Question

Part A

Determine the [OH−] of a solution that is 0.140 M in CO3.

Express your answer using two significant figures.

Part B

Determine pH of this solution.

Express your answer to two decimal places.

Answer 1

Part A:

Carbonate ion undergoes hydrolysis in water solution creating the following equilibrium:

CO₃^2-(aq) + H₂O(l) ↔ HCO₃^-(aq) + OH^-(aq)

The Kb constant for this reaction is 2.0 × 10^-4. The bicarbonate ion also undergoes hydrolysis creating a secondary equilibrium system:

HCO₃^-(aq) + H₂O(l) ↔ H₂CO₃(aq) + OH^-(aq)

The Kb constant for the second reaction is 2.5 × 10^-8.

Since the second equilibrium constant is so much smaller than the first, the concentration of hydroxide ions generated by the second reaction may be considered neglibible. For the purpose of this problem, let us assume that only the first equilibrium takes place. Due to the small size of the equilibrium constant (2.0 × 10^-4), we can assume that the original concentration of CO₃^2-(aq) ion remains virtually the same. The concentrations of both HCO₃^-(aq) and OH^-(aq) ions are equal to each other or to "X". The equilibrium constant expression for this reaction would be:

Kb = {[HCO₃^-][OH^-]} / [CO₃^2-]
2.0 × 10^-4 = {[X][X]} / [0.140]
(2.0 × 10^-4)[0.120] = X²
2.8 × 10^-5 = X²
5.29 × 10^-3 = X
[OH^-] = 5.29 × 10^-3 M

Part B:

pOH = - log [OH^-]
pOH = - log [5.29 × 10^-3]
pOH = - [-2.28]
pOH = 2.28

pH + pOH = 14.00
pH = 14.00 - pOH
pH = 14.00 - 2.28
pH = 11.72