Question

Acetylsalicylic acid (aspirin), \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\), is the most widely used pain reliever and fever reducer. Find the \(\mathrm{pH}\) of \(0.018 \mathrm{M}\) aqueous aspirin at body temperature \(\left(K_{\mathrm{a}}\right.\) at \(\left.37^{\circ} \mathrm{C}=3.6 \times 10^{-4}\right)\)

Short answer

The pH of 0.018 M aqueous aspirin at 37°C is approximately 2.60.

Step-by-step solution

Write the Ionization Equation

Aspirin, \textrm{HC}_9\textrm{H}_7\textrm{O}_4, ionizes in water according to the following equation: \[\textrm{HC}_9\textrm{H}_7\textrm{O}_4 (aq) \leftrightarrow \textrm{H}^+ (aq) + \textrm{C}_9\textrm{H}_7\textrm{O}_4^- (aq)\]

Set Up the Expression for Ka

The expression for the acid dissociation constant is given by: \[K_a = \frac{[\textrm{H}^+][\textrm{C}_9\textrm{H}_7\textrm{O}_4^-]}{[\textrm{HC}_9\textrm{H}_7\textrm{O}_4]}\]Given, \(K_a = 3.6 \times 10^{-4}\) and the initial concentration of aspirin, \( [\textrm{HC}_9\textrm{H}_7\textrm{O}_4]_i = 0.018M \).

Define Change in Concentrations

Assume that the change in concentration upon dissociation of aspirin is \(x\). Thus, \[ [\textrm{H}^+] = x, \quad [\textrm{C}_9\textrm{H}_7\textrm{O}_4^-] = x, \quad [\textrm{HC}_9\textrm{H}_7\textrm{O}_4] \approx 0.018 - x \].

Substitute into Ka expression

Substitute the changes into the \(K_a\) expression: \[3.6 \times 10^{-4} = \frac{x \times x}{0.018 - x} \approx \frac{x^2}{0.018}\]Since \(x\) is small relative to 0.018, \(0.018 - x \approx 0.018\).

Solve for x

Rearrange to solve for \(x\): \[x^2 = 3.6 \times 10^{-4} \times 0.018\] \[x^2 = 6.48 \times 10^{-6}\] \[x = \sqrt{6.48 \times 10^{-6}}\] \[x \approx 2.54 \times 10^{-3} M\] Since \( [\textrm{H}^+] = x \), \( [\textrm{H}^+] = 2.54 \times 10^{-3} M\).

Calculate the pH

The \(\textrm{pH}\) is calculated using: \[\textrm{pH} = -\log[\textrm{H}^+]\] \[\textrm{pH} = -\log(2.54 \times 10^{-3})\] \[\textrm{pH} \approx 2.60\].

Key concepts

acid dissociation constant

The acid dissociation constant, often denoted as \(K_a\), is a crucial concept in understanding the strength of an acid. It measures the extent to which an acid can donate a proton (\(H^+\)) in an aqueous solution. For aspirin, the \(K_a\) value at body temperature \( (37^\text{°}C) \) is given as \(3.6 \times 10^{-4}\). This value indicates that aspirin is a weak acid because its \(K_a\) is much less than 1. The ionization of aspirin in water can be expressed using the chemical equation: \[ \text{HC}_9\text{H}_7\text{O}_4 \rightleftharpoons \text{H}^+ + \text{C}_9\text{H}_7\text{O}_4^- \]At equilibrium, the ratio of the concentrations of the dissociated ions to the concentration of the undissociated acid is represented by: \[ K_a = \frac{[\text{H}^+][\text{C}_9\text{H}_7\text{O}_4^-]}{[\text{HC}_9\text{H}_7\text{O}_4]} \]These expressions help us understand how much of the aspirin dissociates when dissolved in water, which allows for pH calculation.

ionization equation

The ionization equation represents the process by which an acid dissociates into its ions in an aqueous solution. For aspirin, the ionization equation is: \[ \text{HC}_9\text{H}_7\text{O}_4 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{C}_9\text{H}_7\text{O}_4^- (aq) \]This equation shows that one molecule of aspirin splits into one hydrogen ion (\(H^+\)) and one acetylsalicylate ion (\(\text{C}_9\text{H}_7\text{O}_4^-\)). This process is crucial for calculating the concentrations of these ions in the solution. When the initial concentration of aspirin is known, we can set up an expression for the changes in concentration due to ionization. Assuming the change is x, we have: \[ \text{[H}^+\text{]} = x, \text{ [C}_9\text{H}_7\text{O}_4^- \text{]} = x, \text{ and [HC}_9\text{H}_7\text{O}_4 \text{]} \thickapprox 0.018 - x \]Given that \(x\) is generally small, the approximation \(0.018 - x \thickapprox 0.018\) simplifies calculations without significant error.

concentration changes

Understanding concentration changes is fundamental to calculating pH for solutions of weak acids like aspirin. When aspirin dissolves in water, it dissociates into ions, altering their concentrations. Initially, the concentration of aspirin is \(0.018\text{ M}\). Let the change due to dissociation be \(x\). Therefore:

• Hydrogen ion concentration: \( \text{[H}^+\text{]} = x \)
• Acetylsalicylate ion concentration: \( \text{[C}_9\text{H}_7\text{O}_4^- \text{]} = x \)
• Remaining aspirin concentration: \( \text{[HC}_9\text{H}_7\text{O}_4 \text{]} \thickapprox 0.018 - x \)

Substituting these into the \(K_a\) expression, we get: \[ 3.6 \times 10^{-4} = \frac{x \times x}{0.018 - x} \thickapprox \frac{x^2}{0.018} \] Solving for \(x\) yields:
\[ x \thickapprox 2.54 \times 10^{-3} \]This \(x\) is essentially the \( \text{[H}^+ \text{]} \), allowing us to calculate the pH using:
\[ \text{pH} = -\text{log} [\text{H}^+] \thickapprox 2.60 \]