In the laboratory, a general chemistry student measured the pH of a 0.552 M aqueous solution...

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Chemistry

In the laboratory, a general chemistry student measured the pHof a 0.552 M aqueous solution of hydrofluoric acid to be 1.685. Usethe information she obtained to determine the Ka for this acid.Ka(experiment) =

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4.5 Ratings (612 Votes)

HF is a weak acid.

For a weak acid HA,

HAÂ Â Â Â Â Â Â Â Â +Â Â Â Â Â Â Â Â Â Â Â Â H₂OÂ Â Â Â Â Â Â =Â Â Â Â Â Â Â Â Â Â Â Â H₃O⁺Â Â Â Â Â +Â Â Â Â Â Â Â Â Â Â Â Â A^-

CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

C(1-Î±)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Î±CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Î±C

K = [H₃O⁺][ A^-]/[ HA][ H₂O]

Ka = K[H₂O] = [H₃O⁺][ A^-]/[ HA]

Ka = (Î±C)²/ C(1-Î±) = Î±²CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (as Î±<<1)

Î± = âˆš (Ka/C)

[H₃O⁺] = Î±C = C * âˆš (Ka/C) = âˆš (K_aC)

pH = - log[H₃O⁺] = - log âˆš (K_aC)

= 0.5 ( pK_a - logC)

Here, pH = 1.685 = 0.5 ( pK_a - log0.552)

or, 3.37 = pK_a + 0.258

or, pK_a = 3.112

or, K_a = 10^-3.112 = 7.73 * 10^Â-4

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