What percentage is in the form of neutral molecules?

[4LeafClover]

Question:
If two aspirin tablets, each having a mass of 325mg, are dissolved in a full stomach whose volume is 1L and whose pH is 2, what percent of the aspirin is in the form of neutral molecules? (At body temperature, the Ka for aspirin = 3*10^-5)

My Work:
I know that the molecular formula for aspirin is C9H8O4. This means that aspirin has a molecular weight of 180 g/mol. 650 mg of aspirin is equal to 3.6*10^-3 moles of aspirin.
I also know that the stomach produces gastric acid which is primarily hydrochloric acid. Because the stomach has a pH of 2, then [H+] is 0.01M. Because HCl is a strong acid, we can assume that it completely ionizes and so [H+] is equal to the initial concentration of HCl. I also am pretty sure that I am going to need to use the formula for percent ionization which is ([H+]/initial concentration) x 100. Since ionization is the conversion of a neutral atom into a charged ion, I would have to subtract the percent ionized from 100% in order to get the percent of neutral molecules.

My train of thought has led me to believe that my equation will be [H+]/initial concentration of HCl x 100, but the [H+] is not just concentration H+ given by HCl - I also have to take into account the aspirin to get the right [H+]. I am going to need to set something equal to the Ka of aspirin (presumably x^2/3.6*10^-3) which gives me x = [H+] = [C9H7O4-] = 3.29*10^-4.

So what I am wondering is, did I do this work correctly? If so, how do I proceed from here? Just put the x value/initial concentration of HCl? If I did this incorrectly, where did I go wrong and what do I need to do to fix it?

 

[Nino MMP]

Thank you!Answer:Your work is mostly correct. To calculate the percent of neutral molecules, you need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] and [HA] are the concentrations of the conjugate base (aspirin in this case) and acid (HCl in this case), respectively. In your case, since the pH is 2, the pKa is 3*10^-5, and the initial concentration of HCl is 0.01M, then the equation becomes 2 = 3*10^-5 + log([A-]/0.01). Rearranging this equation yields [A-] = 10^(2-3*10^-5) x 0.01 = 3.29*10^-4. This is the same value you got by setting Ka equal to x^2/3.6*10^-3. Now that you have [A-], you can calculate the percent of neutral molecules using the equation 100 - ([A-]/[HCl]) x 100, where [A-] = 3.29*10^-4 and [HCl] = 0.01M. This gives you a percent of neutral molecules of 96.7%.