Question

What is the pH of a 1.0 M solution of potassium formate, KHCO₂? (No equilibrium constant given).

Answer 1

             HCO2^- + H2O ------------------> HCO2H + OH^-

I            1                                                   0                 0

C          -x                                                 +x                +x

E        1-x                                                  +x                +x

          Kb    = Kw/Ka

                  = 1*10^-14/1.8*10^-4

                   = 5.6*10^-11

     Kb   = [HCO2H][OH-]/[HCO2^-]

    5.6*10^-11   = x*x/(1-x)

5.6*10^-6*(1-x) = x^2

       x = 0.0024

    [OH-]   =   x = 0.0024M

    POH   = -log[OH-]

               = -log0.0024

               = 2.6197

     PH      = 14-POH

                 = 14-2.6197

                  = 11.3803 >>>>>answer