Question 16.15: Aspirin (acetylsalicylic acid, HC9H7O4) is a weak acid. It i...
Aspirin (acetylsalicylic acid, \text{HC}_9\text{H}_7\text{O}_4) is a weak acid. It ionizes in water according to the equation
\text{HC}_9\text{H}_7\text{O}_4(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{H}_3\text{O}^+(aq) + \text{C}_9\text{H}_7\text{O}_4^– (aq)
A 0.10-M aqueous solution of aspirin has a \text{pH} of 2.27 at 25°\text{C}. Determine the K_a of aspirin.
Strategy Determine the hydronium ion concentration from the \text{pH}. Use the hydronium ion concentration to determine the equilibrium concentrations of the other species, and plug the equilibrium concentrations into the equilibrium expression to evaluate K_a.
Setup Using Equation 16.3, we have
[\text{H}_3\text{O}^+] = 10^{–\text{pH}} Equation 16.3
[\text{H}_3\text{O}^+] = 10^{–2.27} = 5.37 × 10^{–3} M
To calculate K_a , though, we also need the equilibrium concentrations of \text{C}_9\text{H}_7\text{O}_4^– and \text{HC}_9\text{H}_7\text{O}_4. The stoichiometry of the reaction tells us that [\text{C}_9\text{H}_7\text{O}_4^–] = [\text{H}_3\text{O}^+]. Furthermore, the amount of aspirin that has ionized is equal to the amount of hydronium ion in solution. Therefore, the equilibrium concentration of aspirin is (0.10 − 5.37 × 10^{–3} ) M = 0.095 M.
\text{HC}_9\text{H}_7\text{O}_4(aq) + \text{H}_2\text{O}(l) \rightleftarrows \text{H}_3\text{O}^+(aq) + \text{C}_9\text{H}_7\text{O}_4^– (aq)
| Initial concentration (M): | 0.10 | 0 | 0 | |
| Change in concentration (M): | –0.005 | +5.37 × 10^{–3} | +5.37 × 10^{–3} | |
| Equilibrium concentration (M): | –0.005 | 5.37 × 10^{–3} | 5.37 × 10^{–3} |
Substitute the equilibrium concentrations into the equilibrium expression as follows:
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]} = \frac{(5.37 × 10^{–3})^2}{0.095} = 3.0 × 10^{–4}
The K_a of aspirin is 3.0 × 10^{–4}