M16Q2: Designing and Recognizing Buffer Solutions

Chem 103 Textbook Team; Chem 104 Textbook Team

M16Q2: Designing and Recognizing Buffer Solutions

Learning Objectives

Determine suitable conjugate acid-base pairs to prepare a buffer at a specified pH.
Apply a K_a expression or the Henderson-Hasselbalch equation in order to calculate the pH of a buffer system.
Calculate the amount of conjugate base (or acid) that must be added to a weak acid (or base) to prepare a buffer with a given pH.

| Key Concepts and Summary | End of Section Exercises |

Selection of Suitable Buffer Mixtures

There are three useful guidelines for selecting buffer mixtures:

A buffer system is only effective at stabilizing the pH within ± 1 of the pK_a of the buffer system. Not every buffer is effective at every pH.
A good buffer mixture should have similar concentrations of each conjugate acid/base species. A buffer solution has generally lost its usefulness when one component of the buffer pair (either the acid or base) is less than 10% of the other. Figure 1 shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.

Figure 1. The graph, an illustration of buffering action, shows change of pH as an increasing amount of a 0.10-M NaOH solution is added to 100 mL of a buffer solution in which, initially, [CH₃CO₂H] = 0.10 M and [CH₃CO₂⁻] = 0.10 M.
The higher the concentrations of the conjugate acid/base species, the more effective the buffer will be. For example, if you have 10 moles of both the weak acid and base, you are able to resist pH change from much more strong acid/base compared to if you only have 1 mole of each the weak acid and weak base. This will be discussed more in the next section.

The following table has a few common buffer systems with their effective pH ranges:

Weak Acid	Weak Base	pK_a	Effective pH Range
CH₃COOH (acetic acid)	CH₃COO^– (acetate ion)	4.74	3.74 – 5.74
H₂CO₃ (carbonic acid)	HCO₃^– (hydrogen carbonate ion)	6.37	5.37 – 7.37
NH₄⁺ (ammonium ion)	NH₃ (ammonia)	9.25	8.25 – 10.25
HCO₃^– (hydrogen carbonate ion)	CO₃^2- (carbonate ion)	10.33	9.33 – 11.33

Creating a Buffer System

Buffer systems can be created by:

Directly combining roughly equal amounts of a weak acid and weak base conjugate pair. This approach is demonstrated in Examples 1 and 2.
By adding enough strong acid to a weak base to convert about half of the original weak base to its conjugate weak acid. Note that all of the strong acid is consumed in this reaction so all that will be left is the weak conjugate acid/base pair. This approach is demonstrated in Example 3.
By adding enough strong base to a weak acid to convert about half of the original weak acid to its conjugate weak base. Again, note that all of the strong base will be consumed in this reaction so that all that will be left is the weak conjugate acid/base pair. This approach is demonstrated in Example 4.

Example 1

Choosing and Making an Effective Buffer Solution by Directly Combining a Weak Acid and Weak Base Conjugate Pair

From the table above,

choose which conjugate acid/base pair would be the most effective buffer at a pH of 10.10.
With that conjugate acid/base pair, how many grams of weak base (assume you are adding a sodium salt containing the weak base) must be added to 1.0 L of a 0.75 M solution of weak acid to make a buffer with a pH of 10.10.

Solution

(a) There are two buffers that qualify to as an effective buffer at a pH of 10.10 – both NH₄⁺/NH₃ and HCO₃^–/CO₃^2-. However, a pH of 10.10 is at the very edge of the range for NH₄⁺/NH₃, so HCO₃^–/CO₃^2- will be the more effective buffer at that range.

(b) To calculate the amount of carbonate ion we must add to the buffer, we will use the Henderson-Hasselbalch equation to first calculate what concentration of carbonate ion we will need in the buffer solution:

pH = pK_a + log $\dfrac{[\text{A}^{-}]}{[\text{HA}]}$

10.10 = 10.33 + log $\dfrac{[\text{CO}_{3}^{2-}]}{0.75\;\text{M}}$

$\dfrac{[\text{CO}_{3}^{2-}]}{0.75\;\text{M}}$ = 0.589

[CO₃^2-] = 0.441 M

Next, we will use dimensional analysis to calculate how many grams of Na₂CO₃ we must add to achieve a 1.0 L solution of 0.441 M CO₃^2- ion.

1.0 L × $\dfrac{0.441\ \text{mol CO}_{3}^{2-}}{1\;\text{L}} \times \dfrac{1\ \text{mol Na}_{2}\text{CO}_{3}}{1\ \text{mol CO}_{3}^{2-}} \times \dfrac{105.99\ \text{g Na}_{2}\text{CO}_{3}}{1\ \text{mol Na}_{2}\text{CO}_{3}}$ = 47 g Na₂CO₃

Example 2

Creating a Buffer by Directly Combining a Weak Acid and Weak Base Conjugate Pair

How many grams of NaCH₃COO should be added to 0.500 L of a 1.25 M acetic acid, CH₃COOH, solution to create a buffer with a pH of 4.60? The pK_a of acetic acid is 4.74. Assume the addition of NaCH₃COO does not change the volume of the solution.

Solution

The buffer will consist of CH₃COO^– and CH₃COOH:

CH₃COOH(aq) + H₂O(ℓ) ⇌ CH₃COO^–(aq) + H₃O⁺(aq)

Use the Henderson-Hasselbalch formula and solve for the concentration of base required for the buffer:

pH = pK_a + log $\dfrac{[\text{base}]}{[\text{acid}]}$

log $\dfrac{[\text{base}]}{[\text{acid}]}$ = pH – pK_a

$\dfrac{[\text{base}]}{[\text{acid}]}$ = 10^{(pH – pK_a)}

$\dfrac{[\text{CH}_{3}\text{COO}^{-}]}{1.25 \;\text{M}}$ = 10^{(4.60 – 4.74)}

[CH₃COO^–] = 10^{(4.60 – 4.74)} × (1.25 M) = 0.91 M

Use the concentration to find the moles, and grams of NaCH₃COO:

0.500 L × $\dfrac{0.91\ \text{mol CH}_{3}\text{COO}^{-}}{1\;\text{L}} \times \dfrac{1\ \text{mol NaCH}_{3}\text{COO}}{1\ \text{mol CH}_{3}\text{COO}^{-}} \times \dfrac{82.0343\ \text{g NaCH}_{3}\text{COO}}{1\ \text{mol NaCH}_{3}\text{COO}}$

= 37 g NaCH₃COO

Example 3

Creating a Buffer by Combining a Weak Base and a Strong Acid

Will a buffer solution be created when 15 moles of HCl are added to 5.00 L of a solution containing 32 moles of CH₃COO^–?
If so, what will the pH be of the resulting solution?
The K_b of CH₃COO^– is 5.56 x 10^-10. Assume that there is no volume change upon the addition of the HCl.

Solution

(a) Consider the following reaction and treat this question as a limiting reactant problem:

CH₃COO^–(aq) + HCl(aq) → CH₃COOH(aq) + Cl^–(aq)

CH₃COO^– and HCl react in a 1 mole to 1 mole ratio. There are fewer moles of HCl so HCl is the limiting reactant. Calculate how many moles of CH₃COOH will be produced and how many moles of CH₃COO^– will be left after the reaction:

15 mol HCl × $\dfrac{1\ \text{mol CH}_{3}\text{COOH}}{1\ \text{mol HCl}}$ = 15 mol CH₃COOH produced

15 mol HCl × $\dfrac{1\ \text{mol CH}_{3}\text{COO}^{-}}{1\ \text{mol HCl}}$ = 15 mol CH₃COO^– used

32 mol CH₃COO^– initially – 15 mol CH₃COO^– used = 17 mol CH₃COO^– left over

The mole ratio of 17 moles weak base to 15 moles weak acid is $\dfrac{17 \ \text{moles base}}{15 \ \text{moles acid}}$ = 1.13, which is close to equal amounts weak acid and weak base so this will form a buffer solution.

(b) Calculate the pH using the Henderson-Hasselbalch equation. Remember that even though a K_b is given, use the K_a in the Henderson-Hasselbalch equation:

pH = pK_a + log $\dfrac{[\text{base}]}{[\text{acid}]}$ = 4.74 + log $\dfrac{17 \ \text{mol}}{15 \ \text{mol}}$

Wait! Moles were used in the Henderson-Hasselbalch equation rather than concentrations! Why is this acceptable? Because they share the same volume of solution:

$\dfrac{[\text{base}]}{[\text{acid}]} = \dfrac{\left(\frac{\text{moles base}}{\text{L solution}}\right)}{\left(\frac{\text{moles acid}}{\text{L solution}}\right)} = \dfrac{\text{moles base}}{\text{moles acid}}$

Example 4

Creating a Buffer by Combining a Weak Acid and a Strong Base

Will a buffer solution be created when 0.50 moles of KOH are added to 0.250 L of a solution containing 0.50 moles of HF?
If so, what will the pH be of the resulting solution?
The K_a of HF is 6.8 × 10⁻⁴. Assume no volume change upon the addition of KOH.

Solution

Consider the resulting reaction and treat this question as a limiting reactant problem:

HF(aq) + OH^–(aq) → F^–(aq) + H₂O(ℓ)

Note: KOH is very soluble in water and completely dissociates to form K⁺ and OH^– ions in solution. K⁺ ions are neutral and will not participate in the reaction.

HF and OH^– react in a 1:1 molar ratio. Since the initial moles of HF and OH^– are equal, they will react to completion and form 0.50 moles of F^–.

With only F^– present, a buffer solution will not form.

Question: If the F^– ion is present in water, won’t some HF form? Could that make a buffer?

No. If you use the K_b (from the given K_a) and an ICE table, you will find that the concentration of HF that would form is 5.4 x 10^-6M. A ratio of 2.00 M F^– to 5.4 x 10^-6 M HF is not close enough to form a buffer solution.

Example 5

Identifying Effective Buffer Solutions

Choose which of the following solutions could make effective buffer solutions:

CH₃COOH & HCl
NaCH₃COO & HCl
NH₃ & NH₄Cl
HNO₃ & NO₃^–

Solution

B and C are the correct answers.

A – You cannot make a buffer from a weak acid and a strong acid. Adding a strong acid to a weak acid will not create any of its conjugate base.

B – You can make a buffer from a weak base and strong acid when enough strong acid is added to create the conjugate acid but not so much that all of the original weak base is converted. This is illustrated in the following reaction:

CH₃COO^–(aq) + HCl(aq) → CH₃COOH(aq) + Cl^–(aq)

C – You can make a buffer from a weak base and its conjugate weak acid.

D – You cannot make a buffer from a strong acid with its conjugate base. Both components must be weak species.

Check Your Learning
Choose which of the following solutions could make effective buffer solutions:

H₂CO₃ and NaHCO₃
H₂CO₃ and Na₂CO₃
Na₂CO₃ and HCl
NaHCO₃ and NaOH

Answer:

A, C and D

Key Concepts and Summary

A good buffer system stabilizes the pH within ±1 of the pKa, has similar concentrations of each conjugate acid/base species, and the more moles of buffer, the more capacity the buffer has to resist a pH change. A buffer system can be created in three ways: one, by combining roughly equal amounts of weak acid and weak base; two, by adding enough strong acid to weak base to convert some of the weak base into weak acid; and three, by adding enough strong base to weak acid to convert some of the weak acid into weak base.

Chemistry End of Section Exercises

NaOH is a strong base but it is still possible to create a buffer when combining it with acetic acid, CH₃COOH. Why? Use a chemical equation in your explanation.
A student has been asked to produce a buffer solution with a pH of 9.25 using one of the following weak acids: HA (K_a = 5.60 × 10^-10), HB (K_a = 7.78 × 10^-7) or HC (K_a = 3.29 × 10^-9). Which would be the best choice and why.
Which acid (and conjugate base) would be the best choice to make buffers with the following pHs. Explain your choice. Appendix I will be helpful for solving this problem)
1. 3.45
2. 9.8
3. 12.38
4. 2.12
Which of the following acid/base pairs would be the best for forming a pH = 4.7 buffer?
1. The formate ion (K_b = 3.3 × 10^-11) and its conjugate acid
2. Phenol (K_a = 1.7 × 10^-10) and its conjugate base
3. The hypoiodite ion (K_b = 4.3 × 10^-4) and its conjugate acid
4. Aniline (K_b = 3.9 × 10^-10) and its conjugate acid
5. Ammonia (K_b = 1.8 × 10^-5) and its conjugate acid
Which of the following is not an effective buffer solution?
1. A mixture of 0.10 M NH₃(aq) and 0.25 M NH₄Cl(aq)
2. A mixture of 0.10 M NH₃(aq) and 0.10 M NH₄Br(aq)
3. A mixture of 0.10 M HBr and 0.15 M NaBr(aq)
4. A mixture of 0.10 M CH₃COOH(aq) and 0.25 M CH₃COONa
5. A mixture of 0.10 M formic acid and 0.10 M sodium formate
You are asked to prepare an acetic acid/sodium acetate buffer solution with a pH of 4.00 ± 0.02. What mole ratio of CH₃COONa to CH₃COOH should be used? K_a(CH₃COOH) = 1.8 × 10^-5.
What is the K_a of the weak acid, HA, found in an equimolar weak acid/weak base buffer solution with a pH of 7.740?
A buffer solution with a pH of 3.35 is comprised of 0.15 M HA and 0.10 M A^–. What is the K_a of HA?
How much solid sodium acetate, NaCH₃COO, must be added to 0.300 L of a 0.50 M acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)
What mass of NH₄Cl must be added to 0.750 L of a 0.100 M solution of NH₃ to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.)

Answers to Chemistry End of Section Exercises

It is possible to add enough NaOH to the acetic acid to convert about half of it to its conjugate base, CH₃COO^–, creating a mixture of the conjugate pair: CH₃COOH(aq) + OH^–(aq) → CH₃COO^–(aq) + H₂O(ℓ)
HA is the best choice for the acid/base conjugate pair as it is the one with a pK_a closest to the desired pH.
The best choice for the acid/base conjugate pair would be the one with a pK_a closest to the desired pH.
(a) HCCOH/HCOO^–
(b) HC₆H₅O/C₆H₅O^–
(c) HPO₄^2-/PO₄^3-
(d) H₃PO₄/H₂PO₄^–
D
C
0.18
1.82 × 10^-8
2.98 × 10^-4
22 grams
3.8 grams

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