Question

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if the pH of the solution was 2.61?

Answer 1

Answer:

The Ka is 3.74 *10^-4

Explanation:

Step 1: Data given

Mass of aspirin = 2.00 grams

Volume of water = 0.600 L

pH of solution = 2.61

Molar mass of acetylsalicylic acid = 180 g/mol

Step 2: The balanced equation

C9H8O4 + H2O ⇆ H3O+ + C9H7O4-

Step 3: Calculate moles acetylsalicylic acid

Moles acetylsalicylic acid = mass / molar mass

Moles acetylsalicylic acid = 2.00 grams / 180 g/mol

Moles acetylsalicylic acid = 0.0111 moles

Step 4: The initial concentration

[C9H8O4] = 0.0111 moles / 0.600 L = 0.0185 M

[H3O+] = 0M

[C9H8O4-] = 0M

Step 5: The concentration at the equilibrium

[C9H8O4] =0.0185 - X M

[H3O+] = XM

[C9H8O4-] = XM

Step 6: Calculate Ka

Ka = [H3O+][C9H7O4-] / [C9H8O4]

Ka = x² / (0.0185 - x)

pH = 2.61;  [H3O+] = 10^-pH = 10^-2.61 = 0.00245 = x

Ka = (0.00245)² / (0.0185 - 0.00245) = 3.74 * 10^-4

The Ka is 3.74 *10^-4