131. How many mL of 0.1 M HC1 are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Answer:

Let ‘a’ moles of Na₂CO₃ and a moles of NaHCO₃ are present in 1 g equimolar mixture of two. Then we can write (a x 106) + (a x 84) = 1
(Molar masses are : Na₂CO₃ = 106 and NaHCO₃ = 84)
a = 5.26 x 10^–3 ‘a’ moles of Na₂CO₃ = 2
a equivalent of Na₂CO₃ ‘a’ moles of NaHCO₃ 
=a equivalent of NaHCO₃ (2a x 1000) + (a x 1000) = 0.1 x V
∴                   $3\mathrm{a} = \frac{0.1\mathrm{V}}{1000} = 1\times {10}^{-4}\mathrm{V}$

[where V is volume of HCl (0.1 M)] Substituting the value of a, we have
1 x 10^–4 = 3a = 3 x 5.26 x 10^–3
V = 157.8 mL