Question:
$50 \, cm^3\; of\; 0.2 \, N \, HCl$ is titrated against $0.1 \, N \, NaOH$ solution. The titration is discontinued after adding $50\,cm^3 \; of\; NaOH$. The remaining titration is completed by adding $0.5\, N\, KOH$. The volume of $KOH$ required for completing the titration is
$50 \, cm^3\; of\; 0.2 \, N \, HCl$ is titrated against $0.1 \, N \, NaOH$ solution. The titration is discontinued after adding $50\,cm^3 \; of\; NaOH$. The remaining titration is completed by adding $0.5\, N\, KOH$. The volume of $KOH$ required for completing the titration is
Updated On: Jun 15, 2024
- $10\,cm^3$
- $12\,cm^3$
- $10.5\,cm^3$
- $25\,cm^3$
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The Correct Option is A
Solution and Explanation
When $0.1\, N NaOH$ is used,
$\underset{\text{(For HCl)}}{N_{1} V_{1}}= \underset{\text{(For NaOH)}}{N_{2} V_{2}}$
$0.2\, N \times V_{1}=50 \times 0.1\, N$
$V_{1}=\frac{50 \times 0.1}{0.2}=25\, cm ^{3}$
When $0.5\, N KOH$ is used,
$\underset{\text{(For remaining HCl)}}{N_{1} V_{1}}=\underset{\text{(For KOH)}}{N_{3} V_{3}}$
$0.2\, N \times 25 =0.5\, N \times V_{3}$
$V_{3} =\frac{0.2 \times 25}{0.5}$
$=10\, cm ^{3}$
$\underset{\text{(For HCl)}}{N_{1} V_{1}}= \underset{\text{(For NaOH)}}{N_{2} V_{2}}$
$0.2\, N \times V_{1}=50 \times 0.1\, N$
$V_{1}=\frac{50 \times 0.1}{0.2}=25\, cm ^{3}$
When $0.5\, N KOH$ is used,
$\underset{\text{(For remaining HCl)}}{N_{1} V_{1}}=\underset{\text{(For KOH)}}{N_{3} V_{3}}$
$0.2\, N \times 25 =0.5\, N \times V_{3}$
$V_{3} =\frac{0.2 \times 25}{0.5}$
$=10\, cm ^{3}$
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There are two ways of classifying the matter:
(A) Physical Classification:
Matter can exist in three physical states:
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(B) Chemical Classification:
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