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 = `"x"/106` mol

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

= 84 g mol⁻¹

∴ Number of moles of NaHCO₃ = `(1 - "x")/84` mol

According to the question,

`"x"/106 = (1-"x")/84`

⇒ 84x = 106 − 106x

⇒ 190x = 106

⇒ x = 0.558

Therefore, number of moles of Na₂CO₃ = `0.558/106` mol

= 0.00526 mol

And, number of moles of NaHCO₃ = `(1-0.558)/84`

= 0.00526 mol

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

\[\ce{\underset{2 mol}{2 HCl} + \underset{1 mol}{Na2CO3} -> 2 NaCl + H2O + CO2}\]

\[\ce{\underset{1 mol}{HCl} + \underset{1 mol}{NaHCO3} -> NaCl + H2O + CO2}\]

1 mol of Na₂CO₃ reacts with 2 mol of HCl.

Therefore, 0.00526 mol of Na₂CO₃ reacts with 2 × 0.00526 mol = 0.01052 mol.

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

Therefore, 0.00526 mol of NaHCO₃ reacts with 0.00526 mol of HCl.

Total moles of HCl required = (0.01052 + 0.00526) mol

= 0.01578 mol

In 0.1 M of HCl,

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

Therefore, 0.01578 mol of HCl is present in = `(1000xx0.01578)/0.1` mol

= 157.8 mL of the solution

Hence, 157.8 mL of 0.1 M of HCl is required to react completely with a 1 g mixture of Na₂CO₃ and NaHCO₃, containing equimolar amounts of both.