What is the pH of a mixture that is 0.15 M in HF and 0.15 M in NaF?

Chemistry
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What is the pH of a mixture that is 0.15 M in HF and 0.15 M in NaF?

May 13th, 2015

Equilibrium: HF(aq) + H2O(l) <=> H3O^+(aq) + F^-(aq)
[HF] = 0.15 - x; [H3O^+] = x; [F^-] = 0.15 + x

Literature value of Ka = 7.20 x 10^-4 = [H3O^+][F^-]/[HF] =
(x)(0.15 + x)/(0.15- x) = 7.20 x 10^-4
0.15(x) + x^2 = 7.20 x 10^-4(0.15 - x) = 1.66 x 10^-4 - 7.20 x 10^-4(x)
This can be solved as a quadratic equation. See the 2nd link.
x^2 + 0.15(x) - 1.66 x 10^-4 = 0
x = 1.10 x 10^-3 M = [H3O^+]

pH = -log[H3O^+] = -log(1.10 x 10^-3) = 2.96

May 13th, 2015

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