Question

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Answer 1

Let the amount of Na₂CO₃ in the mixture be x g.

Then, the amount of NaHCO₃ in the mixture is (1 − x) g.

Molar mass of Na₂CO₃ = 2 × 23 + 1 × 12 + 3 × 16

= 106 g mol⁻¹

Number of moles Na₂CO₃

Molar mass of NaHCO₃ = 1 × 23 + 1 × 1 × 12 + 3 × 16

= 84 g mol⁻¹

Number of moles of NaHCO₃

According to the question,

⇒ 84x = 106 − 106x

⇒ 190x = 106

⇒ x = 0.5579

Therefore, number of moles of Na₂CO₃

= 0.0053 mol

And, number of moles of NaHCO₃

= 0.0053 mol

HCl reacts with Na₂CO₃ and NaHCO₃ according to the following equation.

1 mol of Na₂CO₃ reacts with 2 mol of HCl.

Therefore, 0.0053 mol of Na₂CO₃ reacts with 2 × 0.0053 mol = 0.0106 mol.

Similarly, 1 mol of NaHCO₃ reacts with 1 mol of HCl.

Therefore, 0.0053 mol of NaHCO₃ reacts with 0.0053 mol of HCl.

Total moles of HCl required = (0.0106 + 0.0053) mol

= 0.0159 mol

In 0.1 M of HCl,

0.1 mol of HCl is preset in 1000 mL of the solution.

Therefore, 0.0159 mol of HCl is present in

= 159 mL of the solution

Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na₂CO₃ and NaHCO_3, containing equimolar amounts of both.