Aspirin (acetylsalicylic acid, HC9H7O4) is a weak acid. It ionizes in water according to the equation
HC9H7O4(aq) + H2O(l) ⇌ H3O+(aq) + C9H7O4-(aq)
A 0.20-M aqueous solution of aspirin has a pH of 2.57 at 25°C. Determine the Ka of aspirin.
The Correct Answer and Explanation is :
To determine the acid dissociation constant (Ka) of aspirin (acetylsalicylic acid, HC9H7O4), we start with the given information: a 0.20 M solution of aspirin has a pH of 2.57 at 25°C.
Step 1: Calculate the Hydronium Ion Concentration [H3O+].
The pH of a solution is related to the hydronium ion concentration by the equation:
[ \text{pH} = -\log[\text{H}_3\text{O}^+] ]
Rearranging to solve for ([\text{H}_3\text{O}^+]):
[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} ]
Substituting the given pH:
[ [\text{H}_3\text{O}^+] = 10^{-2.57} \approx 2.69 \times 10^{-3} \, \text{M} ]
Step 2: Determine the Degree of Ionization.
In the ionization of aspirin:
[ \text{HC}_9\text{H}_7\text{O}_4(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{C}_9\text{H}_7\text{O}_4^-(aq) ]
Each molecule of HC9H7O4 that ionizes produces one H3O+ ion and one C9H7O4− ion. Therefore, the concentration of C9H7O4− ions at equilibrium is equal to the concentration of H3O+ ions:
[ [\text{C}_9\text{H}_7\text{O}_4^-] = [\text{H}_3\text{O}^+] = 2.69 \times 10^{-3} \, \text{M} ]
The change in concentration of HC9H7O4 due to ionization is also 2.69 × 10−3 M. Thus, the equilibrium concentration of HC9H7O4 is:
[ [\text{HC}9\text{H}_7\text{O}_4]{\text{eq}} = 0.20 \, \text{M} – 2.69 \times 10^{-3} \, \text{M} \approx 0.1973 \, \text{M} ]
Step 3: Calculate the Acid Dissociation Constant (Ka).
The expression for the acid dissociation constant is:
[ K_a = \frac{[\text{H}_3\text{O}^+][\text{C}_9\text{H}_7\text{O}_4^-]}{[\text{HC}_9\text{H}_7\text{O}_4]} ]
Substituting the equilibrium concentrations:
[ K_a = \frac{(2.69 \times 10^{-3} \, \text{M}) \times (2.69 \times 10^{-3} \, \text{M})}{0.1973 \, \text{M}} ]
[ K_a = \frac{7.24 \times 10^{-6} \, \text{M}^2}{0.1973 \, \text{M}} ]
[ K_a \approx 3.67 \times 10^{-5} ]
Therefore, the acid dissociation constant (Ka) of aspirin at 25°C is approximately 3.67 × 10−5.
This value indicates that aspirin is a weak acid, as its Ka is significantly less than 1. The small degree of ionization in water corresponds to the relatively low hydronium ion concentration observed in the solution.
Understanding the Ka of aspirin is important in pharmaceutical contexts, as it influences the drug’s absorption and efficacy. The ionization state can affect how well aspirin is absorbed in different parts of the gastrointestinal tract, impacting its therapeutic action.