Question

A solution is prepared by dissolving 11.6 g of a mixture of sodium carbonate and sodium bicarbonate in 1.00 L of water. A 300.0 cm3sample of the solution is then treated with excess HNO3 and boiled to remove all the dissolved gas. A total of 0.940 L of dry CO2 is collected at 298 K and 0.972 atm.

1. Find the molarity of the carbonate in the solution.

2. Find the molarity of the bicarbonate in the solution.

Answer 1

Let´s start from the end.

The dry gas collected is = 0.940 L at 298 K and 0,972 atm of CO₂ then we can calculate the moles of it.

PV = nRT n =PV/RT R= 0.082 Latm/Kmol

Na2CO3 + 2 HNO3 → 2 NaNO3 + CO2 + H2O
NaHCO3 + HNO3 → NaNO3 + CO2 + H2O

n = PV / RT = (0.972 atm) x (0.940 L) / ((0.08205746 L atm/K mol) x (298 K)) = 0.0373646 mol CO2

(11.6 g) / (1.00 L) x (0.3000 L) = 3.38 g of the mixture in the titrated sample

Let z be the mass (in grams) of Na2CO3 in the titrated sample.
Then 3.38-z is the mass of Na2CO3 in the titrated sample.

z / (105.9884 g Na2CO3/mol) x (1 mol CO2 / 1 mol Na2CO3) = (0.00943499 z) mol CO2 from Na2CO3

(3.38-z) / (84.0066 g NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = (0.040354 - 0.0119038 z) mol CO2 from NaHCO3

Add the moles of CO2 and set the sum equal to the total moles of CO2 as calculated above:
(0.00943499 z) + (0.040354 - 0.0119038 z) = 0.0373646
Solve for z algebraically:
z = 1.21087 g Na2CO3
3.38-z = 2.17913 g NaHCO3

(1.21087 g Na2CO3) / (105.9884 g Na2CO3/mol) / (0.3000 L) = 0.0381 mol/L Na2CO3
(2.17913 g NaHCO3) / (84.0066 g NaHCO3/mol) / (0.3000 L) = 0.0865 mol/L NaHCO3