What is the pH of a 0.16 M solution of Benzoic acid, pKa= 4.202

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Question

Answer 1

Answer #1

Initial concentration of benzoic acid = [C₆H₅COOH]_initial = 0.16 M

pK_a benzoic acid = 4.202

K_a = 10^-pK_a

K_a = 10^-4.202

K_a = 6.28 x 10^-5

K_a = [C₆H₅COO^-]_eq[H⁺]_eq / [C₆H₅COOH]_eq

6.28 x 10^-5 = [(x) * (x)] / (0.16 M - x)

Solving for x, x = 3.14 x 10^-3 M

[H⁺] = x = 3.14 x 10^-3 M

pH = -log[H⁺]

pH = -log(3.14 x 10^-3 M)

pH = 2.50

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Answer 2

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