According to the balanced equation of the reaction:
CO32-(aq) + H2O(l) ↔ HCO3-(aq) + OH-(aq)
no. of moles of HCO3- = molarity * volume
= 0.2 m* 0.065L
= 0.013 mol
[HCO3-] = moles / total volume
= 0.013 /(0.065L +0.075L)
= 0.093 M
no. of moles of CO32- = molarity * volume
= 0.15 m * 0.075L
= 0.01125 Moles
[CO3-2] = moles / total volume
= 0.01125 moles / (0.065L + 0.075L)
= 0.14 M
when the value of Kb of CO32- = 1.8 x 10^-4
∴Pkb = -㏒Kb
= - ㏒ 1.8 x 10^-4
= 3.7
by using H-H equation we can get the POH
∵ POH = Pkb + ㏒[CO32-][HCO3-]
∴POH = 3.7 +㏒(0.14/0.093)
= 3.88
∵PH = 14 - POH
∴ PH = 14 - 3.88
= 10.12
