Question

How many millilitres of 0.1 M $H_{2}S{O}_4$ must be added to 50 mL of 0.1 M $NaOH$ to give a solution that has a concentration of 0.05 M in $H_{2}S{O}_4$?

• A. 400 mL
• B. 50 mL
• C. 200 mL
• D. 100 mL

Answer 1

The correct option is D 100 mL

When $H_{2}S{O}_4$ is mixed with $NaOH$ then neutralization reaction takes place,

and we have formula

$N_1{V}_1-N_2{V}_2=N(V_1+V_2)$

normality = molarity$\times$ n-factor

n-factor for $H_{2}S{O}_4$ = 2

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

so, putting the values we have

$2\times 0.1\times x-50\times 0.1=0.05(x+50)$

on solving the equation we get,

$x=100ml$

so, millilitres of $H_{2}S{O}_4$ added is $100ml.$