Answer to Question #174776 in Inorganic Chemistry for Shallock

Question

 0.3749 g soda ash sample is analyzed by titrating sodium carbonate with the standard 0.2388M HCl solution, requiring 49.38ml. The reaction is:

          CO₃^2-  + 2H⁺     →   H₂O +CO₂

Calculate the percent sodium carbonate in the sample

Answer 1

Molar Mass of Na2CO3 = 105.99g/mol

105.99/2 = 52.995 g/ equivalent

Let amount of Na2CO3 in 0.3749g be (xg)

xg = (x/52.995) g equivalent

49.38ml of 0.2388ml solution will contain (49.38 × 0.2388 M / 1000) equivalent of HCl

x = (49.38 × 0.2388/100) × 52.995 = 0.6249141

= 0.3749/0.6249141 = 59.99% of sodium Carbonate