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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? (K_a for HF = 7.2×10^-4)

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

queen_honey_blossom answered 2 years ago

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