The \rm p \it K_{\rm a} of this reaction is 4.2. In a 0.74\it M solution of benzoic acid, what percentage of the molecules are ionized?
To calculate the percentage of ionized molecules in a solution of benzoic acid, you need to consider the acid dissociation constant (\(K_a\)) and the concentration of the acid in solution.
The acid dissociation constant (\(K_a\)) is a measure of the strength of an acid in aqueous solution. It quantifies the extent to which an acid dissociates into its ions. In this case, the \(K_a\) of the reaction is given as 4.2.
To calculate the percentage of ionized molecules, you can use the following equation:
\[
\text{{% of ionized molecules}} = \frac{{[\text{{ionized form}}]}}{{[\text{{undissociated form}}]}} \times 100
\]
In this case, the "ionized form" refers to the concentration of the dissociated benzoic acid (\([A^-]\)), and the "undissociated form" refers to the concentration of the undissociated benzoic acid (\([HA]\)).
Step 1: Calculate the concentration of undissociated benzoic acid (\([HA]\)).
Given that the concentration of the benzoic acid solution is 0.74 M, the concentration of the undissociated form (\([HA]\)) is also 0.74 M.
Step 2: Calculate the concentration of ionized benzoic acid (\([A^-]\)).
Since the acid ionizes according to the reaction: \(HA \rightarrow H^+ + A^-\), the concentration of the ionized form (\([A^-]\)) is dependent on the \(K_a\).
Using the equation: \(K_a = \frac{{[H^+][A^-]}}{{[HA]}}\), we can rearrange it to solve for \([A^-]\):
\([A^-] = \frac{{K_a \times [HA]}}{{[H^+]}}\)
In this case, since the concentration of \(H^+\) (hydronium ion) is unknown, we can assume it to be negligible, and thus, we take it as 0.
\([A^-] = \frac{{K_a \times [HA]}}{{0}} = 0\)
Step 3: Calculate the percentage of ionized molecules.
\[
\text{{% of ionized molecules}} = \frac{{[A^-]}}{{[HA]}} \times 100 = \frac{{0}}{{0.74}} \times 100 = 0\%
\]
Therefore, in a 0.74 M solution of benzoic acid, the percentage of molecules that are ionized is 0%.