Question

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:

C5H5N+H2O⇌C5H5NH++OH−

a) The pKb of pyridine is 8.75. What is the pH of a 0.440 M solution of pyridine?

Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:

C6H5COOH⇌C6H5COO−+H+

b) The pKa of this reaction is 4.2. In a 0.63 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

a C5H5N+H2O⇌C5H5NH++OH−

0.44M                      0              0

0.44-x                      x                x

pKb=8.75

pKb = -log Kb

Kb = 1.78 x 10^-9

Kb = [OH-] [C5H5NH+] / [C5H5N]

1.78x10^-9 = x^2 / 0.44-x

x is very very small as compared with 0.44

so,we can write

1.78x 10^-9 = x^2 / 0.44

x^2 = 0.44x1.78 x10^-9= 7.832x10^-10

x= 2.798 x10^-5

[OH-] = 2.798x10^-5

pOH = -log [OH-] = 4.553

pH= 14-4.553= 9.447

b) C6H5COOH ⇌ C6H5COO− + H+

                 0.63           0                0      I

          0.63-x                  x              x

pKa = 4.2

Ka = 6.31 x10^-5

For each C6H5COOH that is ionized we have one H^+ ion produced.

Ka = [H+] [ C6H5COO-] / [C6H5COOH]

6.31x10^-5 = [x] [x] / [0.63-x]    [ as x is very small]

6.31x10^-5 x 0.63 = x^2

x = [H+] = 6.305 x10^-3

% ionized = (concentration of H^+/0.63) x 100%

% ionized = .6305 x10^-2/.63 x100 = 1%