The active ingredient in aspirin is acetylsalicylic acid HC9H7O4, a monoprotic acid with a Ka=3.3×10^-4 at 25 degrees C . What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 573 mg of acetylsalicylic acid each, in 258 mL of water?
The Correct Answer and Explanation is :
To determine the pH of a solution made by dissolving two extra-strength aspirin tablets, each containing 573 mg of acetylsalicylic acid (ASA), in 258 mL of water, we can follow these steps:
- Calculate the moles of acetylsalicylic acid:
- Molar mass of ASA (HC₉H₇O₄): 180.16 g/mol
- Total mass of ASA: 573 mg/tablet × 2 tablets = 1146 mg = 1.146 g
- Moles of ASA: 1.146 g ÷ 180.16 g/mol ≈ 0.00636 mol
- Determine the concentration of ASA in the solution:
- Volume of solution: 258 mL = 0.258 L
- Concentration (C): 0.00636 mol ÷ 0.258 L ≈ 0.0247 M
- Set up the ionization equilibrium expression:
- ASA (HC₉H₇O₄) is a monoprotic weak acid that partially ionizes in water:
[ \text{HC}_9\text{H}_7\text{O}_4 \rightleftharpoons \text{H}^+ + \text{C}_9\text{H}_7\text{O}_4^- ] - Acid dissociation constant (Ka): 3.3 × 10⁻⁴
- Expression: [ K_a = \frac{[\text{H}^+][\text{C}_9\text{H}_7\text{O}_4^-]}{[\text{HC}_9\text{H}_7\text{O}_4]} ]
- Assume the degree of ionization is small:
- Let ( x ) represent the concentration of ionized ASA:
- [ [\text{H}^+] = x ]
- [ [\text{C}_9\text{H}_7\text{O}_4^-] = x ]
- [ [\text{HC}_9\text{H}_7\text{O}_4] \approx 0.0247 – x \approx 0.0247 ]
- Since ( x ) is expected to be much smaller than the initial concentration, the approximation holds.
- Solve for ( x ):
- [ K_a = \frac{x^2}{0.0247} ]
- [ x^2 = K_a \times 0.0247 ]
- [ x^2 = 3.3 \times 10^{-4} \times 0.0247 ]
- [ x^2 = 8.151 \times 10^{-6} ]
- [ x = \sqrt{8.151 \times 10^{-6}} ]
- [ x \approx 0.00285 \, \text{M} ]
- Calculate the pH:
- [ \text{pH} = -\log[\text{H}^+] ]
- [ \text{pH} = -\log[0.00285] ]
- [ \text{pH} \approx 2.55 ]
Therefore, the pH of the solution is approximately 2.55.
This calculation assumes that the ionization of acetylsalicylic acid is minimal, allowing the approximation ( [\text{HC}_9\text{H}_7\text{O}_4] \approx 0.0247 \, \text{M} ). Given the relatively small Ka value, this assumption is valid, leading to a slightly acidic solution, as expected for a weak acid like acetylsalicylic acid.