34 Designing and Recognizing Buffer Solutions (M16Q2)
Learning Objectives
- Determine suitable conjugate acid-base pairs to prepare a buffer at a specified pH.
- Apply a Ka 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.
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 pKa 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.
![A graph is shown with a horizontal axis labeled “Added m L of 0.10 M N a O H” which has markings and vertical gridlines every 10 units from 0 to 110. The vertical axis is labeled “p H” and is marked every 1 unit beginning at 0 extending to 11. A break is shown in the vertical axis between 0 and 4. A red curve is drawn on the graph which increases gradually from the point (0, 4.8) up to about (100, 7) after which the graph has a vertical section up to about (100, 11). The curve is labeled [ C H subscript 3 C O subscript 2 H ] is 11 percent of [ C H subscript 3 CO subscript 2 superscript negative].](https://wisc.pb.unizin.org/app/uploads/sites/126/2017/08/CNX_Chem_14_06_buffer.jpg)
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:
Creating a Buffer System
Buffer systems can be created by:
1) Directly combining roughly equal amounts of a weak acid and weak base conjugate pair. This approach is demonstrated in Examples 1 and 2.
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.
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
(a) From the table above, choose which conjugate acid/base pair would be the most effective buffer at a pH of 10.10.
(b) 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–/CO32-. However, a pH of 10.10 is at the very edge of the range for NH4+/NH3, so HCO3–/CO32- 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 = pKa + log[latex]\frac{[A^{-}]}{[HA]}[/latex]
10.10 = 10.33 + log[latex]\frac{[CO_{3}^{2-}]}{[0.75]}[/latex]
[latex]\frac{[CO_{3}^{2-}]}{[0.75]}[/latex] = 0.589
[CO32-] = 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 CO32- ion.
1.0 L × [latex]\frac{0.441\ mol\ CO_{3}^{2-}}{1 L} \times \frac{1\ mol\ Na_{2}CO_{3}}{1\ mol\ CO_{3}^{2-}} \times \frac{105.99\ g\ Na_{2}CO_{3}}{1\ mol\ Na_{2}CO_{3}}[/latex] = 46.74 g Na2CO3
= 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 pKa 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 = pKa + log[latex]\frac{[base]}{[acid]}[/latex]
log[latex]\frac{[base]}{[acid]}[/latex] = pH – pKa
[latex]\frac{[base]}{[acid]}[/latex] = 10(pH – pKa)
[latex]\frac{[CH_{3}COO^{-}]}{1.25 M}[/latex] = 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 × [latex]\frac{0.91\ mol\ CH_{3}COO^{-}}{1 L} \times \frac{1\ mol\ NaCH_{3}COO}{1\ mol\ CH_{3}COO^{-}} \times \frac{82.0343\ g\ NaCH_{3}COO}{1\ mol\ NaCH_{3}COO}[/latex]
= 37 g NaCH3COO
Example 3
Creating a Buffer by Combining a Weak Base and a Strong Acid
(a) 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–?
(b) If so, what will the pH be of the resulting solution?
The Kb 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 [latex]\times \ \frac{1\ mol\ CH_{3}COOH}{1\ mol\ HCl}[/latex] = 15 mol CH3COOH produced
15 mol HCl [latex]\times \ \frac{1\ mol\ CH_{3}COO^{-}}{1\ mol\ HCl}[/latex] = 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 [latex]\frac{17 \ moles \ base}{15 \ moles \ acid}[/latex] = 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 Kb is given, use the Ka in the Henderson-Hasselbalch equation:
pH = pKa + log[latex]\frac{[base]}{[acid]}[/latex] = 4.74 + log [latex]\frac{17 \ mol}{15 \ mol}[/latex]
Wait! Moles were used in the Henderson-Hasselbalch equation rather than concentrations! Why is this acceptable? Because they share the same volume of solution:
[latex]\frac{[base]}{[acid]}[/latex] = [latex]\frac{\left(\begin{array}{c}\frac{moles \ base}{L \ solution}\end{array}\right)}{\left(\begin{array}{c}\frac{moles \ acid}{L \ solution}\end{array}\right)}[/latex] = [latex]\frac{moles \ base}{moles \ acid}[/latex]
Example 4
Creating a Buffer by Combining a Weak Acid and a Strong Base
(a) 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?
(b) If so, what will the pH be of the resulting solution?
The Ka 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 Kb (from the given Ka) and an ICE table, you will find that the concentration of HF that would form is 5.4 x 10-6 M. 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:
(a) CH3COOH & HCl
(b) NaCH3COO & HCl
(c) NH3 & NH4Cl
(d) 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 base 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:
(a) H2CO3 and NaHCO3
(b) H2CO3 and Na2CO3
(c) Na2CO3 and HCl
(d) 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 (Ka = 5.60 × 10-10), HB (Ka = 7.78 × 10-7) or HC (Ka = 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)
- 3.45
- 9.8
- 12.38
- 2.12
- What is the Ka 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 Ka 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 pKa closest to the desired pH.
- The best choice for the acid/base conjugate pair would be the one with a pKa closest to the desired pH.
- HCCOH/HCOO–
- HC6H5O/C6H5O–
- HPO42-/PO43-
- H3PO4/H2PO4–
- 1.82 × 10-8
- 2.98 × 10-4
- 22 grams
- 3.8 grams
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