaCl + H2O
LTL: \(0,9a< \dfrac{2a}{2}\rightarrow\) HCl dư
Theo pthh: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{NaOH}=2.0,9a=1,8a\left(mol\right)\\n_{NaCl}=n_{NaOH}=0,9a\left(mol\right)\end{matrix}\right.\)
\(V_{dd\left(sau.pư\right)}=a+3a=4a\left(l\right)\)
=> \(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{2a-1,8a}{4a}=0,05M\\C_{M\left(NaCl\right)}=\dfrac{0,9a}{4a}=0,225M\end{matrix}\right.\)
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