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?

Part B

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

Answer 1

A)

Molar mass of C9H8O4,

MM = 9*MM(C) + 8*MM(H) + 4*MM(O)

= 9*12.01 + 8*1.008 + 4*16.0

= 180.154 g/mol

mass(C9H8O4)= 2.00 g

use:

number of mol of C9H8O4,

n = mass of C9H8O4/molar mass of C9H8O4

=(2 g)/(1.802*10^2 g/mol)

= 1.11*10^-2 mol

volume , V = 0.6 L

use:

Molarity,

M = number of mol / volume in L

= 1.11*10^-2/0.6

= 1.85*10^-2 M

use:

pH = -log [H+]

2.62 = -log [H+]

[H+] = 2.399*10^-3 M

HA dissociates as:

HA -----> H+ + A-

1.85*10^-2 0 0

1.85*10^-2-x x x

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

Ka = x*x/(c-x)

Ka = 2.399*10^-3*2.399*10^-3/(0.0185-2.399*10^-3)

Ka = 3.574*10^-4

Answer: 3.57*10^-4

B)

use:

pH = -log [H+]

11.87 = -log [H+]

[H+] = 1.349*10^-12 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(1.349*10^-12)

[OH-] = 7.413*10^-3 M

C2H5NH2 dissociates as:

C2H5NH2 +H2O -----> C2H5NH3+ + OH-

0.1 0 0

0.1-x x x

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

Kb = x*x/(c-x)

Kb = 7.413*10^-3*7.413*10^-3/(0.1-7.413*10^-3)

Kb = 5.935*10^-4

Answer: 5.94*10^-4