The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in $200 \mathrm{~mL}$ of water?

To convert the mass of acetylsalicylic acid (\(100 \mathrm{mg}\)) into moles, we need the molar mass of acetylsalicylic acid, which can be calculated as follows: Molar mass of C\(_9\)H\(_8\)O\(_4\) = 9 (Molar mass of C) + 8 (Molar mass of H) + 4 (Molar mass of O) = 9(12.01 g/mol) + 8(1.01 g/mol) + 4(16.00 g/mol) = 180.16 g/mol Now, convert the mass to moles: 100 mg = 0.1 g (1 mg = 0.001 g) Moles of acetylsalicylic acid = mass/molar mass = 0.1 g / 180.16 g/mol ≈ 0.000555 mol

To find the molar concentration of acetylsalicylic acid, we will divide the moles of acetylsalicylic acid by the volume of the solution in liters: Volume of the solution = 200 mL = 0.2 L (1 mL = 0.001 L) Molar concentration of acetylsalicylic acid = Moles / Volume = 0.000555 mol / 0.2 L ≈ 0.00278 M

The dissociation of acetylsalicylic acid can be represented by the following equation: \(\mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4}\rightleftharpoons\mathrm{H}^{+}+\mathrm{C}_{9}\mathrm{H}_{7}\mathrm{O}_{4}^{-}\) Since \(K_a\) of acetylsalicylic acid is given (\(3.3 \times 10^{-4}\)), we can write: \[K_a = \frac{[\mathrm{H}^{+}][\mathrm{C}_{9}\mathrm{H}_{7}\mathrm{O}_{4}^{-}]}{[\mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4}]}\] Let the concentration of \(\mathrm{H}^{+}\) ions at equilibrium be x: \[\begin{array}{l} 3.3 \times 10^{-4} = \frac{x^{2}}{(0.00278 - x)} \end{array}\] However, since the value of \(K_a\) is small, we can assume that x is much smaller compared to 0.00278. Therefore, we can approximate the equation as follows: \[\begin{array}{l} 3.3 \times 10^{-4} = \frac{x^{2}}{0.00278} \end{array}\] Solve for x: \[x = \sqrt{3.3 \times 10^{-4} \times 0.00278} \approx 0.000295\]

Now that we have found the concentration of \(\mathrm{H}^{+}\) ions, we can calculate the pH of the aspirin solution using the pH formula: \[pH = -\log_{10}[\mathrm{H}^{+}]\] pH = -log10(0.000295) ≈ 3.53 The pH of the aspirin solution is approximately 3.53.