Question

Sodium fluoride is added to many municipal water supplies to reduce tooth decay. Calculate the pH of a 0.00307 M solution of NaF, given that the K_a of HF= 6.80 x 10^-4 at 25 degree celcius.

Answer 1

                 NaF -------> Na⁺   + F^-

               0.00307M                0.00307M

             F^- + H2O --------> HF + OH^-

I           0.00307                  0       0

C           -x                          +x     +x

E        0.00307-x                +x     +x

Kb   = Kw/Ka

          = 1*10^-14/6.8*10^-4

            = 1.47*10^-11

     Kb   = [HF]{OH-]/[HF]

      1.47*10^-11    = x*x/0.00307-x

    1.47*10^-11 *(0.00307-x) = x²

           x = 2.12*10^-7

      [OH-]   = x   = 2.12*10^-7 M

     POH   = -log[OH-]

                = -log2.12*10^-7

                = 6.6736

      PH    = 14-POH

               = 14-6.6736   = 7.3264