Calculate the pH of a 0.0164 M aqueous solution
of acetylsalicylic acid (aspirin)
(HC9H7O4,
Ka = 3.0×10-4).
pH =
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aspirin is weak acid dissociate in water as
HC9H7O4 + H2O C9H7O-4 + H3O+
Ka = [C9H7O-4] [H3O+] / [HC9H7O4]
but [C9H7O-4] = [H3O+]
consider [C9H7O-4 ] = [ H3O+] = X
then Ka = [X] [X] / [HC9H7O4]
Ka = X2 / [HC9H7O4 ]
X2 = Ka X [HC9H7O4]
substitute the value
X2 = (3.0 X 10-4) X (0.0164) = 4.92 X 10-6
X = 0.0022181 M
[C9H7O-4 ] = [H3O+] = X = 0.0022181 M
[H3O+] = 0.0022181 M
pH = -log [H3O+] = -log 0.0022181 = 2.65
pH of 0.0164 M HC9H7O4 = 2.65
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