Question

Benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(\mathrm{p} K_{\mathrm{a}} 4.19\right)\), is only slightly soluble in water, but its sodium salt, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\), is quite soluble in water. In which solution(s) will benzoic acid dissolve? (a) Aqueous \(\mathrm{NaOH}\) (b) Aqueous \(\mathrm{NaHCO}_{3}\) (c) Aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)

Short answer

Answer: Benzoic acid will dissolve in (a) aqueous NaOH and (b) aqueous NaHCO3, but not in (c) aqueous Na2CO3.

Step-by-step solution

Understand the Reactants

The reactants given in the problem are Benzoic acid (\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\)) and three possible solutions: aqueous \(\mathrm{NaOH}\), aqueous \(\mathrm{NaHCO}_{3}\), and aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). The first solution is a strong base, while the second and third solutions are weak bases.

Determine the \(\mathrm{p}K_{\mathrm{a}}\) of the Bases

To see if these solutions will react with benzoic acid, we need to compare their basicity. A common way of doing this is by looking at the \(\mathrm{p}K_{\mathrm{b}}\) values, which represent the basicity of the base. For \(\mathrm{NaOH}\), since it is a strong base, it does not have a \(\mathrm{p}K_{\mathrm{b}}\) value, but we can infer that it is a strong base due to its classification. For the bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)), its conjugate acid is \(\mathrm{H}_{2} \mathrm{CO}_{3}\), which has a \(\mathrm{p}K_{\mathrm{a}}\approx 6.37\). The carbonate ion (\(\mathrm{CO}_{3}^{2-}\)) has a conjugate acid of \(\mathrm{HCO}_{3}^{-}\), with a \(\mathrm{p}K_{\mathrm{a}}\approx 10.33\). To find the \(\mathrm{p}K_{\mathrm{b}}\) values for these weak bases, we can use the following relationship: \(\mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} = 14\). Thus, for \(\mathrm{HCO}_{3}^{-}\), \(\mathrm{p}K_{\mathrm{b}} \approx 7.63\), and for \(\mathrm{CO}_{3}^{2-}\), \(\mathrm{p}K_{\mathrm{b}} \approx 3.67\).

Compare the pKa and pKb values

In order to determine whether benzoic acid will dissolve in these given solutions, we must compare the \(\mathrm{p}K_{\mathrm{a}}\) of benzoic acid (\(4.19\)) to the \(\mathrm{p}K_{\mathrm{b}}\) of the weak bases. Should the \(\mathrm{p}K_{\mathrm{b}}\) be higher than the \(\mathrm{p}K_{\mathrm{a}}\), benzoic acid will dissolve in the weak base solution as it forms more water-soluble products.

Identify the Solubility of Benzoic Acid in Each Solution

(a) Aqueous \(\mathrm{NaOH}\): Since \(\mathrm{NaOH}\) is a strong base, it will react with benzoic acid and dissolve it, forming a soluble benzoate ion and water. (b) Aqueous \(\mathrm{NaHCO}_{3}\): The \(\mathrm{p}K_{\mathrm{b}}\) of \(\mathrm{HCO}_{3}^{-}\) is \(7.63\), which is greater than the \(\mathrm{p}K_{\mathrm{a}}\) of benzoic acid. Since the \(\mathrm{p}K_{\mathrm{b}}\) is higher than the \(\mathrm{p}K_{\mathrm{a}}\), benzoic acid will dissolve in \(\mathrm{NaHCO}_{3}\) solution. (c) Aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\): The \(\mathrm{p}K_{\mathrm{b}}\) of \(\mathrm{CO}_{3}^{2-}\) is \(3.67\), which is less than the \(\mathrm{p}K_{\mathrm{a}}\) of benzoic acid. Therefore, benzoic acid will not dissolve in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution because the \(\mathrm{p}K_{\mathrm{b}}\) value is lower than \(\mathrm{p}K_{\mathrm{a}}\). Final answer: Benzoic acid will dissolve in (a) Aqueous \(\mathrm{NaOH}\) and (b) Aqueous \(\mathrm{NaHCO}_{3}\), but not in (c) Aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).