Calculate the concentration of F- ions in a 2 M solution of hydrogen fluoride, HF. The Ka of HF is 6.8 x 10-4.
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Correct Answer: D. 3.7 x 10-2 M
The dissociation of HF is given by
HF -> (H+) + (F-)
Let [H+] = [F-] = x. Although the actual equilbrium concentration of HF is (2-x) M due to dissociation, we assume that x is very small and ignore the difference. That is, we take [HF] = 2 M at equilibrium. Since the equilbrium expression for this reaction is Ka = [H+][F-]/[HF], we have Ka = x2/2, so
x2 = 2Ka = 2(6.8 x 10-4) = 13.6 x 10-4, so x = 3.7 x 10-2.
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