Question

Sulphuric acid reacts with sodium hydroxide as follows
$H_{2}S{O}_4$ + $2NaOH$ $\to$ $N{a}_{2}S{O}_4$+ $2H_{2}O$
When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is ?

• A. $0.1mol{L}^{-1}$
• B. $7.10g$
• C. $0.025mol{L}^{-1}$
• D. $3.55g$

Answer 1

The correct option is B $7.10g$

Moles of $H2S{0}_4$ taken = 0.1 moles

Moles of $NaOH$ taken = 0.1 moles

As $H2S{0}_4$ and $NaOH$ react in ratio 1:2, so 0.1 moles of $H2S{0}_4$ reacts with 0.2 mole of $NaOH$ which we don’t have.

0.1 mole of $NaOH$ reacts with 0.05 mole of $H2S{0}_4$, so $NaOH$ is Limiting Reactant. Product is calculated w.r.t limiting reactant so Number of moles of $N{a}_{2}S{0}_4$ formed will also be equal to 0.05.

Mass of $N{a}_{2}S{0}_4=0.05\times 142=7.1g$