Question

$50c{m}^{3}of0.2NHCl$ is titrated against $0.1NNaOH$ solution. The titration is discontinued after adding $50c{m}^{3}ofNaOH$. The remaining titration is completed by adding $0.5NKOH$. The volume of $KOH$ required for completing the titration is

• A. $10c{m}^3$
• B. $12c{m}^3$
• C. $10.5c{m}^3$
• D. $25c{m}^3$

Answer 1

Answer: (A)

When $0.1NNaOH$ is used,

$\underset{\text{(For HCl)}}{N_1{V}_1}=\underset{\text{(For NaOH)}}{N_2{V}_2}$

$0.2N\times V_1=50\times 0.1N$

$V_1=\frac{50\times 0.1}{0.2}=25c{m}^3$

When $0.5NKOH$ is used,

$\underset{\text{(For remaining HCl)}}{N_1{V}_1}=\underset{\text{(For KOH)}}{N_3{V}_3}$

$0.2N\times 25=0.5N\times V_3$

$V_3=\frac{0.2\times 25}{0.5}$

$=10c{m}^3$