Answer:
69.3%
Explanation:
The question should read as follows:
A weak acid, HA, has a pKa of 4.357. If a solution of this acid has a pH of 4.005, what percentage of the acid is not ionized? Assume all H⁺ in the solution came from the ionization of HA.
The Henderson-Hasselbalch equation relates the pKa and pH of a solution to the ratio of ionized (A⁻) and unionized (HA) forms of a weak acid:
pH = pKa + log ([A⁻]/[HA])
Substituting and solving for [A⁻]/[HA]:
4.005 = 4.3574 = log([A⁻]/[HA])
-0.3524 = log([A⁻]/[HA])
[A⁻]/[HA] = 0.444/1
The percentage of acid that is not ionized (i.e. the percentage of acid in the HA form) is calculated:
[HA]/([A⁻] + [HA]) x 100% = 1/(1+0.444) x 100% = 69.3%
