Question

Aspirin has the structural formula CC(=O)Oc1ccccc1C(=O)O At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{d}\) 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 pH is 2 , what percent of the aspirin is in the form of neutral molecules?

Short answer

Approximately \(92.8\%\) of the aspirin is in the form of neutral molecules in a full stomach with a volume of \(1 \ \mathrm{L}\) and a pH of 2 after dissolving two aspirin tablets, each having a mass of \(325 \ \mathrm{mg}\).

Step-by-step solution

Convert the mass of aspirin tablets to moles

To convert the mass of aspirin tablets to moles, we need the molar mass of aspirin. The structural formula of aspirin is C\(_9\)H\(_8\)O\(_4\) with the corresponding molar mass of \(180.16 \ \mathrm{g/mol}\). We are given that there are two aspirin tablets, each having a mass of \(325 \ \mathrm{mg}\), which is equal to \(650 \ \mathrm{mg}\). Now, convert the mass of aspirin to moles: \[ \text{moles of aspirin} = \frac{650 \ \mathrm{mg}}{180.16 \ \mathrm{g/mol}} \times \frac{1 \ \mathrm{g}}{1000 \ \mathrm{mg}} = 0.00361 \ \mathrm{mol} \]

Calculate the concentration of aspirin in the stomach

Now we'll find the concentration of aspirin in the stomach, which has a volume of \(1 \ \mathrm{L}\). \[ \text{Concentration of aspirin} = \frac{0.00361 \ \mathrm{mol}}{1 \ \mathrm{L}} = 0.00361 \ \mathrm{M} \]

Write an expression for total aspirin concentration

The dissociation of aspirin in water can be represented as: \[AH \rightleftharpoons A^- + H^+\] where \(AH\) is the neutral form of the aspirin, and \(A^-\) is the dissociated form with H+ as the proton. At equilibrium, we can write the expression for \(K_d\): \[K_d = \frac{[A^-][H^+]}{[AH]}\] We are given the pH of the stomach, pH=2, which enables us to calculate the concentration of H+ ions: \[[H^+] = 10^{-\text{pH}} = 10^{-2} = 0.01 \ \mathrm{M}\]

Solve the expression to find the concentration of neutral molecules

Now we are given \(K_d = 3 \times 10^{-5}\), and we have the concentration of H+ ions. We can use these values in the equilibrium expression to find the concentration of neutral molecules [AH]: \[ 3 \times 10^{-5} = \frac{[A^-][0.01]}{[AH]} \] We know that the total aspirin concentration, \(0.00361 \ \mathrm{M}\), equals the sum of the concentrations of \(AH\) and \(A^-\): \[0.00361 = [AH] + [A^-]\] Now we have two equations with two unknowns, which can be solved simultaneously to find \([AH]\): \[ \begin{cases} 3 \times 10^{-5} = \frac{[A^-][0.01]}{[AH]}\\ 0.00361 = [AH] + [A^-] \end{cases} \] Solving the above equations simultaneously, we get \([AH] = 0.00335 \ \mathrm{M}\).

Calculate the percentage of aspirin in the form of neutral molecules

To find the percentage of aspirin in the form of neutral molecules, we'll divide the concentration of \(AH\) by the total aspirin concentration and multiply by 100: \[ \text{Percentage of aspirin as neutral molecules} = \frac{[AH]}{0.00361 \ \mathrm{M}} \times 100 = \frac{0.00335 \ \mathrm{M}}{0.00361 \ \mathrm{M}} \times 100 \approx 92.8\% \] Hence, approximately \(92.8\%\) of the aspirin is in the form of neutral molecules.