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In the laboratory, a general chemistry student measured the pH of a 0.579 M aqueous solution...


In the laboratory, a general chemistry student measured the pH of a 0.579 M aqueous solution of hydrocyanic acid to be 4.833.
In the laboratory, a general chemistry student measured the pH of a 0.579 M aqueous solution of hydrocyanic acid to be 4.833. Use the information she obtained to determine the K, for this acid. K(experiment)
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use:

pH = -log [H+]

4.833 = -log [H+]

[H+] = 1.469*10^-5 M

HA dissociates as:

HA -----> H+ + A-

0.579 0 0

0.579-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 1.469*10^-5*1.469*10^-5/(0.579-1.469*10^-5)

Ka = 3.727*10^-10

Answer: 3.73*10^-10

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