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 1L of 0.1M sulphuric acid solutoin is allowed to react with 1L of 0.1M sodium hydroxide solutoin, 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

(b)

For the reaction, $H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

1L of 0.1 M $M{H}_{2}S{O}_4$ contains = 0.1 mole of $H_{2}S{O}_4$

1L of 0.1 M NaOH contains = 0.1 mole of NaOH

According to the reaction, 1 mole of $H_{2}S{O}_4$ reacts with 2 moles of NaOH. Hence 0.1 mole of NaOH will react with 0.05 mole of $H_{2}S{O}_4$ (and 0.05 mole of $H_{2}S{O}_4$ will be left unreacted), i.e., NaOH is the limiting reactant. Since, 2 moles of NaOH produce 1 mole of $N{a}_{2}S{O}_4$.

Hence, 0.1 mole of NaOH will produce 0.05 mole of $N{a}_{2}S{O}_4$.

Mass of $N{a}_{2}S{O}_4=moles\times molarmass$

$=0.5\times (46+32+64)g$

= 7.10 g

Volume of solution after mixing = 2L

Since, only 0.05 mole of $H_{2}S{O}_4$ is left behind as NaOH completely used in the reaction.

Therefore, molarity of the given solution is calculated from moles of $H_{2}S{O}_4$.

$H_{2}S{O}_4$ left unreacted in the solution = 0.05 mole

$\therefore$ Molarity of the solution $=\frac{0.05}{2}=0.025mol{L}^{-1}$