Question

You are given 10.00mL of a solution of an unknown acid. The pH of this solution is exactly 5.62. You determine that the concentration of the unknown acid was 0.1224 M. You also determined that the acid was monoprotoic (HA). What is the K_a and pK_a of your unknown acid?

Answer 1

Given, 10.0 mL of 0.1224 M monoprotic acid

So, [HA] = 0.1224 M

pH = -log [H+] = 5.62

[H+] = 10^(-5.62)

[H+] = 2.4 x 10^-6 M

Now, let us put ICE table for the acid.

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

I          0.1224                 0         0

C          -x                       +x       +x

E         0.1224-x               +x       +x

But, [H+] = x = 2.4 x 10^-6

So, [HA] = 0.1224 - (2.4 x 10^-6) = 0.1223 M

[A-] = 2.4 x 10^-6 M

[H+] = 2.4 x 10^-6 M

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

Ka = (2.4 x 10^-6)^2 / 0.1224

Ka = 4.70 x 10^-11

Pka = -log Ka

= -log (4.70 x 10^-11)

= 10.33