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)

$N{a}_{2}C{O}_3+2HCl\to 2NaCl+H_{2}O+C{O}_2$ $1mol2mol$

${\text{Na}}_2{\text{CO}}_3=\text{1 g}=\frac{1}{106}\text{m}\text{o}\text{l}.\text{N}{\text{a}}_2\text{C}{\text{o}}_3\equiv \frac{2}{106}\text{m}\text{o}\text{l}.\text{H}\text{C}\text{l}\frac{1}{106}\text{m}\text{o}\text{l}.\text{N}{\text{a}}_2\text{C}{\text{o}}_3\equiv \frac{2}{106}\text{m}\text{o}\text{l}.\text{H}\text{C}\text{l}$

${\text{NaHCO}}_3+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2$

${\text{NaHCO}}_3=\text{1 g}=\frac{1}{84}\text{mol}\equiv \frac{1}{84}molHCl$

Total HCl required $=\left(\frac{1}{53}+\frac{1}{84}\right)molHCl$

Let volume of $0.1NHCl=VmL$

then, $\text{Moles}=\frac{\text{0.1}\times \text{V}}{1000}=\frac{\text{V}}{10000}$

$\therefore \frac{\text{V}}{10000}=\left(\frac{1}{53}+\frac{1}{84}\right)$

$=0.0308mol$

$\text{V}=\text{0.0308}\times 10000$

$=308mL$