Calculate the ph of a solution formed by mixing 65 ml of 0.20 m nahco3 with 75 ml of 0.15 m na2co3.

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