Question

Determine the [OH−] of a solution that is 0.140 M in CO3 -2.

Answer 1

The hydrolysis of carbonate ion is

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

The Kb constant for this reaction is 1.8 X 10^-4. The bicarbonate ion formed in the above reaction also undergoes hydrolysis

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

The Kb constant for the reaction is 2.27 X 10^-8.

We may ignore the second equilibrium as the Kb value is very low and it will not product significant amount of hydroxide ions

For first equilibrium

Kb = {[HCO3-][OH-]} / [CO₃^2-]
2.0 X 10^-4 = {[x][x]} / [0.140-x ], we can ignore x in denominator
2.0 X 10^-4 = {[x][x]} / [0.140
x = 5.29 X 10^-3 M
pOH = - log [OH-] = 2.277
pH + pOH = 14.00
pH = 14.00 - pOH
pH = 14.00 - 2.277
pH = 11.723