Question

Mixture of 1 g each of Na₂CO₃ and NaHCO₃ is reacted with 0.1 N HCl. The quantity of 0.1 M HCl required to react completely with the above mixture is

• A. 15.78 mL
• B. 157.8 mL
• C. 198.4 mL
• D. 308 mL

Answer 1

Answer: (D)

$Na _{2} CO _{3}+2 HCl \rightarrow 2 NaCl + H _{2} O + CO _{2}$ $1 mol \quad 2 mol$

$\text{Na}_{2} \text{CO}_{3} = \text{1 g} = \frac{1}{106} \text{m} \text{o} \text{l} . \text{N} \text{a}_{2} \text{C} \text{o}_{3} \equiv \frac{2}{106} \text{m} \text{o} \text{l} . \text{H} \text{C} \text{l}\frac{1}{106} \text{m} \text{o} \text{l} . \text{N} \text{a}_{2} \text{C} \text{o}_{3} \equiv \frac{2}{106} \text{m} \text{o} \text{l} . \text{H} \text{C} \text{l}$

$\text{NaHCO}_{3} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_{2} \text{O} + \text{CO}_{2}$

$\text{NaHCO}_{3} = \text{1 g} = \frac{1}{8 4} \text{mol} \equiv \frac{1}{8 4} mol HCl$

Total HCl required $= \left(\frac{1}{5 3} + \frac{1}{8 4}\right) mol HCl$

Let volume of $0.1 \,N\, HCl = V\, mL$

then, $\text{Moles} = \, \frac{\text{0.1} \times \text{V}}{1 0 0 0} = \frac{\text{V}}{1 0 0 0 0}$

$∴ \, \, \frac{\text{V}}{1 0 0 00} = \left(\frac{1}{5 3} + \frac{1}{8 4}\right)$

$= \, 0.0308 mol$

$\text{V} = \text{0.0308} \times 1 0 0 0 0$

$= \, 308\, mL$