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 2.62?

Express your answer numerically using two significant figures.

Ka =

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.

Kb =

Answer 1

we know that

moles = mass / molar mass

so

moles of aspirin = 2 / 180 = 0.01111

now

concentration = moles / volume (L)

so

conc of apsirin = 0.0111 / 0.6 = 0.0185185

let aspirin be the weak acid

HA ---> H+ + A-

using ICE table

at equilibrium

[HA] = 0.0185185 - y

[H+] = [A-] = y

given

pH = 2.62

we know that

pH = -log [H+]

so

-log [H+] = 2.62

[H+] = 0.0024

so

y = 0.0024

so

[HA] = 0.0185185 - 0.0024 = 0.0161185

now

Ka = [H+] [A-] / [HA]

Ka = [0.0024] [0.0024] / [0.0161185]

Ka = 3.574 x 10-4

so

Ka value is 3.574 x 10-4

B)

we know that

pOH = 14 - pH

so

pOH = 14 - 11.87 = 2.13

now

-log [OH-] = 2.13

[OH-] = 0.007413

C2H5NH2 + H20 --- C2H5NH3+ + OH-

so

at equilibrium

[OH-] = [C2H5NH3+] = 0.007413

[C2H5Nh2] = 0.1 - 0.007413 = 0.092587

now

Kb = [C2H5NH3+] [OH-] / [C2H5NH2]

Kb = [0.007413] [0.007413] / [0.092587]

Kb = 5.94 x 10-4

so

Kb value is 5.94 x 10-4