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

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

Expert Solution

HF is a weak acid.

For a weak acid HA,

HA + H₂O = H₃O⁺ + A^-

C 0 0

C(1-α) αC αC

K = [H₃O⁺][ A^-]/[ HA][ H₂O]

Ka = K[H₂O] = [H₃O⁺][ A^-]/[ HA]

Ka = (αC)²/ C(1-α) = α²C (as α<<1)

α = √ (Ka/C)

[H₃O⁺] = αC = C * √ (Ka/C) = √ (K_aC)

pH = - log[H₃O⁺] = - log √ (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^-4

queen_honey_blossom answered 2 years ago

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