Question

At ${25}^{\circ }\mathrm{C}$, a saturated solution of benzoic acid $(K_{\mathrm{a}}=6.4\times {10}^{-5})$ has a pH of $2.80$. Calculate the water solubility of benzoic acid in moles per liter.

Short answer

The solubility of benzoic acid in water at 25°C can be calculated using the given Ka value and the pH of the solution. First, determine the concentration of hydronium ions ($[{\mathrm{H}}_3{\mathrm{O}}^{+}]$) using the pH value: $[{\mathrm{H}}_3{\mathrm{O}}^{+}]={10}^{-2.80}$. Then, use the Ka expression $\frac{[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}][{\mathrm{H}}_3{\mathrm{O}}^{+}]}{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}]}$ and the relationship $[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}]=[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to find the solubility (S) in moles per liter: $6.4\times {10}^{-5}=\frac{[{\mathrm{H}}_3{\mathrm{O}}^{+}{]}^2}{S-[{\mathrm{H}}_3{\mathrm{O}}^{+}]}$.

Step-by-step solution

Write the chemical equation

Write down the equilibrium reaction for the benzoic acid (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$) in water. The benzoic acid donates a proton to the water, resulting in the formation of benzoate ion (${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}$) and hydronium ion (${\mathrm{H}}_3{\mathrm{O}}^{+}$): $${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$$

Relate pH to the concentration of hydronium ions

Calculate the concentration of hydronium ions ($[{\mathrm{H}}_3{\mathrm{O}}^{+}]$) to find the concentration of benzoic acid in the solution using the pH value: $$[{\mathrm{H}}_3{\mathrm{O}}^{+}]={10}^{-\mathrm{pH}}$$ Given a pH of 2.80, the concentration of hydronium ions can be calculated as follows: $$[{\mathrm{H}}_3{\mathrm{O}}^{+}]={10}^{-2.80}$$

Use the Ka expression

Write down the Ka expression for benzoic acid and look for relationships between the variables: $$K_a=\frac{[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}][{\mathrm{H}}_3{\mathrm{O}}^{+}]}{[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}]}$$ Since the benzoic acid, when dissociated, forms a 1:1 ratio between the benzoate ion ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}$ and the hydronium ion ${\mathrm{H}}_3{\mathrm{O}}^{+}$, we can write the concentrations as follows: $$60$$ $$[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}]=[{\mathrm{H}}_3{\mathrm{O}}^{+}]$$ Let the solubility of benzoic acid in water be S mol/L. Then, $$[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}]=S-[{\mathrm{H}}_3{\mathrm{O}}^{+}]$$ Substitute these expressions in the Ka equation: $$K_a=\frac{[{\mathrm{H}}_3{\mathrm{O}}^{+}{]}^2}{S-[{\mathrm{H}}_3{\mathrm{O}}^{+}]}$$

Solve for benzoic acid solubility

Use the given Ka value of $6.4\times {10}^{-5}$ and the calculated $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ value to solve for the solubility (S) in moles per liter: $$6.4\times {10}^{-5}=\frac{[{\mathrm{H}}_3{\mathrm{O}}^{+}{]}^2}{S-[{\mathrm{H}}_3{\mathrm{O}}^{+}]}$$ Once calculated, the value of S will represent the solubility of benzoic acid in moles per liter.

Key concepts

Benzoic Acid

Benzoic acid is a simple aromatic carboxylic acid with the chemical formula ${\text{C}}_6{\text{H}}_5\text{COOH}$. It is a white crystalline solid that is slightly water-soluble.
This acid is found naturally in several plants and serves as the starting material for the synthesis of many other chemical substances. In water, benzoic acid can partially dissociate, which leads to its ability to donate a proton (${\text{H}}^{+}$) and form the benzoate ion ${\text{C}}_6{\text{H}}_5{\text{COO}}^{-}$. This reaction is reversible and can be written as:

• ${\text{C}}_6{\text{H}}_5\text{COOH (aq)}\rightleftharpoons {\text{C}}_6{\text{H}}_5{\text{COO}}^{-}(aq)+{\text{H}}_3{\text{O}}^{+}(aq)$

The extent to which benzoic acid dissociates is linked to its water solubility and the pH of the solution.

pH

pH is a numeric scale or unit used to specify the acidity or basicity of an aqueous solution. It is based on the concentration of hydrogen ions (${\text{H}}^{+}$) and is calculated using the formula:

• $\text{pH}=-{\log }_{10}[{\text{H}}_3{\text{O}}^{+}]$

In the context of benzoic acid dissolving in water at a pH of 2.80, this pH value indicates a notably acidic solution, meaning there's a high concentration of ${\text{H}}_3{\text{O}}^{+}$ ions over ${\text{OH}}^{-}$ ions.
Knowing the pH helps us infer how much benzoic acid dissociates into benzoate ions and hydronium ions. With a pH of 2.80, we can determine the concentration of ${\text{H}}_3{\text{O}}^{+}$ present in the solution through the use of logarithmic relationships.

Acid Dissociation Constant

The acid dissociation constant, denoted as $K_a$, is crucial for understanding the strength of an acid in solution. It provides a quantitative measure of the extent of acid dissociation, that is, how much the acid breaks into its ions in a solution.
The expression for the $K_a$ of benzoic acid is given by:

• $K_a=\frac{[{\text{C}}_6{\text{H}}_5{\text{COO}}^{-}][{\text{H}}_3{\text{O}}^{+}]}{[{\text{C}}_6{\text{H}}_5\text{COOH}]}$

For benzoic acid, $K_a$ is $6.4\times {10}^{-5}$, which indicates a weak acid. This constant helps establish the relationship between the concentrations of the different species in the solution, namely the benzoate ion, hydronium ion, and undissociated benzoic acid.
This calculation is vital in determining solubility since the higher the $K_a$, the more the benzoic acid will dissociate, suggesting greater solubility under the same conditions.

Hydronium Ion Concentration

Hydronium ion concentration, symbolized as $[{\text{H}}_3{\text{O}}^{+}]$, is a cornerstone in the calculation of pH and the dissociation of acids in solution. It can be directly derived from the pH value using the formula:

• $[{\text{H}}_3{\text{O}}^{+}]={10}^{-\text{pH}}$

Given that the pH of the benzoic acid solution is 2.80, the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions becomes critical to determining both the state of equilibrium and the solubility of benzoic acid.
This concentration is used in the $K_a$ expression to directly correlate with how much benzoic acid solubilizes and dissociates in the aqueous environment.
Once computed, $[{\text{H}}_3{\text{O}}^{+}]$ not only guides the pH but also anchors other calculations, especially when solving for the equilibrium conditions of weak acid solutions like that of benzoic acid.