Question

Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid ionizes in water: C6H5COOH⇌C6H5COO−+H+ The pKa of this reaction is 4.2. In a 0.51 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

1.1 percent

Step-by-step explanation:

C6H5COOH⇌C6H5COO+H

Ka = [C6H5COO-][H+]/[C6H5COOH]

From pKa of 4.2, Ka = 6.3x10^-5

6.3x10^-5 = (x)(x)/0.51-x and assuming x is small...

6.3x10^-5 = x^2/0.51

x^2 = 3.213x10^-5

x =5.67 x10^-3

(5.67x10^-3/0.51)x 100 =1.1 percent