Question

Acetylsalicylic acid, also known as Aspirin, has a formula of C9H8O4 and is a weak acid with a Ka of 3.0 x 10-4. It has been suggested that a regular low-dose regimen of “baby aspirin”, approximately 81 mg, may help reduce the risk of a heart attack.

a. Determine the molarity of acetylsalicylic acid in a solution of “baby aspirin” if a tablet containing 81.0 mg of acetylsalicylic acid is dissolved in a cup of water (8.0 fluid ounces). Note: 1 fl oz = 29.5735 mL. (Note: this is not the way to ingest aspirin, please do not do this yourself)

b. Calculate the approximate pH and percent ionization of the solution made in part a.

c. Calculate the exact pH and percent ionization of the solution made in part a.

d. Calculate the equivalence point if the solution made in part a is titrated with 0.10 M NaOH.

Answer 1

Solution-
a) Let's find , the moles of acid present are calculated:

n Acid = 81 mg * (1 g/1000 mg) * (1 mol/180.16 g)

= 4.5x10^-4 mol

So the volume can be calculated as below

V = 8 oz * (29.5735 mL/1 oz) * (1 L / 1000 mL) = 0.237 L

Now the molarity can be calculated as below

M = n / V = ​​4.5x10 ^ -4 mol / 0.237 L
= 0.0019 M

b) Now for the hydrolysis:

HA + H2O = H3O + + A-

Ka expression can be written as:

Ka = [H3O +] * [A-] / [HA]

= X ^ 2 / 0.0019 - X = 3x10 ^ -4

we are assuming that - X is negligible and clears therefore

X = [H3O +] = √3x10 ^ -4 * 0.0019 = 7.55x10 ^ -4 M

The pH and % ionization both can be calculated as below:

pH = - log [H3O +] = - log 7.55x10 ^ -4 = 3.12

% ion = X * 100/[HA]

= 7.55x10 ^ -4 * 100 / 0.0019

= 39.74%

c) Now in this case we have the expression of Ka and hence it is not neglected - X

X ^ 2 + 3x10 ^ -4 * X - 5.7x10 ^ -7 = 0

Applying second degree equation

X = [H3O +] = 6.2x10 ^ -4 M

So the calculated pH and % ionization:

pH = - log 6.2x10 ^ -4
= 3.21

% ion = 6.2x10 ^ -4 * 100 / 0.0019 = 32.63%

d) At the equivalence point is calculated at the base volume :

Vb = Ca * Va / Cb

= 0.0019 M * 0.237 L / 0.1 M

= 0.0045 L
= 4.5 mL