Given
C = 0.1 M
pH = 4.50
\(\therefore\) pH = – log [H+]
or [H+] = 10–pH = 10–4.50
= 3.16 × 10–5
[H+] = [A–] = 3.16 × 10–5
For the reaction,
HA ⇌ H+ + A–
Ka = \(\frac{[H^+][A^-]}{[HA]}\)
= \(\frac{(3.16\times10^{-15})(3.16\times10^{-5})}{0.1}\)
[∵ [HA]equi = 0.1 – (3.16 × 10–5 ) ~ 0.1]
= 1.0 × 10–8
\(\because\) pKa = – log [Ka]
= – log (1 × 10–8 )
= 8 log 10
= 8