How many mL of 0.1 M HCl are required to react completely with 1 g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of both?

Options

A. 157.9 ml
B. 152.6 ml
C. 200 ml
D. 98.5 ml

Answer

A

Explanation

Let amount of `Na_(2)CO_(3)` in 1 gm of mixture be ‘a’ and since `Na_(2)CO_(3)` and `NaHCO_(3)` are in equimolar amounts.
`(a)/(106) = ((1-a))/(84) :. A = 0.558 gm :. "Weight of "NaHCO_(3) = (1-0.558) = 0.442 gm`
Meq of `Na_(2)CO_(3) + "Meq of " NaHCO_(3) = "Meq of HCl"`
`:. (0.558)/(53) xx 1000 + (0.442)/(84) xx 100 = 0.1 xx V " " :. " "V = 157.9 mL`