Question

1. The active ingredient in aspirin is acetylsalicylic acid (pKa = 3.48). A 2.51 g sample of acetylsalicylic acid was mixed with 40.00 mL of 0.5106 M NaOH. What is the pH of this solution?
2. NOTE: ANSWER KEY SAYS pH = 13.2, THIS IS A PRACTICE TEST QUESTION, HOW AM I SUPPOSED TO DO THIS WITH NOTHING BUT A PERIODIC TABLE, I DO NOT KNOW THE pH IS 13.2 IN THIS SCENARIO. THIS IS NOT ORGANIC CHEM AND ORGANIC CHEM IS NOT A PREREQ I AM NOT SUPPOSED TO KNOW THE STRUCTURES. SO HOW THE HECK AM I TO FIND THE MOLAR MASS OF THE ACID?????

Answer 1

formula of  acetylsalicylic acid = C₉H₈O₄

thus

molar mass of acid = (12 * 9 + 8 * 1.0 + 4 * 16) = 180 g / mole.

2.51 g sample of acetylsalicylic acid = mass / molar mass = 2.51 g / 180 g / mole = 0.0139 mole.

40.00 mL of 0.5106 M NaOH = volume of solution in L * molarity = 0.040 L * 0.5106 mole / L = 0.02042 mole.

one mole acetyl salicylic acid is neutralized by one mole NaOH.

Here

moles of excess NaOH = (0.02042 - 0.0139) = 6.524 * 10^-3 mole.

thus

concentration of excess NaOH = moles of NaOh / volume of solution in L = 6.524 * 10^-3 mole / 0.040 L = 0.1631 M

NaOH is strong base.

Hence

[NaOh] = [OH-]

we know,

POH = - log [OH-]

or

POH = - log(0.1631)

or

POH = 0.78

PH + POH = 14

or

PH = 14 - 0.78

or

PH = 13.2 (answer)