Oh no! A chemist has bumped their head into a doorway in the lab and has developed a mild headache. Annoyed by this headache, they decide to take two aspirin tablets to relieve the pain. Aspirin is also known as acetylsalicylic acid (C9H8O4), which is a weak monoprotic acid with a pKa of 3.5. Each aspirin tablet contains 500 mg of aspirin. The two tablets are dissolved into 390 mL of water in a glass - to two decimal places, what is the pH of this solution?

To solve this problem, we need to follow these steps:

1. Convert the mass of aspirin to moles. The molar mass of aspirin (C9H8O4) is approximately 180.16 g/mol. So, 1000 mg of aspirin (500 mg/tablet x 2 tablets) is 1 g, which is 1/180.16 = 0.00555 mol.

2. Calculate the concentration of the aspirin solution. The volume of the solution is 390 mL, which is 0.39 L. So, the concentration (M) is 0.00555 mol / 0.39 L = 0.01423 M.

3. Use the pKa to find the Ka. The formula to convert pKa to Ka is Ka = 10^-pKa. So, Ka = 10^-3.5 = 0.000316.

4. Set up an ICE (Initial, Change, Equilibrium) table. Initially, we have 0.01423 M of aspirin and no products. The change is -x for aspirin and +x for the products. At equilibrium, we have 0.01423 - x M of aspirin and x M of the products.

5. Write the expression for Ka. For a weak acid HA dissociating into H+ and A-, the expression is Ka = [H+][A-] / [HA]. Substituting the equilibrium concentrations from the ICE table, we get 0.000316 = x^2 / (0.01423 - x).

6. Since aspirin is a weak acid, we can assume that x is small compared to 0.01423. So, the equation simplifies to 0.000316 = x^2 / 0.01423. Solving for x, we get x = sqrt(0.000316 * 0.01423) = 0.000747.

7. The concentration of H+ ions, [H+], is equal to x, which is 0.000747 M. To find the pH, use the formula pH = -log[H+]. So, pH = -log(0.000747) = 3.13.

So, the pH of the aspirin solution is approximately 3.13.