Question

Sulphuric acid reacts with sodium hydroxide as follows:

\[\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}\]

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:

(i) 0.1 mol L^–1

(ii) 7.10 g

(iii) 0.025 mol L^–1

(iv) 3.55 g

Answer 1

(ii) 7.10 g

(iii) 0.025 mol L^–1 

Explanation:

0.1 mole \[\ce{H2SO4}\] reacts with 1 mole of \[\ce{NaOH}\].

0.1 mole of \[\ce{NaOH}\] will react with `0.1/2` mole of \[\ce{H2SO4}\].

Here, \[\ce{NaOH}\] is the limiting reagent.

2 mole of \[\ce{NaOH}\] produces 1 mole of \[\ce{Na2SO4}\].

0.1 mole of \[\ce{NaOH}\] will give `0.1/2` mole of \[\ce{Na2SO4}\].

No. of mole = `"Given mass"/("Molar mass" (Na_2SO_4))`

On substituting the value in tha above equation, the mass can be calculated as 0.05 mol = `"Given mass"/(142  g  mol^-1)`

Given mass = 7.10 g

Volume of solution after mixing is 2 L.

So, the molarity of \[\ce{Na2SO4}\] is,

Molarity = `0.05/2` = 0.025 mol L^–1