Question

A solution is prepared by dissolving 11.3 g of a mixture of sodium carbonate and sodium bicarbonate in 1.00 L of water. A 300.0 cm3 sample 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. Find the molarity of the carbonate in the solution and the molarity of the bicarbonate in the solution.

Answer 1

Sodium bicarbonate reaction

NaHCO₃ + HNO₃ -> NaNO₃ + H₂O + CO₂

Sodium carbonate reaction:

Na₂CO₃ + 2HNO₃ -> 2NaNO₃ + H₂O + CO₂

We have 0.940L of CO2, which we can turn to moles assuming ideal gas behavior:

n = PV/RT

n = 0.972 atm * 0.94 L / 0.082 Latm/Kmol * 298K

n = 0.03739 moles of CO2

As we have a one on one mol relationship on stoichiometry of both reactions, we can assume the production is 50-50% for both reactions, hence, the molarity of both will be, first, in the 0.3L:

M = 0.18695 moles/0.3L

M = 0.0623 M

Now we apply M1V1 = M2V2

0.0623M * 0.3L = M2 * 1L

M2 = 0.01869M