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Calculate the pH of a buffer that is 0.040 M HF and 0.020 M LiF. The Ka for HF is 3.5*10^-4.?

December 4, 2022 0

HF ===> H+ + F-

Ka = [H+][F-]/[HF] = 3.5 x10^-4

Let [H+] = x. Then [F-] = 0.020 + x, and [HF] = 0.040. But x is so small that we can forget it and say that [F-] = 0.020 M.

(x)(0.020)/(0.040) = 3.5 x 10^-4

x = 7.0 x 10^-4 = [H+]

pH = -Log[H+], and Log(7.0×10^-4) = -3.15, so pH = +3.15
pH = pKa + log [LiF]/[HF]

pH = 3.46 + log [0.02]/[0.04]

pH = 3.46 + log 0.5 = 3.46 + (-0.3) = 3.16
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