Question

A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to remove CO2, and then back-titrating the excess acid with 0.100 M NaOH. If 5.6 mL NaOH is required for the back-titration, what is the percent Na2CO3 in the sample?

Answer 1

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added =  0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 ×  0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 =  0.005 moles -  0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3  in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

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