[Sis290025]

A 0.062 M solution of an unknown acid (HA) has a pH of 3.78. Calculate the pKa of this acid.


I don’t know if this is the correct way of solving this, so please point me in the right direction.

HA <-> A- + H+, where

[HA] = 0.062 M – x
[A-] = x
[H+] = x

x = [H+] = antilog(-3.78) = 1.66*10^-4 M

[HA] = 0.062 M – (1.66*10^-4) = 0.06183 M

pH = pKa + log[A-]/[HA]

pKa = pH – log[A-]/[HA] = 3.78 – log [(1.66*10^-4M)]/[0.06183] = 6.35



OR

Ka = [H+][A-]/[HA] = (1.66*10^-4 M)^2/(0.06183 M) = 4.46*10^-7

pKa = -log (4.46*10^-7) = 6.35

Thank you.

 

[Borek]

Twice correct.

Note that in fact you have not used two different methods, as Henderson-Hasselbalch equation is nothing else but rearranged dissociation constant.