\(50 \mathrm{~cm}^{3}\) of \(0.2 \mathrm{~N} \mathrm{HCl}\) is titrated against \(0.1 \mathrm{~N}\) \(\mathrm{NaOH}\) solution. The titration is 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

When \(0.1 \mathrm{~N} \mathrm{NaOH}\) is used,
\[
\begin{gathered}
\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\text { (For HCl) } \\
\text { (For NaOH) } \\
0.2 \mathrm{~N} \times \mathrm{V}_{1}=50 \times 0.1 \mathrm{~N} \\
\mathrm{~V}_{1}=\frac{50 \times 0.1}{0.2}=25 \mathrm{~cm}^{3}
\end{gathered}
\]
When \(0.5 \mathrm{~N} \mathrm{KOH}\) is used,
\[
\begin{aligned}
\mathrm{N}_{1} \mathrm{~V}_{1}=& \mathrm{N}_{3} \mathrm{~V}_{3} \\
\text { (For remaining HCl) } &(\text { For KOH}) \\
0.2 \mathrm{~N} \times 25 &=0.5 \mathrm{~N} \times \mathrm{V}_{3} \\
\mathrm{~V}_{3} &=\frac{0.2 \times 25}{0.5} \\
=& 10 \mathrm{~cm}^{3}
\end{aligned}
\]