± Determining the pH of a Weak Base and the Percent Ionization of a Weak Acid Pa

Question

± Determining the pH of a Weak Base and the Percent Ionization of a Weak Acid

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+H2OC5H5NH++OH

The pKb of pyridine is 8.75. What is the pH of a 0.225 M solution of pyridine? (Assume that the temperature is 25 C .)

Express the pH numerically to two decimal places.

pH=???

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 ionizes in water:

C6H5COOHC6H5COO+H+

The pKa of this reaction is 4.2. In a 0.56 M solution of benzoic acid, what percentage of the molecules are ionized?

Express your answer to two significant figures and include the appropriate units.

% Ionization = ???

Explanation / Answer

Part A:

C5H5N +   H2O    C5H5NH+    +   OH

At equilibrium [C5H5NH+]= [OH] = x

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

At equilibrium [C5H5N] = 0.225 - x

pKb = -log(Kb)

Kb = 10-8.75

10-8.75 = x2 / 0.225 - x

x2 = 1.77 x 10-9(0.225 - x)

x2 = 4.001 x 10-10 - 1.77 x 10-9x

x2 + 1.77 x 10-9x - 4.001 x 10-10 = 0

x =2.000 x 10-5 M

pOH = -log(2.000 x 10-5) = 4.6989

pH = 14 - 4.6989

pH = 9.301

Part B)

C6H5COOH        C6H5COO +   H+

Ka = x2 / 0.56 -1

10-4.2 = x2 / 0.56 -1

6.309 x 10-5 = x2 / 0.56 -1

x2 = 6.309 x 10-5 (0.56 -x)

x2 = 3.533x10-5 -6.309x10-5x

x2 + 6.309x10-5x - 3.533x10-5 = 0

x = 0.0059M =>[H+]

Percentage of ionization is

= [H+] / Concentration of Benzoic acid x 100

= 0.0059 / 0.56

= 0.01053 x 100

Percentage of ionization is 1.053

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