What will be the volume of $0.1 M H_{2} S O_{4}$ solution required to neutralize $40 m L$ of $0.1 M K O H$ solution ?

40 m L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

39 m L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 m L

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10 m L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $20 m L$
$H_{2} S O_{4}$ is dibasic.
$∴$ n-factor for $H_{2} S O_{4} = 2$
n-factor for $K O H = 1$

The condition of neutralization is :
Equivalents of $H_{2} S O_{4} =$ Equivalents of $K O H$
$∴ N_{H_{2} S O_{4}} \times V_{H_{2} S O_{4}} = N_{K O H} \times V_{K O H}$

$Normality = Molarity \times n_{f}$

$∴ M_{H_{2} S O_{4}} \times {n_{f}}_{H_{2} S O_{4}} \times V_{H_{2} S O_{4}} = M_{K O H} \times {n_{f}}_{K O H} \times V_{K O H}$

Putting the values,

$0.1 \times 2 \times V_{H_{2} S O_{4}} = 0.1 \times 1 \times 40$
$∴ V_{H_{2} S O_{4}} = 20 m L$

Suggest Corrections

Similar questions

(H_{3} P O_{3}) .

To neutralize completely 20 ml of 0.1 M aqueous solution of $H_{3} P O_{3}$ , the volume of 0.1 M aqueous KOH solution required is:

To neutralize completely 20 mL of 0.1molar aqueous of H₃PO₄,what volume of 0.1M KOH solution will be required?

0.1 M H_{2} S O_{4}

40 m L

0.1 M K O H

Join BYJU'S Learning Program