Question

If a weak acid is 25% deprotonated at ph 4, what is the pka?if a weak acid is 25% deprotonated at ph 4, what is the pka?

Answer 1

4.48
pH=pKa+log([A-/HA])

25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.

4 = pKa + log$\frac{.25}{.75}$
4 - log$\frac{.25}{.75}$ = pKa
4.48=pKa

Answer 2

Answer:

pKa = 4.5

Explanation:

The weak acid can be represented by the general formula, HA and the dissociation equilibrium given as:

$HA \rightleftharpoons H^{+}+A^{-}$

where HA = protonated form

A- = deprotonated form

The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of the concentrations of HA and A- as;

$pH = pKa + log\frac{[A-]]}{[HA]]}-----(1)$

It is given that:

% deprotonated i.e. A- = 25%

Therefore, %protonated i.e. HA = 100 -25 = 75%

pH = 4

Based on equation (1)

$4 = pKa + log\frac{[25]]}{[75]]} = pKa-0.477$

pKa = 4.477 i.e. around 4.5