Question

The volume of $0.2N$ $H_{2}S{O}_4$ mixed with $0.5NKOH$ (remaining volume) to prepare $150c{m}^3$ of $0.2NKOH$ solution is:

• A. $50c{m}^3$
• B. $60c{m}^3$
• C. $70c{m}^3$
• D. $80c{m}^3$

Answer 1

Answer: (A)

$50c{m}^3$

For neutralisation of acid and bases we have

$N_1{V}_1-N_2{V}_2=NV$

$0.5\ast 1(50-x)-0.2\ast x=0.2\ast 150$

x = volume of $H_{2}S{O}_4$

therefore x = 50 ml = 50 $c{m}^3$