Question

A 0.30-molar solution of a weak monoprotic acid HA is 1.5% ionized. What is the value of K_a for this acid?

4.8 x 10^-6

6.9 x 10^-5

5.2 x 10^-3

3.7 x 10^-4

1.1 x 10^-5

PLEASE ANSWER ASAP MUCH APPRECIATED THANK YOU !!!!!!!!!

Answer 1

acid is weak acid.

concentration of weak acid = 0.30 M

and

1.5% ionized.

HA ..............> H+ + A-

thus

[H+] = [A-] = 0.30 * 1.5 / 100 = 4.5 * 10^-3 M

and

[HA] = (0.30 - 4.5 * 10^-3) = 0.2955 M

Now,

Ka = [H+][A-] / [HA]

or

Ka = (4.5 * 10^-3) * (4.5 * 10^-3) / 0.2955

or

Ka = 6.9 * 10^-5

option 6.9 x 10^-5 is the answer.