Determine the ph of a 0.22 m naf solution at 25°c. the ka of hf is 3.5 × 10-5.

When NaF is in aqueous solution it dissociates into ions and reacts with water forming NaOH and HF. The solution would be a mixture of a strong base and a weak acid. Both of these substances contribute to the pH of the solution. We calculate pH as follows:

Ka + Kb = 1x10^-14
Kb = 1x10^-14 - 3.5x 10 ^-5
Kb = 6.5 x10^-5Kb = [Na+] [OH-] / [NaF]

We let x be the concentration of Na in equilbrium,

Kb = (x) (x) /0.22
6.5 x10^-5 = x^2 /0.22
x = 3.78x10^-3 = [OH]pOH = -log [OH]
pOH = 2.42

pH + pOH = 14
pH = 14 - pOH 
pH = 14 - 2.42 
pH = 11.58

Therefore, the pH of the solution would be 11.58.

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