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$50 \mathrm{~cm}^{3}$ of $0.2 \mathrm{~N} \mathrm{HCl}$ is titrated against $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ solution. The titration was discontinued after adding $50 \mathrm{~cm}^{3}$ of $\mathrm{NaOH}$. The remaining titration is completed by adding $0.5 \mathrm{~N} \mathrm{KOH}$. The volume of $\mathrm{KOH}$ required for completing the titration is
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Verified Answer
The correct answer is:
$10 \mathrm{~cm}^{3}$
No. of equivalent of $\mathrm{HCl}$ remaining after adding
$$
\begin{aligned}
50 \mathrm{~cm}^{3} \text { of } 0.1 \mathrm{~N} \mathrm{NaOH} &=\frac{0.2 \times 50-0.1 \times 50}{100} \\
&=\frac{0.5}{100}
\end{aligned}
$$
$\therefore$ Volume of $0.5 \mathrm{~N} \mathrm{KOH}$ required $\frac{0.5}{100} \mathrm{eq}$
$$
\begin{aligned}
& \equiv \frac{V \times 0.5}{1000} \\
V &=\frac{0.5}{100} \times \frac{1000}{0.5} \\
&=10 \mathrm{~cm}^{3}
\end{aligned}
$$
$$
\begin{aligned}
50 \mathrm{~cm}^{3} \text { of } 0.1 \mathrm{~N} \mathrm{NaOH} &=\frac{0.2 \times 50-0.1 \times 50}{100} \\
&=\frac{0.5}{100}
\end{aligned}
$$
$\therefore$ Volume of $0.5 \mathrm{~N} \mathrm{KOH}$ required $\frac{0.5}{100} \mathrm{eq}$
$$
\begin{aligned}
& \equiv \frac{V \times 0.5}{1000} \\
V &=\frac{0.5}{100} \times \frac{1000}{0.5} \\
&=10 \mathrm{~cm}^{3}
\end{aligned}
$$
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