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?
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
$\therefore$ Number of moles $\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{x}{106} \mathrm{~mol}$
Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16
= 84 g mol−1
$\therefore$ Number of moles of $\mathrm{NaHCO}_{3}=\frac{1-x}{84} \mathrm{~mol}$
According to the question,
$\frac{x}{106}=\frac{1-x}{84}$
⇒ 84x = 106 − 106x
⇒ 190x = 106
⇒ x = 0.5579
Therefore, number of moles of $\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{0.5579}{106} \mathrm{~mol}$
= 0.0053 mol
And, number of moles of $\mathrm{NaHCO}_{3}=\frac{1-0.5579}{84}$
= 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, $0.0159 \mathrm{~mol}$ of $\mathrm{HCl}$ is present in $\frac{1000 \times 0.0159}{0.1} \mathrm{~mol}$
= 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.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.