Question

Part A

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 2.60?

Express your answer numerically using two significant figures.

Part B

A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate the Kb for ethylamine.

Express your answer numerically using two significant figures.

Answer 1

Ans. Part A: Moles of aspirin = Mass / Molar mass

= 2.00 g/ (180.16012 g/ mol)

= 0.01110 M

# Given, pH of solution = 2.60

Now,

            [H₃O⁺] = antilog (-pH) = antilog (-2.60) = 0.0025 M

The pH of aspirin indicates that equilibrium [H₃O⁺] = 0.0025 M

# Create an ICE table with initial [AH] = 0.0111 M and equilibrium [H₃O⁺] = 0.0025 M as shown in figure-

Now,

            Equilibrium concentrations of-

            [H₃O⁺] = [H⁺] = X = 0.0025 M         - as calculated above

            [A^-] = X = 0.0025 M

[AH] = 0.0111 – X = 0.0111 – 0.0025 = 0.0086 M          

Now,

            Ka of aspirin = [A^-] [H₃O⁺] / [AH]               - all concentrations at equilibrium

            Or, Ka = (0.0025 x 0.0025) / 0.0086 = 7.27 x 10^-4

Therefore, Ka of aspirin = 7.3 x 10^-4

# Part B. pOH of solution = 14.-00 – pH = 14.00 – 11.87 = 2.13

[OH^-] of solution = antilog (-pOH) = antilog (-2.13) = 0.0074 M

The pH of ethylamine (C₂H₅NH₂) indicates that equilibrium [OH^-] = 0.0074 M

## Create an ICE table with initial [C₂H₅NH₂] = 0.100 M and equilibrium [OH^-] = 0.0074 M as shown in figure-

Now,

            Equilibrium concentrations of-

            [OH^-] = X = 0.0074 M         - as calculated above

            [C₂H₅NH₃⁺] = X = 0.0074 M

[C₂H₅NH₂] = 0.100 – X = 0.100 – 0.0074 = 0.0926 M     

Now,

            Kb of C₂H₅NH₂ = [C₂H₅NH₃⁺] [OH^-] / [C₂H₅NH₂]             - all [Conc.] at equilibrium

            Or, Kb = (0.0074 x 0.0074) / 0.0926 = 5.914 x 10^-4

Therefore, Kb of ethylamine = 5.9 x 10^-4