Question

Part A 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− The pKb of pyridine is 8.75. What is the pH of a 0.260 M solution of pyridine? Part B 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+ The pKa of this reaction is 4.2. In a 0.64 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

A)
pKb = -log Kb
8.75 = -log Kb
Kb = 1.78*10^-9

C5H5N+H2O⇌C5H5NH++OH−
0.26 0 0 (initial)
0.26-x x x (at equilibrium)

Kb = x*x/(0.26-x)
since Kb is small, x will be small and can be ignored as compared to 0.26
Above expression thus becomes
Kb = x*x/(0.26-x)
1.78*10^-9 = x^2/0.26
x = 2.15*10^-5 M

So,
[OH-] = 2.15*10^-5

pOH = -log [OH-]
= -log (2.15*10^-5)
=4.67

pH = 14 - pOH
= 14 - 4.67
= 9.33

b)
pKa = -log Ka
4.2 = -log Ka
Ka= 6.31*10^-5

C6H5COOH⇌C6H5COO−+H+
0.64 0 0 (initial)
0.64-x x x (at equilibrium)

Ka = x*x/(0.64-x)
since Ka is small
6.31*10^-5 = x^2/0.64
x = 6.35*10^-3

% dissociation = 6.35*10^-3 * 100/0.64
= 0.99 %