PART 1: Pyridine is a weak base that is used in the manufacture of pesticides an

Question

PART 1:
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 0.410M solution of pyridine?

PART 2:

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 0.77M solution of benzoic acid, what percentage of the molecules are ionized?

Explanation / Answer

Part 1

Assume P = Pyridine

Then

P + H2O <-> PH+ ad OH-

Kb = [PH+][OH-] / [P]

Kb = 10^-pka = 10^-8.75 =1.78*10^-9

[PH+] = [OH-] = x

[P] = 0.41 - x

Substitute and solve for X!

Kb = [PH+][OH-] / [P]

1.78*10^-9 = x*x /(0.41-x)

x = 2.634*10^-5

Since x = [Oh-]

[OH-] = 2.634*10^-5

But we need pH so:

pOH = -log(2.634*10^-5) = 4.58

pH = 14-pOH = 14-4.58 = 9.42

pH = 9.42 which is basic as expected

Part 2

HB is a weak acid, it will form an equilibrium when in solution:

HB <-> H+ and B-

Note that there is 1 mol of H+ for every mol of B-

Since Ka = 10^-pka = 10^-4.2

Ka = 6.31*10^-5

The equilibrium expression is given by:

Ka = [H+][B-] / [HB]

Let us assume: [H+]= [B-] = x ... since 1 mol of H+ is present for every mol of B-, the concentrations must be the same

Not only that. Since we are ionizing some acid, we must correct the concentration of [HB] as follows:

[HB] = 0.77 -x where the "-x" corrects the dissociated acid. Since it is a weak acid, in general, you may ignore this. For this proble I will not ignore it but you may do it to simplify math

Substitute all in Ka expression

Ka = [H+][B-] / [HB]

6.31*10^-5 = x*x/(0.77-x)

Solve for x

x = 0.0069 and -0.0070.... ignore negative value since there are no negative concentrations

and since x = [H+] = 0.0069

pH = -log[H+] = -log(0.0069) = 2.16

And the % of ionization is = ionied / total *100% = 0.0069 / 0.77 *100% = 0.8961 % which is a very small amount

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