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Asked by Datta Last Modified
Answer 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 NaHCO3, containing equimolar amounts of both.
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