Question

$50 \, cm^3$ of $0.2 \, N \, HCl$ is titrated against $0.1 \, N \, NaOH$ solution. The titration was 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. $16.2 \, cm^3$
• D. $21.0\, cm^3$

Answer 1

Answer: (A)

$10\, cm^3$

Answer 2

No. of equivalent of $HCl$ remaining after adding $50\, cm ^{3}$ of $0.1\, N\, NaOH =\frac{0.2 \times 50-0.1 \times 50}{100}$ $=\frac{0.5}{100}$ $\therefore$ Volume of $0.5\, N$ KOH required $\frac{0.5}{100}$ eq $\equiv \frac{V \times 0.5}{1000}$ $V =\frac{0.5}{100} \times \frac{1000}{0.5}$ $=10\, cm ^{3}$