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.
           Â
HAÂ Â <-----------> A- +Â Â H+
IÂ Â Â Â Â Â Â Â Â
0.1224Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â 0
CÂ Â Â Â Â Â Â Â Â
-x                     Â
+x      +x
EÂ Â Â Â Â Â Â Â
0.1224-x             Â
+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
