Question

What is the pH of the solution which results from mixing 25mL of 0.2M CH3CO2 and 25ml of 0.20 KOH? ka= CH3CO2H=1.8E-5

additional questions: why do you use 5mmoles rather than 10? Why do you use kb rather than ka?

answer: 8.87

Thanks!

Answer 1

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

no of moles of CH3COOK = molarity * volume in L

                                            = 0.2*0.025   = 0.005 moles

CH3COOK ---------------> CH3COO^- + K^+

               CH3COO^-   + H2O ---------------> CH3COOH + OH^-

I            0.005                                                      0                0

C            -x                                                           +x               +x

E          0.005-x                                                     +x               +x

            Kb   = Kw/Ka

                     = 1*10^-14/1.8*10^-5    = 5.6*10^-10

     Kb   = [CH3COOH][OH-]/[CH3COO-]

     5.6*10^-10 = x*x/(0.005-x)

    5.6*10^-10 *(0.005-x) =x^2

   x= 1.67*10^-6

[OH-]   = x= 1.67*10^-6M

POH   = -log[OH-]

           = -log1.67*10^-6

          = 5.7772

PH    = 14-POH

          = 14-5.7772    = 8.2228