Problem 50 - Chemistry and Chemical Reactivity

Question

Phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right),\) commonly called carbolic acid, is a weak organic acid. $$\begin{aligned} &\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\\\ &&K_{\mathrm{a}}=1.3 \times 10^{-10}\end{aligned}$$ If you dissolve \(0.195 \mathrm{g}\) of the acid in enough water to make 125 mL of solution, what is the equilibrium hydronium ion concentration? What is the \(\mathrm{pH}\) of the solution?

Step-by-Step Solution

Answer

[H₃O⁺] ≈ 1.467 × 10⁻⁶ M; pH ≈ 5.83.

Step 1: Calculate Molarity of Phenol

First, determine the moles of phenol by using its molar mass. The molar mass of phenol (C₆H₅OH) is approximately 94.11 g/mol. Thus, the moles of phenol are calculated as follows:\[ \text{Moles of phenol} = \frac{0.195 \text{ g}}{94.11 \text{ g/mol}} \approx 0.002072 \text{ mol} \]Then, calculate the molarity (M) by dividing the moles by the volume of the solution in liters. The solution volume is 125 mL or 0.125 L:\[ \text{Molarity} = \frac{0.002072 \text{ mol}}{0.125 \text{ L}} \approx 0.016576 \text{ M} \]

Step 2: Set-up the Equilibrium Expression

Phenol is a weak acid and partially dissociates in water. The equilibrium expression is based on the acid dissociation constant \( K_a \):\[ K_a = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^-][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}]} = 1.3 \times 10^{-10} \]At equilibrium, let \( [\mathrm{H}_3\mathrm{O}^+] = x \). Since phenol is in such a low concentration, we assume that the change in its concentration is negligible, i.e., \( 0.016576 - x \approx 0.016576 \). Thus:\[ 1.3 \times 10^{-10} = \frac{x^2}{0.016576} \]

Step 3: Solve for Hydronium Ion Concentration

To find \( x \), the concentration of \( \mathrm{H}_3\mathrm{O}^+ \), substitute into the equilibrium expression:\[ x^2 = 1.3 \times 10^{-10} \times 0.016576 \]\[ x^2 = 2.15488 \times 10^{-12} \]Taking the square root gives:\[ x = \sqrt{2.15488 \times 10^{-12}} \approx 1.467 \times 10^{-6} \text{ M} \]Thus, \( [\mathrm{H}_3\mathrm{O}^+] = 1.467 \times 10^{-6} \text{ M} \).

Step 4: Calculate the pH

The pH of the solution is calculated using the formula:\[ \mathrm{pH} = -\log([\mathrm{H}_3\mathrm{O}^+]) \]Substitute the value from Step 3:\[ \mathrm{pH} = -\log(1.467 \times 10^{-6}) \approx 5.83 \]

Key Concepts

Weak Acid

A weak acid is a type of acid that does not fully dissociate in water. This means that only a small fraction of the acid molecules break apart into ions. In our context, phenol, or carbolic acid \( \text{C}_{6}\text{H}_{5}\text{OH} \), serves as a classic example of a weak acid. When dissolved in water, only a few of its molecules donate protons to water molecules.
Here are a few key characteristics of weak acids:

• They have low concentrations of hydrogen ions (\(\text{H}^+\) ) in solution.
• They have higher \(\text{pH}\) values relative to strong acids of the same concentration.
• They exhibit a reversible reaction when dissolved in water, establishing an equilibrium between ionized and non-ionized forms.

Understanding weak acids involves recognizing their partial dissociation and how this influences their behavior in solutions. Knowing that phenol is a weak acid helps us understand why its \(\text{pH} \) is not extremely low, even though it is categorized as an acid.

Dissociation Constant

The dissociation constant, often symbolized as \(K_a\), is fundamental when studying acid-base equilibrium. It quantifies the degree to which an acid dissociates in solution. For our phenol example, the given equilibrium reaction is:\[\text{C}_6 \text{H}_5 \text{OH}\left(\text{aq}\right)+\text{H}_2 \text{O}\left(\ell\right)\rightleftarrows \text{C}_6 \text{H}_5 \text{O}^-\left(\text{aq}\right)+\text{H}_3 \text{O}^+\left(\text{aq}\right)\]Here, the \(K_a\) is \(1.3 \times 10^{-10}\). This small value tells us that the equilibrium strongly favors the reactants, signifying little ionization.
So how do we use \(K_a\)? It helps us write an equilibrium expression:

• It is calculated from concentrations at equilibrium: \[K_a = \frac{[\text{C}_6 \text{H}_5 \text{O}^-][\text{H}_3 \text{O}^+]}{[\text{C}_6 \text{H}_5 \text{OH}]}\]
• A large \(K_a\) value indicates a strong tendency to dissociate, typical for strong acids, while a small \(K_a\) suggests weak dissociation.

By understanding \(K_a\), you gain insight into the strength and behavior of an acid in solution, which is crucial when predicting reaction outcomes and determining solution \(\text{pH}\).

pH Calculation

Calculating the \(\text{pH}\) of a solution is essential for understanding its acidity or basicity, a crucial part of acid-base equilibrium studies. In our example, we were tasked with determining the \(\text{pH}\) of a phenol solution. This involves two main steps:

• **Calculate Hydronium Ion Concentration:** We first determined the equilibrium concentration of \(\text{H}_3 \text{O}^+\) ions using the formula:\[x^2 = K_a \times [\text{Initial Concentration of Phenol}]\]Where \(x\) turns out to be the concentration of hydronium ions.
• **Determine pH:** With \(x = [\text{H}_3 \text{O}^+]\), we can find \(\text{pH}\) with:\[\text{pH} = -\log([\text{H}_3 \text{O}^+])\]

In the exercise, the calculation segment found \(\mathrm{pH} \approx 5.83\). This number represents that the solution is weakly acidic, confirming phenol does not significantly alter the water's natural \(\text{pH}\). Understanding these steps and the logic behind \(\text{pH}\) calculations allows students to deeply explore how weak acids behave in solutions and predict the \(\text{pH}\) of various acidic solutions proficiently.