Question

Calculate the pH of a solution formed by mixing 65 mL of 0.16 M NaHCO3 with 75 mL of 0.29 M Na2CO3. Express your answer using two decimal places.

Answer 1

The species concerned with are HCO₃^- and CO₃^2-. CO₃^2- is a proton(H⁺) acceptor(basic) while HCO₃^- is amphoteric since it can behave as a proton donor(acidic) and a proton acceptor(basic). Since a base is present, HCO₃^- will behave as an acid in this case. We shall consider the dissociation of CO₃^2-:
CO₃^2-(aq)+H₂O(l)↔HCO₃^- (aq)+OH^-(aq)

Kb of CO₃^2-= 1.8x10^-4
pKb of CO₃^2-
= -lg(Kb)
= -lg(1.8x10^-4)
= 3.74

Kb= [HCO₃^- ][OH^-]/[CO₃^2-]
[OH^-]= (Kb)([CO₃^2-]/[HCO₃^- ]
pOH= pKb+lg([HCO₃^- ]/[CO₃^2-])

No of moles of HCO₃^-
= no of moles of NaHCO₃
= [NaHCO₃]x(volume of the NaHCO₃ solution in L)
= 0.16x65x10^-3
= 0.0104
[HCO₃^- ] in the resulting solution
= no of moles of HCO₃^- /volume of the resulting solution in L
= 0.0104/((75+65)x10^-3)
= 0.0743M

No of moles of CO₃^2-
= [Na₂CO₃]x(volume of the Na₂CO₃ solution in L)
= 0.29x75x10^-3
= 0.02175
[CO₃^2-] in the resulting solution
= no of moles of CO₃^2-/volume of the resulting solution in L
= 0.02175/((75+65)x10^-3)
= 0.1553M

pOH= pKb+lg([HCO₃^- ]/[CO₃^2-])
= 3.74+lg((0.0743/0.1553))
= 3.1003

pH= 14-pOH
= 14-3.1
= 10.89