Aspirin has the structural formula At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg},\) are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?

First, we need to convert the mass of aspirin to moles. The total mass of two tablets is \(325 \times 2 = 650\,\text{mg}\). The molar mass of aspirin is \(180.16\,\text{g/mol}\). We will convert the mass to grams and then to moles: \[650\,\text{mg} \times \frac{1\,\text{g}}{1000\,\text{mg}}\times \frac{1\,\text{mol}}{180.16\,\text{g}} = 0.00361\,\text{mol}\]

Next, we need to calculate the initial concentration of H+ ions in the stomach. We will use the given pH value: \[\text{pH} = -\log_{10}[\text{H}^+] \Rightarrow [\text{H}^+] = 10^{-\text{pH}}\] For an initial pH of \(2\), we get: \[[\text{H}^+] = 10^{-2} = 0.01\,\text{M}\]

Now we will set up an equilibrium expression for the reaction between aspirin and water, using the given \(K_a\) value: {\small\[K_a = \frac{[\text{Aspirin}^-][\text{H}^+]}{[\text{Aspirin}]} = 3 \times 10^{-5}\]} Let x be the change in concentration of aspirin, its conjugate base, and H+ ions at equilibrium. Therefore, we have: \[[\text{Aspirin}] = 0.00361 - x\] \[[\text{Aspirin}^-] = x\] \[[\text{H}^+] = 0.01 + x\] Substituting these values into the equilibrium expression, we get: \[\frac{x(0.01+x)}{0.00361-x} = 3 \times 10^{-5}\] Since \(K_a\) is relatively small, the change in concentration (x) will be negligible compared to the initial concentration. Thus, we can approximate: \[\frac{x \times 0.01}{0.00361} = 3 \times 10^{-5}\]

Now, we will solve for x and calculate the percentage of aspirin in the form of neutral molecules: \[x = 0.00361 \times 3 \times 10^{-5} \div 0.01 = 1.083\times 10^{-6}\,\text{mol}\] Since x represents the concentration of aspirin's conjugate base, the concentration of neutral aspirin molecules is \(0.00361-x\): \[[\text{Aspirin}] = 0.00361 - 1.083 \times 10^{-6} = 0.003608917\,\text{mol}\] To find the percentage, we will use the following formula: \[\text{Percent Neutral Aspirin} = \frac{\text{Amount of Neutral Aspirin}}{\text{Total Amount of Aspirin}} \times 100\%\] \[\text{Percent Neutral Aspirin} = \frac{0.003608917}{0.00361} \times 100\% \approx 99.97\%\] So, approximately \(99.97\%\) of the aspirin is in the form of neutral molecules.