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First recognize that NaF is the salt of a strong base (NaOH) and a weak acid (HF), so the pH will be alkaline. Next, look at the hydrolysis of NaF: NaF + H2O ---> NaOH + HF, or looking at it another way.... F^- + H2O ---> HF + OH- and here F^- acts as a base, so we need the Kb for NaF and that will be the inverse of the Ka for HF. The Ka for HF is 6.6x10^-4, so Kb = 1x10^-14/6.6x10^-4 = 1.5x10^-11.

Now, Kb = [HF][OH-]/[F-] = (x)(x)/(0.89) = 1.5x10^-11

x^2 = 1.3x10^-11

x = 3.6x10^-6 = [OH-]

pOH = -log 3.6x10^-6 = 5.44

pH = 8.6 (note the pH is alkaline, as expected)

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