Question

What will be the volume of $0.1M{H}_{2}S{O}_4$​ solution required to neutralize $40mL$ of $0.1MKOH$ solution ?

• A. $40mL$
• B. $39mL$
• C. $20mL$
• D. $10mL$

Answer 1

The correct option is C $20mL$

$H_{2}S{O}_4$ is dibasic.
$\therefore$ n-factor for $H_{2}S{O}_4=2$
n-factor for $KOH=1$

The condition of neutralization is :
Equivalents of $H_{2}S{O}_4=$ Equivalents of $KOH$
$\therefore N_{H_{2}S{O}_4}\times V_{H_{2}S{O}_4}=N_{KOH}\times V_{KOH}$

$\text{Normality}=\text{Molarity}\times n_f$

$\therefore M_{H_{2}S{O}_4}\times {n_f}_{H_{2}S{O}_4}\times V_{H_{2}S{O}_4}=M_{KOH}\times {n_f}_{KOH}\times V_{KOH}$

Putting the values,

$0.1\times 2\times V_{H_{2}S{O}_4}=0.1\times 1\times 40$
$\therefore V_{H_{2}S{O}_4}=20mL$