Question

Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 9.0 ounces of water.

Common aspirin is acetylsalicylic acid, which has the structure shown below and a pKa of 3.5

Answer 1

The concentration of aspirin in water will be--

9 fl. oz = 0.266162 L. , 640 mg = 0.640 g , molar mass of aspirin = 180.157 g/mol

moles of aspirin = 0.640 g /180.157 g/mol = 3.55 x 10^-3 mol aspirin

molarity = 3.55 x10^-3 mol/ 0.266162L = 0.0133 M

Next, we need to set up an ice chart,

K_a = x²/(0.0133-x) ,with x being the concentration of H⁺ in solution

pKa 3.5

=> Ka = 10^-pk_a = 10^-3.5 = 3.16 x10^-4,  

Now, 3.16 x10^-4 = x²/(0.0133-x)

=> x² + 0.000316 x -0.0000042028 = 0

=> x = -0.000316 + 0.004112305436 / 2

Using quadratic formula we get,

x = 1.898 x10^-3 M = [H⁺]

pH = -log [H⁺] = -log (1.898 x10^-3) = - (log 1.898 -3 log 10) = -( 0.28-3) =2.72

Therefore. pH = 2.72