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  • Sulphuric acid reacts with sodium hydroxide as follows : H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0. 1M sodium hydroxide solution

Sulphuric acid reacts with sodium hydroxide as follows:

H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O

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 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

Answers (1)

The answers are options (ii) and (iii)

H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O

Given (0.1 M), Moles of H2SO4 = 0.1 moles

Given (0.1M), Moles of NaOH = 0.1 moles

The ratio of H2SO4 ­and NaOH is 1:2, using the ratio, we figure for 0.1 mole of H2SO4 , 0.2 moles of NaOH is used, which is not available.

Therefore, if 0.1 moles of NaOH reacts, then only 0.05(0.1/2) moles of H2SO4 is used. NaOH is the limiting reactant, and the reaction and products are evaluated according to the limiting reactant.

Therefore, the number of moles formed of Na2SO4 will be 0.05 as well.

This is because 2 moles of NaOH produce 1 mole of H2SO4, so 1 mole of NaOH will logically produce 0.05 moles of Na2SO4.

Therefore, mass of Na2SO4 = 0.05 × (molar mass)

0.05 × (23 ×2 + 32 + 16× 4)

0.05 × 142 = 7.1 g

Molarity = Number of moles / Volume in L

Molarity = 0.05/ 2 L = 0.023 mol/L

 

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