pH of Aspirin given pKa

Question

Aspirin pKa of 3.5

Molar Mass = 180.15 g

32oz = 1.0567L

Calculate the pH of a solution in which one normal adult dose of aspirin 6.50X 10² mg is dissolved in 8.0 oz of water.

I have attempted this problem multiple times and get a different answer each time. Please help, maybe I got the real answer at some point. Thank you!

Answer 1

To calculate the pH, we need to find the [H+], then take the negative log of that value.

We have 6.50x10² mg of the weak acid (aspirin) in 8 oz of solution. We need to convert this to moles / liter.

6.50x10² mg = 650 mg

650 mg x 1 g/1000 mg = 0.650 g

moles = 0.650 g / 180.15 g/mol = 0.003608 moles

Now that we have the moles, we need to convert 8 oz into liters

8 oz x 1.0567 L/32 oz = 0.264 L

Molarity of the aspirin = 0.003608 mols / 0.265 L = 0.0136 M

Let aspirin be represented by HA

HA ==> H⁺ + A^-

Ka = [H⁺] + [A^-] / [HA] and Ka = 1x10^3.5 = 3.16x10^-4

3.16x10^-4 = (x)(x) / 0.0136-x

x² = 4.30x10^-6 - 3.16x10^-4x

Solve using the quadratic formula

x = 0.00192 M = [H⁺]

pH = -log 0.00192

pH = 2.72