Buffers, Acid-Base Equilibria | Chemistry 2e

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 solution containing appreciable amounts of a weak conjugate acid-base pair 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 (CNX_Chem_14_06_compare). 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)).

View original figure (external; image not reproduced in this text edition)

Original figure description: Two images are shown. Image a on the left shows two beakers that each contain yellow solutions. The beaker on the left is labeled “Unbuffered” and the beaker on the right is labeled “p H equals 8.0 buffer.” Image b similarly shows 2 beakers. The beaker on the left contains a bright orange solution and is labeled “Unbuffered.” The beaker on the right is labeled “p H equals 8.0 buffer.”

(a) The unbuffered solution on the left and the buffered solution on the right have the same pH (pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH. (b) After the addition of 1 mL of a 0.01-M HCl solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)

How Buffers Work

To illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, strong base added to this solution will neutralize hydronium ion, causing the acetic acid ionization equilibrium to shift to the right and generate additional amounts of the weak conjugate base (acetate ion):

{CH}_{3} {CO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {CH}_{3} {CO}_{2}^{−} (a q)

Likewise, strong acid added to this buffer solution will shift the above ionization equilibrium left, producing additional amounts of the weak conjugate acid (acetic acid). CNX_Chem_14_06_bufferchrt provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH, therefore, changes much less drastically than it would in an unbuffered solution.

View original figure (external; image not reproduced in this text edition)

Original figure description: This figure begins with a chemical reaction at the top: C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O superscript positive sign ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ). Below the text is a figure that resembles a bar graph. In the middle are two bars of equal height. One is labeled, “C H subscript 3 C O O H,” and the other is labeled, “C H subscript 3 C O O superscript negative sign.” There is a dotted line at the same height of the bars which extends to the left and right. Above these two bars is the phrase, “Buffer solution equimolar in acid and base.” There is an arrow pointing to the right which is labeled, “Add O H superscript negative sign.” The arrow points to two bars again, but this time the C H subscript 3 C O O H bar is shorter than that C H subscript 3 C O O superscript negative sign bar. Above these two bars is the phrase, “Buffer solution after addition of strong base.” From the middle bars again, there is an arrow that points left. The arrow is labeled, “Add H subscript 3 O superscript positive sign.” This arrow points to two bars again, but this time the C H subscript 3 C O O H bar is taller than the C H subscript 3 C O O superscript negative sign bar. These two bars are labeled, “Buffer solution after addition of strong acid.”

Buffering action in a mixture of acetic acid and acetate salt.

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 affect 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.

(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.

(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.

Solution

(a) Following the ICE approach to this equilibrium calculation yields the following:

View original figure (external; image not reproduced in this text edition)

Original figure description: This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of “[ C H subscript 3 C O subscript 2 H ] [ H subscript 2 O ] equilibrium arrow H subscript 3 O superscript plus sign [ C H subscript 3 C O subscript 2 superscript negative sign ].” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10, negative x, 0.10 minus sign x. The second column is blank. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0.10, positive x, 0.10 plus sign x.

Substituting the equilibrium concentration terms into the K_a expression, assuming x << 0.10, and solving the simplified equation for x yields

x = 1.8 \times 10^{−5} M

[H_{3} O^{+}] = 0 + x = 1.8 \times 10^{−5} M

pH = −log [H_{3} O^{+}] = −log (1.8 \times 10^{−5})

= 4.74

(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.

Adding strong base will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.

0.0010 L \times (\frac{0.10 mol NaOH}{1 L}) = 1.0 \times 10^{−4} mol NaOH

The initial molar amount of acetic acid is

0.100 L \times (\frac{0.100 mol {CH}_{3} {CO}_{2} H}{1 L}) = 1.00 \times 10^{−2} mol {CH}_{3} {CO}_{2} H

The amount of acetic acid remaining after some is neutralized by the added base is

(1.0 \times 10^{−2}) - (0.01 \times 10^{−2}) = 0.99 \times 10^{−2} mol {CH}_{3} {CO}_{2} H

The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of

(1.0 \times 10^{−2}) + (0.01 \times 10^{−2}) = 1.01 \times 10^{−2} mol {NaCH}_{3} {CO}_{2}

Compute molar concentrations for the two buffer components:

[{CH}_{3} {CO}_{2} H] = \frac{9.9 \times 10^{−3} mol}{0.101 L} = 0.098 M

[{NaCH}_{3} {CO}_{2}] = \frac{1.01 \times 10^{−2} mol}{0.101 L} = 0.100 M

Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).

(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.

The amount of hydronium ion initially present in the solution is

[H_{3} O^{+}] = 10^{- 4.74} = 1.8 \times 10^{- 5} M

mol H_{3} O^{+} = (0.100 L) (1.8 \times 10^{−5} M) = 1.8 \times 10^{−6} mol H_{3} O^{+}

The amount of hydroxide ion added to the solution is

mol {OH}^{-} = (0.0010 L) (0.10 M) = 1.0 \times 10^{−4} mol {OH}^{-}

The added hydroxide will neutralize hydronium ion via the reaction

H_{3} O^{+} (a q) + {OH}^{-} (a q) ⇋ 2 H_{2} O (l)

The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).

The amount of hydroxide ion remaining is

1.0 \times 10^{−4} mol - 1.8 \times 10^{−6} mol = 9.8 \times 10^{−5} mol {OH}^{-}

corresponding to a hydroxide molarity of

9.8 \times 10^{−5} mol {OH}^{-} / 0.101 L = 9.7 \times 10^{−4} M

The pH of the solution is then calculated to be

pH = 14.00 - pOH = 14.00 - - log (9.7 \times 10^{−4}) = 10.99

In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75).

Check Your Learning

Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8

\times

10⁻⁵ M HCl solution from 4.74 to 3.00.

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (CNX_Chem_14_06_exhaust). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.

View original figure (external; image not reproduced in this text edition)

Original figure description: Three Vernier LabQuest devices measure pH: 8.00 (yellow solution), 7.66 (yellow solution), and 1.53 (red solution), demonstrating a range of pH values.

The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)

The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.

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. CNX_Chem_14_06_buffer shows how pH changes for 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.

View original figure (external; image not reproduced in this text edition)

Original figure description: 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].

Change in 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 $[{CH}_{3} {CO}_{2}^{−}] = 0.10 M .$ Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.
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, ${HCO}_{3}^{−} .$ When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:

H_{3} O^{+} (a q) + {HCO}_{3}^{−} (a q) ⟶ H_{2} {CO}_{3} (a q) + H_{2} O (l)

An added hydroxide ion is removed by the reaction:

{OH}^{−} (a q) + H_{2} {CO}_{3} (a q) ⟶ {HCO}_{3}^{−} (a q) + H_{2} O (l)

The added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair (H₃O⁺ is converted to H₂CO₃ and OH^- is converted to HCO₃^-). The pH of human blood thus remains very near the value determined by the buffer pairs pKa, in this case, 7.35. Normal variations in blood pH are usually less than 0.1, and pH changes of 0.4 or greater are likely to be fatal.

The Henderson-Hasselbalch Equation

The ionization-constant expression for a solution of a weak acid can be written as:

K_{a} = \frac{[H_{3} O^{+}] [A^{−}]}{[HA]}

Rearranging to solve for [H₃O⁺] yields:

[H_{3} O^{+}] = K_{a} \times \frac{[HA]}{[A^{−}]}

Taking the negative logarithm of both sides of this equation gives

−log [H_{3} O^{+}] = −log K_{a} − log \frac{[HA]}{[A^{−}]},

which can be written as

pH = p K_{a} + log \frac{[A^{−}]}{[HA]}

where pK_a is the negative of the logarithm of the ionization constant of the weak acid (pK_a = −log K_a). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair 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.

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.

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: The Buffer System in Blood

The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:

{CO}_{2} (g) + 2 H_{2} O (l) ⇌ H_{2} {CO}_{3} (a q) ⇌ {HCO}_{3}^{−} (a q) + H_{3} O^{+} (a q)

The concentration of carbonic acid, H₂CO₃ is approximately 0.0012 M, and the concentration of the hydrogen carbonate ion, ${HCO}_{3}^{−},$ is around 0.024 M. Using the Henderson-Hasselbalch equation and the pK_a of carbonic acid at body temperature, we can calculate the pH of blood:

pH = p K_{a} + log \frac{[base]}{[acid]} = 6.4 + log \frac{0.024}{0.0012} = 7.7

The fact that the H₂CO₃ concentration is significantly lower than that of the ${HCO}_{3}^{−}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.

Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the ${HCO}_{3}^{−}$ ion, producing H₂CO₃. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO₂ from the blood through the lungs driving the equilibrium reaction such that [H₃O⁺] is lowered. If the blood is too alkaline, a lower breath rate increases CO₂ concentration in the blood, driving the equilibrium reaction the other way, increasing [H⁺] and restoring an appropriate pH.

Key Concepts and Summary

Solutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one conjugate partner and preventing further buffering action.

Key Equations

pK_a = −log K_a

pK_b = −log K_b

pH = p K_{a} + log \frac{[A^{−}]}{[HA]}

Explain why a buffer can be prepared from a mixture of NH₄Cl and NaOH but not from NH₃ and NaOH.

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.

Excess H₃O⁺ is removed primarily by the reaction: $H_{3} O^{+} (a q) + H_{2} {PO}_{4}^{−} (a q) ⟶ H_{3} {PO}_{4} (a q) + H_{2} O (l)$
Excess base is removed by the reaction: ${OH}^{−} (a q) + H_{3} {PO}_{4} (a q) ⟶ H_{2} {PO}_{4}^{−} (a q) + H_{2} O (l)$

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?
${CH}_{3} {CO}_{2} H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {CH}_{3} {CO}_{2}^{−} (a q) K_{a} = 1.8 \times 10^{−5}$

[H₃O⁺] = 1.5 $\times$ 10⁻⁴ M

What is [H₃O⁺] in a solution of 0.075 M HNO₂ and 0.030 M NaNO₂?
${HNO}_{2} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {NO}_{2}^{−} (a q) K_{a} = 4.5 \times 10^{−5}$

What is [OH⁻] in a solution of 0.125 M CH₃NH₂ and 0.130 M CH₃NH₃Cl?
${CH}_{3} {NH}_{2} (a q) + H_{2} O (l) ⇌ {CH}_{3} {NH}_{3}^{+} (a q) + {OH}^{−} (a q) K_{b} = 4.4 \times 10^{−4}$

[OH⁻] = 4.2 $\times$ 10⁻⁴ M

What is [OH⁻] in a solution of 1.25 M NH₃ and 0.78 M NH₄NO₃?
${NH}_{3} (a q) + H_{2} O (l) ⇌ {NH}_{4}^{+} (a q) + {OH}^{−} (a q) K_{b} = 1.8 \times 10^{−5}$

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

(d) KOH

(e) CH₃CO₂H

(a) The added HCl will increase the concentration of H₃O⁺ slightly, which will react with ${CH}_{3} {CO}_{2}^{−}$ and produce CH₃CO₂H in the process. Thus, $[{CH}_{3} {CO}_{2}^{−}]$ decreases and [CH₃CO₂H] increases. (b) The added KCH₃CO₂ will increase the concentration of $[{CH}_{3} {CO}_{2}^{−}]$ which will react with H₃O⁺ and produce CH₃CO₂ H in the process. Thus, [H₃O⁺] decreases slightly and [CH₃CO₂H] increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce OH⁻ ions, which will react with the H₃O⁺, thus reducing [H₃O⁺]. Some additional CH₃CO₂H will dissociate, producing $[{CH}_{3} {CO}_{2}^{−}]$ ions in the process. Thus, [CH₃CO₂H] decreases slightly and $[{CH}_{3} {CO}_{2}^{−}]$ increases. (e) The added CH₃CO₂H will increase its concentration, causing more of it to dissociate and producing more $[{CH}_{3} {CO}_{2}^{−}]$ and H₃O⁺ in the process. Thus, [H₃O⁺] increases slightly and $[{CH}_{3} {CO}_{2}^{−}]$ increases.

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

(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?

pH = 8.95

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.)

37 g (0.27 mol)

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 $\times$ 10⁻⁵ as K_a for acetic acid.

(a) What is the pH of the solution?

(b) Is the solution acidic or basic?

(a) pH = 5.222; (b) The solution is acidic. (c) pH = 5.220

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?

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 appreciable amounts of a weak acid-base pair. The pH of a buffer resists change when small amounts of acid or base are added
Henderson-Hasselbalch equation: logarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions

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