$75 \mathrm{~mL}$ 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?

Question

Aicharya · Accepted Answer

$$ \begin{aligned} V_1 & =75 \mathrm{~mL}, V_2=25 \mathrm{~mL} \ M_1 & =0.2 \mathrm{M}, M_2=1 \mathrm{M} \ V_3 & =300 \mathrm{~mL}, \mathrm{pH}=? \ V & =V_1+V_2=75+25=100 \mathrm{~mL} \ M & =\frac{V_1 M_1+V_2 M_2}{V_1+V_2}=\frac{75 	imes 0.2 	imes 25+1}{75+25}=0.4 \mathrm{M} \end{aligned} $$ $	herefore$ Total volume $$ \begin{aligned} V^{\prime}=V_1+V_2+V_3 & =75+25+300=400 \mathrm{~mL} \ M V & =M^{\prime} V^{\prime} \ 0.4 	imes 100 & =M^{\prime} 	imes 400 \ M & =0.1 \mathrm{M} \end{aligned} $$ $$ 	ext { Concentration of } \mathrm{HCl}=0.1=10^{-1} $$ $$ 	herefore \quad \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}ight]=1 $$ ,

$75 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{HCl}$ is mixed with $25 \mathrm{~mL}$ of $1 \mathrm{M}$…

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