Question

In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 15.06 mL of 0.100 M of KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10-5 for acetic acid, CH3COOH.

Answer 1

no of moles of CH3COOH = molarity * volume in L

                                              = 0.1*0.025 = 0.0025 moles

no of moles of KOH             = molarity * volume in L

                                             = 0.1*0.01506   = 0.001506 moles

              CH3COOH + KOH -----------------> CH3COOK + H2O

I            0.0025             0.001506                      0

C          -0.001506       - 0.001506                     0.001506

E          0.000994            0                                 0.001506

         PKa = -logKa

                  = -log1.8*10^-5

                  = 4.75

    PH   = PKa + log[CH3COOK]/[CH3COOH]

          = 4.75 + log0.001506/0.000994

          = 4.75 + 0.1804

          = 4.9304 >>>>>>>>>>>>>>answer