The mass of an equimolar mixture of Na₂CO₃ and NaHCO₃ is 1.0 g. The volume of 0.1 M solution of HCl required to react completely with this mixture is

Let n be the amount of each ofNa₂CO₃ and NaHCO₃ in the given 1.0 g of mixture. We will have

$n{M}_{{\mathrm{Na}}_2{\mathrm{CO}}_3}+n{M}_{{\mathrm{NaHCO}}_3}=1.0\mathrm{g}$

$i. e. , n\left(108\mathrm{g}{\mathrm{mol}}^{-1}+84\mathrm{g}{\mathrm{mol}}^{-1}\right)=1.0\mathrm{g}$

or,  $n=\frac{1.0\mathrm{g}}{192\mathrm{g}{\mathrm{mol}}^{-1}}=\frac{1.0}{192}\mathrm{mol}$

From the reaction

 ${\mathrm{Na}}_2{\mathrm{CO}}_{3 } + 2\mathrm{HCl} \to 2\mathrm{NaCl}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

 (1.0/192) mol       (2.0/192) mol                             

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

 (1.0/192) mol     (1.0/192) mol

we find that Amount of HCL required $=\frac{2.0}{192}\mathrm{mol}+\frac{1.0}{192}\mathrm{mol}=\frac{3.0}{192}\mathrm{mol}$

Volumeof0.1 M HCl required $V=n/M=\left(\frac{3.0}{192}\mathrm{mol}\right)/\left(0.1{\mathrm{molL}}^{-1}\right)=0.1563\mathrm{L}=156.3\mathrm{mL}$