Question

Propanoic acid, ch3ch2cooh, has a pka =4.9. draw the structure of the conjugate base of propanoic acid and give the ph above which 90% of the compound will be in this conjugate base form.

Answer 1

Conjugate base of Propanoic acid ($CH_{3}CH_{2}COOH$ is propanoate where -COOH group gets converted to -CO$O^{1-}$. The structure of conjugate base of Propanoic acid is shown in the diagram.

The $p^{H}$ above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.

$p^{H}$= $p^{k_{a}$+ log$\frac{[Conjugate base]}{[acid]}$=4.9+log$\frac{90}{10}$=5.85

As 90% conjugate base is present, so propanoic acid present 10%.