Question

$100ml$ of $0.2M{H}_{2}S{O}_4$ is reacted with $100ml$ of $0.5MNaOH$ solution. what is the normality of the solution

• A. 0.3N
• B. 0.8N
• C. 0.1N
• D. 1N

Answer 1

Answer: (C)

0.1N

$1M-H_{2}S{O}_4=2N-H_{2}S{O}_4$

100ml of 0.2M 0.2M $H_{2}S{O}_4\equiv 100\times 0.2$ml of 1M $H_{2}S{O}_4$

$\equiv 20ml$ of 2N $H_{2}S{O}_4$

$\equiv 40ml$ of 2N $H_{2}S{O}_4$

$1MNaOH=1NNaOH$

100ml of 0.2M 0.2M $NaOH\equiv 100\times 0.2$ml of 1M $NaOH$

$\equiv 20ml$ of 1N $NaOH$

neutralisation occurs when acid and base are mixed due to the formation of salt and water.

20ml of 1N $NaOH\equiv$ 20ml of 1N $H_{2}S{O}_4$

$20ml\times 1N=200ml\times$ final strength of acid

therefore the normality of solution is $0.1N$