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