Titration of Na_2CO_3 with HCI. Part 1. 1 gram of sodium carbonate is dissolved

Question

Titration of Na_2CO_3 with HCI. Part 1. 1 gram of sodium carbonate is dissolved and diluted to a total volume of 50 mL. A buret was filled with HCI, and the volume needed to reach the equivalence point was 19.5 mL. Determine the molarity of the HCI solution. Part 2. An unknown soda ash sample has a mass 0.2500 g. It is dissolved and diluted to 50 mL. Using your answer from part 1, determine the percent of sodium carbonate in the sample if 22 mL of HCI are required to reach the equivalence point.

Explanation / Answer

Na2CO3 + 2HCl ----> H2O + CO2 + 2NaCl
M1V1/n1 = M2V2/n2
(1/106*1000/50)*50/1 = M2*19.5/2
M2 = concentration of HCl = 0.9676 mol/L

Na2CO3 + 2HCl ----> H2O + CO2 + 2NaCl
M1V1/n1 = M2V2/n2
M1*50/1 = (0.9676 * 22)/2
M1 = 0.2129 mol/L
But molarity from Part 1 M1 = 1/106*1000/50 = 0.1887 mol/L
Hence percentage of Na2CO3 = 0.1887/0.2129*100 = 88.63 %

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