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What is the pH of a 0.186 M aqueous solution of sodium fluoride, NaF? (Ka for...

What is the pH of a 0.186 M aqueous solution of sodium fluoride, NaF? (Ka for HF = 7.2×10-4)

Solutions

Expert Solution

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

          0.186M                            0.186M

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

I      0.186                                0       0

C    -x                                      +x      +x

E    0.186-x                               +x     +x

Kb    = [HF][OH-]/[F-]

Kb   = KW/Ka

          = 1*10^-14/7.2*10^-4   = 1.4*10^-11

Kb    = [HF][OH-]/[F-]

1.4*10^-11    = x*x/0.186-x

1.4*10^-11*(0.186-x) = x^2

x = 1.6*10^-6

[OH-] = x = 1.6*10^-6M

POH   = -log[OH-]

          = -log1.6*10^-6

         = 5.7958

PH   = 14-POH

         = 14-5.7958 = 8.2042>>>>>.answer


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