KCET · Chemistry · Redox Reactions
\(50 \mathrm{~cm}^{3}\) 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
- A \(12 \mathrm{~cm}^{3}\)
- B \(10 \mathrm{~cm}^{3}\)
- C \(21.0 \mathrm{~cm}^{3}\)
- D \(16.2 \mathrm{~cm}^{3}\)
Answer & Solution
Correct Answer
(B) \(10 \mathrm{~cm}^{3}\)
Step-by-step Solution
Detailed explanation
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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