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$75 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{HCl}$ is mixed with $25 \mathrm{~mL}$ of $1 \mathrm{M}$ $\mathrm{HCl}$. To this solution, $300 \mathrm{~mL}$ of distilled water is added. What is the $\mathrm{pH}$ of the resultant solution?
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Verified Answer
The correct answer is:
$1$
$$
\begin{aligned}
V_1 & =75 \mathrm{~mL}, V_2=25 \mathrm{~mL} \\
M_1 & =0.2 \mathrm{M}, M_2=1 \mathrm{M} \\
V_3 & =300 \mathrm{~mL}, \mathrm{pH}=? \\
V & =V_1+V_2=75+25=100 \mathrm{~mL} \\
M & =\frac{V_1 M_1+V_2 M_2}{V_1+V_2}=\frac{75 \times 0.2 \times 25+1}{75+25}=0.4 \mathrm{M}
\end{aligned}
$$
$\therefore$ Total volume
$$
\begin{aligned}
V^{\prime}=V_1+V_2+V_3 & =75+25+300=400 \mathrm{~mL} \\
M V & =M^{\prime} V^{\prime} \\
0.4 \times 100 & =M^{\prime} \times 400 \\
M & =0.1 \mathrm{M}
\end{aligned}
$$
$$
\text { Concentration of } \mathrm{HCl}=0.1=10^{-1}
$$
$$
\therefore \quad \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=1
$$
\begin{aligned}
V_1 & =75 \mathrm{~mL}, V_2=25 \mathrm{~mL} \\
M_1 & =0.2 \mathrm{M}, M_2=1 \mathrm{M} \\
V_3 & =300 \mathrm{~mL}, \mathrm{pH}=? \\
V & =V_1+V_2=75+25=100 \mathrm{~mL} \\
M & =\frac{V_1 M_1+V_2 M_2}{V_1+V_2}=\frac{75 \times 0.2 \times 25+1}{75+25}=0.4 \mathrm{M}
\end{aligned}
$$
$\therefore$ Total volume
$$
\begin{aligned}
V^{\prime}=V_1+V_2+V_3 & =75+25+300=400 \mathrm{~mL} \\
M V & =M^{\prime} V^{\prime} \\
0.4 \times 100 & =M^{\prime} \times 400 \\
M & =0.1 \mathrm{M}
\end{aligned}
$$
$$
\text { Concentration of } \mathrm{HCl}=0.1=10^{-1}
$$
$$
\therefore \quad \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=1
$$
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