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# Buffers

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Sections
Arrhenius Acids and Bases
Bronsted-Lowry Acids and Bases
Lewis Acids and Bases
Auto-Ionization
Ka and Kb of compounds
Weak Acid-Base Equilibria
Ionic Salts of Weak Acids and Bases
Buffers

A buffer is a solution composed of a weak acid with its conjugate base.

###### Buffers

Concept #1: A buffer resists drastic changes in pH by keeping constant the acidic and basic ions found in a solution.

Concept #2: The weak acid and conjugate base can be different from one another by up to a magnitude of 10.

Concept #3: The more concentrated the weak acid and conjugate base then the better the buffer can counteract strong acid or strong base added.

###### Buffer Synthesis

Concept #4: A buffer is comprised of a weak acid and its conjugate base that can be formed in 3 different ways.

Example #1: Which of the following combinations can result in the formation of a buffer?

a)  75 mL of 0.10 M HClO3 with 50 mL of 0.10 M CH3NH2.

b)  25 mL of 0.10 M H2SO3 with 40.0 mL of 0.10 M NaOH.

c)  50 mL of 0.10 M NH4Cl with 50 mL of 0.05 M Sr(OH)2.

d)  50 mL of 0.20 M HF with 40 mL of 0.20 M NaOH.

Example #2: Calculate the pH of a solution formed by mixing 130.0 mL of a 0.300 M C2H5NH2 solution with 70.0 mL of a 0.500 M C2H5NH3+ solution. (Kb of C2H5NH2 is 5.0 x 10-4).

Practice: Which of the following molar ratios is the correct equilibrium ratio of BASE : ACID for a solution made of aniline (Kb = 3.8 x 10-10) and anilinium nitrate where the pH is 4.80?

Example #3: You are asked to go into the lab and prepare a buffer solution with a pH of 6.40 ± 0.2.  Which weak acid would be the best choice?

a)  carbonic acid                                         Ka = 4.2 x 10-7

b)  phenol                                                    Ka = 1.3 x 10-10

c)  ascorbic acid                                        Ka = 8.0 x 10-5

d)  hydrosulfuric acid                                Ka = 9.5 x 10-8

e)  potassium hydrogen phthalate            Ka = 3.1 x 10-6

Example #4: Calculate the pH of a solution made by mixing 8.627 g of sodium butanoate in enough 0.452 M butanoic acid, HC4H7O2, to make 250.0 mL of solution. The Ka of butanoic acid is 1.5 x 10-5.

a) 4.75

b) 4.82

c) 5.00

d) 2.58

e) 4.65