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In the laboratory, a general chemistry student measured the pH of a 0.532 M aqueous solution...

In the laboratory, a general chemistry student measured the pH of a 0.532 M aqueous solution of acetylsalicylic acid (aspirin), HC9H7O4 to be 1.885. Use the information she obtained to determine the Ka for this acid.

Ka(experiment) =

Expert Solution

pH = 1.885

concentration is given by

[HA] = 0.532 M

note that aspirin is a weak acid

HC9H7O4

will dissociate partially as follows

HC9H7O4(aq) <--> C9H7O4-(aq) + H+(aq)

now, this has an equilibirum value

Ka = [C9H7O4-][H+]/[HC9H7O4]

due to sotichioemtry:

[C9H7O4-] = 0

[H+] = 0

[HC9H7O4] = 0.532

in equilibirum:

[C9H7O4-] = 0 + x

[H+] = 0 + x

[HC9H7O4] = 0.532 - x

and we know

[H+] = 10^-pH = 10^-1.885 = 0.013031 M

now, substitute

[C9H7O4-] = 0 + x = 0.013031

[H+] = 0 + x = 0.013031

[HC9H7O4] = 0.532 - 0.013031 = 0.518969

Substitute in Ka

Ka = [C9H7O4-][H+]/[HC9H7O4]

Ka = (0.013031 )(0.013031 )/0.518969

Ka = 0.000327200

KA = 3.27*10^-4

queen_honey_blossom answered 2 years ago

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