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

• A. $10\,cm^3$
• B. $12\,cm^3$
• C. $10.5\,cm^3$
• D. $25\,cm^3$

Answer 1

Answer: (A)

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,

$\underset{\text{(For remaining HCl)}}{N_{1} V_{1}}=\underset{\text{(For KOH)}}{N_{3} V_{3}}$

$0.2\, N \times 25 =0.5\, N \times V_{3}$

$V_{3} =\frac{0.2 \times 25}{0.5}$

$=10\, cm ^{3}$