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).