Determine [OH] of solution?

Carbonate ion undergoes hydrolysis in water solution creating the following equilibrium:CO32-(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 CO32-(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-]} / [CO32-]2.0 × 10^-4 = {[X][X]} / [0.120](2.0 × 10^-4)[0.120] = X^20.24 × 10^-4 = X^20.49 × 10^-2 = X[OH-] = 4.9 × 10^-3 MpOH = – log [OH-]pOH = – log [4.9 × 10^-3]pOH = – [-2.30980392]pOH = 2.31pH + pOH = 14.00pH = 14.00 – pOHpH = 14.00 – 2.31pH = 11.69Answer: The [OH-] of a solution that is .120 M in CO3^2- is 4.9 × 10^-3 M. The pH of this solution is 11.69.The pOH of this solution is 2.31.... Show More

I don’t know where you got the 2.0×10^-4 but that is NOT the Kb for this reaction. The Kb for CO3^2- is 1.8×10-4... Show More

Kb is wrong. Kb is actually 1.8×10^-4... Show More