7.6 Buffers
Learning Objectives
By the end of this section, you will be able to:
- Describe the composition and function of acid–base buffers
- Calculate the pH of a buffer before and after the addition of added acid or base
A mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure 1). A solution of acetic acid and sodium acetate (CH3COOH + CH3COONa) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride (NH3(aq) + NH4Cl(aq)).
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How Buffers Work
A mixture of acetic acid and sodium acetate is acidic because the Ka of acetic acid is greater than the Kb of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. Then more of the acetic acid reacts with water, restoring the hydronium ion concentration almost to its original value:
CH3CO2H(aq)+ H2O(l) → H3O+(aq) + CH3CO2–(aq)
The pH changes very little. If we add an acid such as hydrochloric acid, most of the hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid molecules:
H3O+(aq)+ CH3CO2–(aq) → CH3CO2H(aq) + H2O(l)
Thus, there is very little increase in the concentration of the hydronium ion, and the pH remains practically unchanged (Figure 2).
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Example 1: pH Changes in Buffered and Unbuffered Solutions
Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds.
(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
Solution:
To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples):
- Determine the direction of change. The equilibrium in a mixture of H3O+, CH3CO2–, and CH3CO2H is:
CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) +CH3CO2–(aq)
The equilibrium constant for CH3CO2H is not given, so we look it up in Appendix H: Ka = 1.8 HOLD EQ = 0.10 M and [H3O+] = ~0 M, the reaction shifts to the right to form H3O+.
- Determine x and equilibrium concentrations. A table of changes and concentrations follows:
- Solve for x and the equilibrium concentrations. We find:
x = 1.8 × 10-5 M
and
[H2O+] = 0 + x = 1.8 × 10–5 M
Thus:
pH = -log [H3O+] = -log (1.8 × 10–5)
= 4.74 - Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = Ka.
(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL.
First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium:
- Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains:
- Determine the moles of CH2CO2H. Before reaction, 0.100 L of the buffer solution contains:
- Solve for the amount of NaCH3CO2 produced. The 1.0 × 10−4 mol of NaOH neutralizes 1.0 × 10-4 mol of CH3CO2H, leaving:
(1.0 × 10-2) – (0.01 × 10-2) = 0.99 × 10-2 mol CH3CO2H
and producing 1.0 × 10−4 mol of NaCH3CO2. This makes a total of:
(1.0 × 10-2) – (0.01 × 10-2) = 1.01 × 10-2 mol NaCH3CO2 - Find the molarity of the products. After reaction, CH3CO2H and NaCH3CO2 are contained in 101 mL of the intermediate solution, so:
Now we calculate the pH after the intermediate solution, which is 0.098 M in CH3CO2H and 0.100 M in NaCH3CO2, comes to equilibrium. The calculation is very similar to that in part (a) of this example:
This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution.
(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 × 10−5–M solution of HCl). The volume of the final solution is 101 mL.
Solution:
This 1.8 × 10−5–M solution of HCl has the same hydronium ion concentration as the 0.10-M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains:
As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 × 10−4 mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is:
(1.0 × 10-4) – (1.8 × 10-6) = 9.8 × 10-5M
The concentration of NaOH is:
The pOH of this solution is:
pOH = -log [OH–] = -log (9.7 × 10-4) = 3.01
The pH is:
pH = 14.00 – pOH = 10.99
The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).
Check Your Learning:
Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 × 10−5M HCl solution from 4.74 to 3.00.
Answer:
Moles of H3O+ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L × 0.0010 L = 1.0 × 10−4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:
If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.
Buffer Capacity
Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 3). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.
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Selection of Suitable Buffer Mixtures
There are two useful rules of thumb for selecting buffer mixtures:
- A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure 4 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 4. 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−]=0.10M. - Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.
Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H2CO3, and the bicarbonate ion, H2CO3(aq) + When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction:
H3O+(aq) + HCO3–(aq) → H2CO3(aq) + H2O(l)
When an excess of the hydroxide ion is present, it is removed by the reaction:
OH–(aq) + H2CO3(aq) → HCO3–(aq) + H2O(l)
The pH of human blood thus remains very near 7.35, that is, slightly basic. Variations are usually less than 0.1 of a pH unit. A change of 0.4 of a pH unit is likely to be fatal.
The Henderson-Hasselbalch Equation
The ionization-constant expression for a solution of a weak acid can be written as:
Rearranging to solve for [H3O+], we get:
Taking the negative logarithm of both sides of this equation, we arrive at:
,
which can be written as
where pKa is the negative of the common logarithm of the ionization constant of the weak acid (pKa = −log Ka). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation.
Note: LAWRENCE JOSEPH HENDERSON AND KARL ALBERT HASSELBALCH
Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.
In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born.
Medicine: MEDICINE: THE BUFFER SYSTEM IN BLOOD
View information on the buffer system encountered in natural waters. |
Key Concepts and Summary
A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).
Key Equations
- pKa = −log Ka
- pKb = −log Kb
7.6 Buffers Exercises
- Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H3PO4 and a salt of its conjugate base NaH2PO4.
- Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH3 and a salt of its conjugate acid NH4Cl.
- What is [H3O+] in a solution of 0.25 M CH3CO2H and 0.030 M NaCH3CO2?
- What is [H3O+] in a solution of 0.075 M HNO2 and 0.030 M NaNO2?
- What is [OH−] in a solution of 0.125 M CH3NH2 and 0.130 M CH3NH3Cl?
- What is [OH−] in a solution of 1.25 M NH3 and 0.78 M NH4NO3?
- What concentration of NH4NO3 is required to make [OH−] = 1.0 × 10−5 in a 0.200-M solution of NH3?
- What concentration of NaF is required to make [H3O+] = 2.3 × 10−4 in a 0.300-M solution of HF?
- What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:
(a) HCl
(b) KCH3CO2
(c) NaCl
(d) KOH
(e) CH3CO2H - What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:
(a) KI
(b) NH3
(c) HI
(d) NaOH
(e) NH4Cl - What will be the pH of a buffer solution prepared from 0.20 mol NH3, 0.40 mol NH4NO3, and just enough water to give 1.00 L of solution?
- Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH2PO4, and enough water to make 0.500 L of solution.
- How much solid NaCH3CO2•3H2O 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.)
- A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use 1.80 HOLD EQ 10−5 as Ka for acetic acid.
(a) What is the pH of the solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to 0.200 L of the original buffer? - A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution
diluted to 0.100 L.
(a) What is the pH of this buffer solution?
(b) Is the solution acidic or basic?
(c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution? - Which acid in Figure 1 of the Relative Strengths of Acids and Bases is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.
- Which acid in Figure 1 of the Relative Strengths of Acids and Bases is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.
- Which base in Figure 1 of the Relative Strengths of Acids and Bases is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.
- Which base in Figure 1 of the Relative Strengths of Acids and Bases is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.
Glossary
buffer capacity
amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)
buffer
mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added
Henderson-Hasselbalch equation
equation used to calculate the pH of buffer solutions