A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC₉H₇O₄)was dissolved in enough water to make 100 mL of solution.Given that K_a = 3.0 × 10^-4 for aspirin,what is the pH of the solution?

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To find the pH of the aspirin solution, we first calculate the molarity of aspirin in the solution. Since we have 500.0 mg of aspirin dissolved in 100 mL of solution, we convert the mass of aspirin to grams (500.0 mg = 0.500 g) and then to moles (molar mass of aspirin ≈ 180 g/mol), resulting in approximately 0.00278 moles of aspirin. This gives us a molarity of 0.00278 M in a 0.1 L solution. Using the given $K_a = 3.0 \times 10^{-4}$, we can set up the equation for the dissociation of aspirin (HA) in water: $K_a = \frac{[H^+][A^-]}{[HA]}$. Assuming $x$ is the concentration of $H^+$ ions (and also of $A^-$ since they are produced in a 1:1 ratio), we have $3.0 \times 10^{-4} = \frac{x^2}{0.00278}$. Solving for $x$ gives a $H^+$ concentration of approximately $2.9 \times 10^{-3}$ M. The pH is calculated using $-\log[H^+]$, which gives a pH close to 2.54.

A Tablet Containing 500