Question

19. A. In the laboratory, a general chemistry student measured the pH of a 0.464 M aqueous solution of benzoic acid, C₆H₅COOH to be 2.254.

Use the information she obtained to determine the K_a for this acid.  

K_a(experiment) = _____

B. In the laboratory, a general chemistry student measured the pH of a 0.498 M aqueous solution of acetylsalicylic acid (aspirin), HC₉H₇O₄to be 1.897.

Use the information she obtained to determine the K_a for this acid.  

K_a(experiment) = _____

C. In the laboratory, a general chemistry student measured the pH of a 0.503 M aqueous solution of formic acid, HCOOH to be 2.038.

Use the information she obtained to determine the K_a for this acid.  

K_a(experiment) = _____

Answer 1

A)

use:

pH = -log [H+]

2.254 = -log [H+]

[H+] = 5.572*10^-3 M

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.464 0 0

0.464-x x x

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

Ka = x*x/(c-x)

Ka = 5.572*10^-3*5.572*10^-3/(0.464-5.572*10^-3)

Ka = 6.772*10^-5

Answer: 6.77*10^-5

B)

use:

pH = -log [H+]

1.897 = -log [H+]

[H+] = 1.268*10^-2 M

HC9H7O4 dissociates as:

HC9H7O4 -----> H+ + C9H7O4-

0.498 0 0

0.498-x x x

Ka = [H+][C9H7O4-]/[HC9H7O4]

Ka = x*x/(c-x)

Ka = 1.268*10^-2*1.268*10^-2/(0.498-1.268*10^-2)

Ka = 3.311*10^-4

Answer: 3.31*10^-4

C)

use:

pH = -log [H+]

2.038 = -log [H+]

[H+] = 9.162*10^-3 M

HCOOH dissociates as:

HCOOH -----> H+ + HCOO-

0.503 0 0

0.503-x x x

Ka = [H+][HCOO-]/[HCOOH]

Ka = x*x/(c-x)

Ka = 9.162*10^-3*9.162*10^-3/(0.503-9.162*10^-3)

Ka = 1.7*10^-4

Answer: 1.70*10^-4