Q. $50\ {cm}^{3}$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 NKOH. The volume of KOH required for completing the titration is
A) $12\ {cm}^{3}$
B) $10\ {cm}^{3}$
C) $25\ {cm}^{3}$
D) $10.5\ {cm}^{3}$
Explanation
(b)
When 0.1 n NaOH is used,
${\ \ \ \ \ N}_{1}V_{1} = N_{2}V_{2}$
(For HCl) (For KOH)
$0.2N \times \ V_{1} = \ 50\ \times 0.1N$
${\ \ V}_{1} = \frac{50\ \times 0.1}{0.2} = 25{cm}^{3}$
When 0.5 N KOH is used,
${\ \ \ \ \ \ \ \ \ \ \ \ N}_{1}V_{1}\ = {\ N}_{3}V_{3}$
(For remaining HCl) (for KOH)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2N \times \ 25\ = 0.5\ N \times V_{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{3} = \frac{0.2\ \times 25}{0.5}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10{cm}^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.