Sulphuric acid reacts with sodium hydroxide as follows :

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

Sulphuric acid is H₂SO₄ and sodium hydroxide is NaOH. In the acid-base reaction, Na₂SO₄ is the salt formed as a product.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

H₂SO₄ and NaOH react in the ratio 1:2 for this reaction as H₂SO₄ is a dibasic acid. 1 mole of H₂SO₄ reacts with 2 moles of NaOH, so according to this reaction, 0.05 moles of H₂SO₄ reacts with 0.1 moles of NaOH. Since the reaction can go ahead only with the minimum amount of NaOH being 0.1 moles, NaOH is the limiting reagent. The limiting reagent is known as the reactant, which gets consumed first and limits the amount of product formed. The product formed will be equal to the amount of the other reactant used i.e. H₂SO₄ and it will be 0.05 moles. The mass of Na₂SO₄ formed will be 0.05x142 = 7.10g. The molarity of the solution formed will be

= 0.05/2 = 0.025M.

The correct answers are (ii) and (iii).