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?

Short Answer

Expert verified
In the given problem, we find that the concentration of aspirin in the stomach is approximately 0.00361 M, and the [H+] concentration is 0.01 M. Using the dissociation constant (Ka) of aspirin (3 x 10^(-5)), we calculate the concentration of deprotonated aspirin as 1.08 x 10^(-6) M. This allows us to determine the percentage of aspirin in the form of neutral molecules, which is about 99.97%.

Step by step solution

01

Calculate the concentration of aspirin in the stomach

First, we need to calculate the total mass of aspirin in the stomach. We are given that there are two tablets, and each has a mass of 325 mg. Therefore, the total mass of aspirin in the stomach is: Total mass of aspirin = 2 * 325 mg = 650 mg Now, we will convert the mass of aspirin to moles by dividing it by the molar mass of aspirin (C9H8O4), which is about 180.16 g/mol: Moles of aspirin = (650 mg) * (1 g / 1000 mg) * (1 mol / 180.16 g) Moles of aspirin = 0.00361 mol Finally, we will divide the moles of aspirin by the volume of the stomach to get the concentration of aspirin: Concentration of aspirin = 0.00361 mol / 1 L Concentration of aspirin ≈ 0.00361 M
02

Calculate the [H+] concentration from pH

pH is given as 2, so we will find the [H+] concentration using the pH formula: pH = -log([H+]) We can find [H+] by taking the inverse log of the pH: [H+] = 10^(-2) = 0.01 M
03

Use Ka to find the ratio of protonated to deprotonated aspirin

Ka represents the ratio of protonated and deprotonated aspirin. The reaction can be written as: \(HA \rightleftharpoons H^+ + A^-\) We are given Ka for aspirin at body temperature: Ka = [H+][A-] / [HA] We can find the concentration of A- using the concentration of H+ and HA, which we calculated in the previous steps: Ka = (0.01 M) * ([A-]) / (0.00361 M) We know that Ka = 3 x 10^(-5), so we can rearrange the equation and find the concentration of A-: [A-] = (Ka) * ([HA])/([H+]) [A-] = (3 x 10^(-5)) * (0.00361 M) / (0.01 M) [A-] ≈ 1.08 x 10^(-6) M
04

Calculate the percentage of aspirin in the form of neutral molecules

Now that we have the concentrations of both protonated (HA) and deprotonated (A-) aspirin, we can calculate the percentage of aspirin in the form of neutral molecules: Percentage of neutral aspirin = ([HA] / ([HA] + [A-])) * 100 Percentage of neutral aspirin = (0.00361 M / (0.00361 M + 1.08 x 10^(-6) M)) * 100 Percentage of neutral aspirin ≈ 99.97% Thus, about 99.97% of the aspirin in the stomach is in the form of neutral molecules.

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