Answer to Question #174743 in Inorganic Chemistry for Jazel

Question #174743

A 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:

          CO32-  + 2H    →   H2O +CO2

Calculate the percent sodium carbonate in the sample


1
Expert's answer
2021-03-24T01:10:23-0400

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%


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