To 1.00 L of a 0.100 M aqueous solution of benzoic acid (C₆H₅COOH) is added 1.00 mL of 12.0 M HCl. What is the percentage ionization of the benzoic acid in the resulting solution [K_a(C₆H₅COOH) = 6.5 * 10^-5]

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Quizplus · Accepted Answer

The addition of a strong acid (HCl) to the solution of benzoic acid will suppress the ionization of the benzoic acid due to the common ion effect. The concentration of HCl added is very small compared to the initial concentration of benzoic acid (0.012 mol/L vs. 0.100 mol/L), so the change in pH will be minimal, but it will still decrease the ionization percentage of benzoic acid. Given the Ka value for benzoic acid (6.5 x 10^-5), the percentage ionization in the presence of added HCl will be significantly lower than in its absence, leading to a very small percentage ionization, which is closest to the option D (0.52%).

To 100 L of a 0