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Karnataka Board PUCPUC Science 2nd PUC Class 12

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?

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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?

Numerical
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Solution

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 of 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 = `106/190`

⇒ x = 0.558

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

= 0.00526 mol

And, number of moles of NaHCO= `(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 present in 1000 mL of the solution.

Therefore, volume of solution = `(1000 xx 0.01578)/0.1` mL

= 157.8 mL of the solution

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

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Chapter 1: Solutions - Exercises [Page 28]

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NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 1 Solutions
Exercises | Q 1.6 | Page 28
Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.6 | Page 123

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