Question
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?
Solution
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
∴
[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
