Question

Calculate the pH of a 0.0164 M aqueous solution of acetylsalicylic acid (aspirin) (HC₉H₇O₄, K_a = 3.0×10^-4).

pH =

Submit Answer

Answer 1

aspirin is weak acid dissociate in water as

HC₉H₇O₄ + H₂O    $\rightleftharpoons$   C₉H₇O^-₄ + H₃O⁺

K_a = [C₉H₇O^-₄] [H₃O⁺] / [HC₉H₇O₄]

but [C₉H₇O^-₄] = [H₃O⁺]

consider [C₉H₇O^-₄ ] = [ H₃O⁺] = X

then K_a = [X] [X] / [HC₉H₇O₄]

K_a = X² / [HC₉H₇O₄ ]

X² = K_a 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