Question

The volume of the solution of a sample of 1.881g Na2CO3 with impunities was made 250 cm3. 25 cm3 of this solution neutralize 24.05 cm3 M/10 HCl solutions. Find out the percentage of impunity in Na2CO3

Answer 1

Na2CO3 + 2HCl = 2NaCl + H2CO3
From the equation: n (HCl) = 2 · n (Na2CO3)
CM = n/V n (HCl) = CM · V = 0.01 · (24.05/1000) = 0.002405 moles
n (Na2CO3) = 1/2 · n (HCl) = 0.002405/2 = 0.0012025 moles
n = m/M m = n·M M (Na2CO3) = 58 g/mol
m (Na2CO3) = 0.0012025 · 58 = 0.07 g
m (Na2CO3) in 1 ml= 0.07/25 = 0.003 g
m (Na2CO3) in 250 ml= 0.003 ·250 = 0.75 g
% (Na2CO3) = (0.75/1.881) · 100 = 39.87%