Question

The pKb of pyridine is 8.75. What is the pH of a 0.340M solution of pyridine?

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 0.340M solution of pyridine?

B)

What percentage of the molecules are ionized?

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 0.72M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

A) having the pKb you can find the Kb.

Kb= 10^-pKb

Kb= 1.78x10^-9

Now we have

	C5H5N	C5H5NH+	OH-
initial	0.340 M	0	0
change	-X	+X	+X
equilibrium	0.340 - X	X	X

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

Kb= X*X / 0.340-X we assume 0.340-X= 0.340 because of the low Kb value

Kb= X²/0.340

X=Kb*0.340

X= 1.78x10^-9 * 0.340

X= 2.46x10^-5 = [OH-] = [ C5H5NH+]

pOH= -log [OH-]

pOH = -log ( 2.46x10^-5 )

pOH = 4.6

pH = 14- pOH= 14-4.6

pH= 9.4

B) pKa= 4.2

Ka= 10^-pKa

Ka = 6.31x10^-5

	C6H5COOH	C6H5COO-	H+
initial	0.72	0	0
change	-X	+X	*X
equilibrium	0.72 - X	X	X

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

Ka= X*X / 0.72 - X we assume 0.72 - X = 0.72 because of the low Ka value

6.31x10^-5 = X² / 0.72

X= 6.31x10^-5 * 0.72

X= 0.0067 M =  [C6H5COO-] = [H+]

% ionized = [H+] / [C6H5COOH] * 100 = 0.0067 / 0.72 * 100

%ionized = 0.94 %