Question

Calculate the pH of a 0.770 L solution of a 0.5220 M aqueous solution of acid HA at 25oC. Enter the answer with two decimal places. Ka = 4.71x10-6

Answer 1

HA dissociates as:

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

0.522 0 0

0.522-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.71*10^-6)*0.522) = 1.568*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.568*10^-3 M

So, [H+] = x = 1.568*10^-3 M

use:

pH = -log [H+]

= -log (1.568*10^-3)

= 2.8047

Answer: 2.80