Em tham khảo nha :
\(\begin{array}{l}
17)\\
HCl\,pH = 1\\
{C_M} = 0,1M\\
{V_1}(l)\\
{n_{HCl}} = 0,1 \times {V_1}(mol)\\
HCl\,pH = 4\\
{C_M} = 0,0001M\\
{V_2}(l)\\
HCl\,pH = 3\\
{C_M} = 0,001M\\
\Rightarrow \dfrac{{0,1{V_1} + 0,0001{V_2}}}{{{V_1} + {V_2}}} = 0,001\\
\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{1}{{110}}\\
19)\\
{V_1},{V_2}\\
NaOH + HCl \to NaCl + {H_2}O\\
{C_{{M_{HCl}}}} = 0,1M\\
{n_{HCl}} = 0,1{V_1}\\
{C_{{M_{NaOH}}}} = 0,1M\\
{n_{NaOH}} = 0,1 \times {V_2}\\
pH = 11\\
\Rightarrow pOH = 14 - 11 = 3\\
{\rm{[}}O{H^ - }{\rm{]}} = 0,001M\\
\Rightarrow NaOH\text{ dư}\\
\Rightarrow \dfrac{{0,1{V_2} - 0,1{V_1}}}{{{V_1} + {V_2}}} = 0,001\\
\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{0,1 - 0,001}}{{0,1 + 0,001}} = \dfrac{{99}}{{101}}
\end{array}\)
