Search any question & find its solution
Question:
Answered & Verified by Expert
$50 \mathrm{~cm}^{3}$ 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
Options:
Solution:
1270 Upvotes
Verified Answer
The correct answer is:
$10 \mathrm{~cm}^{3}$
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,
$$
\begin{aligned}
\mathrm{N}_{1} \mathrm{~V}_{1}=& \mathrm{N}_{3} \mathrm{~V}_{3} \\
\text { (For remaining HCl) } &(\text { For KOH}) \\
0.2 \mathrm{~N} \times 25 &=0.5 \mathrm{~N} \times \mathrm{V}_{3} \\
\mathrm{~V}_{3} &=\frac{0.2 \times 25}{0.5} \\
=& 10 \mathrm{~cm}^{3}
\end{aligned}
$$
$$
\begin{gathered}
\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\text { (For HCl) } \\
\text { (For NaOH) } \\
0.2 \mathrm{~N} \times \mathrm{V}_{1}=50 \times 0.1 \mathrm{~N} \\
\mathrm{~V}_{1}=\frac{50 \times 0.1}{0.2}=25 \mathrm{~cm}^{3}
\end{gathered}
$$
When $0.5 \mathrm{~N} \mathrm{KOH}$ is used,
$$
\begin{aligned}
\mathrm{N}_{1} \mathrm{~V}_{1}=& \mathrm{N}_{3} \mathrm{~V}_{3} \\
\text { (For remaining HCl) } &(\text { For KOH}) \\
0.2 \mathrm{~N} \times 25 &=0.5 \mathrm{~N} \times \mathrm{V}_{3} \\
\mathrm{~V}_{3} &=\frac{0.2 \times 25}{0.5} \\
=& 10 \mathrm{~cm}^{3}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.