Buffers

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 ([link]). 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)).

How Buffers Work

A mixture of acetic acid and sodium acetate is acidic because the K_a of acetic acid is greater than the K_b 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. The decrease in hydronium ion concentration causes the acetic acid hydrolysis equilibrium to shift to the right, restoring the hydronium ion concentration almost to its original value, and yielding a relatively modest increase in pH:

If we add an acid such as hydrochloric acid, the resultant increase in hydronium ion concentration shifts the equilibrium to the left. This effectively converts the added strong acid to a much weaker acid (acetic acid), and the buffer solution thus experiences only a slight decrease in pH.

A mixture of ammonia and ammonium chloride is basic because the K_b for ammonia is greater than the K_a for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value:

If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value:

The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid.

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):

Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth is labeled “Check the math.”

Determine the direction of change. The equilibrium in a mixture of H₃O⁺, ${CH}_{3} {CO}_{2}^{−},$
and CH₃CO₂H is:

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

The equilibrium constant for CH₃CO₂H is not given, so we look it up in Appendix H: K_a = 1.8
$\times$
10⁻⁵. With [CH₃CO₂H] =
$[{CH}_{3} {CO}_{2}^{−}]$
= 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:

$x = 1.8 \times 10^{−5} M$

and

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

Thus:

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

$= 4.74$
Check the work. If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = K_a.

(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:

$0.0010 L \times (\frac{0.10 mol NaOH}{1 L}) = 1.0 \times 10^{−4} mol NaOH$
Determine the moles of CH₂CO₂H. Before reaction, 0.100 L of the buffer solution contains:

$0.100 L \times (\frac{0.100 mol {CH}_{3} {CO}_{2} H}{1 L}) = 1.00 \times 10^{−2} mol {CH}_{3} {CO}_{2} H$
Solve for the amount of NaCH₃CO₂ produced. The 1.0 $\times$
10⁻⁴ mol of NaOH neutralizes 1.0
$\times$
10⁻⁴ mol of CH₃CO₂H, leaving:

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

and producing 1.0
$\times$
10⁻⁴ mol of NaCH₃CO₂. This makes a total of:

$(1.0 \times 10^{−2}) + (0.01 \times 10^{−2}) = 1.01 \times 10^{−2} mol {NaCH}_{3} {CO}_{2}$
Find the molarity of the products. After reaction, CH₃CO₂H and NaCH₃CO₂ are contained in 101 mL of the intermediate solution, so:

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

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

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 ([link]).

(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 $\times$

10⁻⁵-M solution of HCl). The volume of the final solution is 101 mL.

Solution This 1.8 $\times$

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:

0.100 L \times (\frac{1.8 \times 10^{−5} mol HCl}{1 L}) = 1.8 \times 10^{−6} mol HCl

As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 $\times$

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 \times 10^{−4}) - (1.8 \times 10^{−6}) = 9.8 \times 10^{−5} M

The concentration of NaOH is:

\frac{9.8 \times 10^{−5} M NaOH}{0.101 L} = 9.7 \times 10^{−4} M

The pOH of this solution is:

pOH = −log [{OH}^{−}] = −log (9.7 \times 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 $\times$

10⁻⁵ M HCl solution from 4.74 to 3.00.

Answer:

Initial pH of 1.8 $\times$

10⁻⁵ M HCl; pH = −log[H₃O⁺] = −log[1.8 $\times$

10⁻⁵] = 4.74* * *

Moles of H₃O⁺ in 100 mL 1.8 $\times$

10⁻⁵ M HCl; 1.8 $\times$

10⁻⁵ moles/L $\times$

0.100 L = 1.8 $\times$

10⁻⁶* * *

Moles of H₃O⁺ added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L $\times$

0.0010 L = 1.0 $\times$

10⁻⁴ moles; final pH after addition of 1.0 mL of 0.10 M HCl:* * *

pH = −log [H_{3} O^{+}] = −log (\frac{total moles H_{3} O^{+}}{total volume}) = −log (\frac{1.0 \times 10^{−4} mol + 1.8 \times 10^{−6} mol}{101 mL (\frac{1 L}{1000 mL})}) = 3.00

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 ([link]). 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.

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

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:

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:

where pK_a is the negative of the common 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 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.

Key Concepts and Summary

A solution containing a mixture of a weak acid and its conjugate base, or of a weak base and its conjugate acid, is called a buffer solution. The presence of a weak conjugate acid-base pair provides reactants that may neutralize small additions of strong acid or base, yielding weaker conjugate partners. The hydronium ion concentration of a buffer solution therefore does not change significantly when a small amount of acid or base is added.

Key Equations

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 concentration of NH₄NO₃ is required to make [OH⁻] = 1.0 $\times$

10⁻⁵ in a 0.200-M solution of NH₃?

[NH₄NO₃] = 0.36 M

What concentration of NaF is required to make [H₃O⁺] = 2.3 $\times$

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

(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.* * *

(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.* * *

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?

Which acid in [link] is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice.

To prepare the best buffer for a weak acid HA and its salt, the ratio $\frac{[H_{3} O^{+}]}{K_{a}}$

should be as close to 1 as possible for effective buffer action. The [H₃O⁺] concentration in a buffer of pH 3.1 is [H₃O⁺] = 10^−3.1 = 7.94 $\times$

10⁻⁴ *M** * *

We can now solve for K_a of the best acid as follows:* * *

\begin{array}{l} \frac{[H_{3} O^{+}]}{K_{a}} = 1 \\ K_{a} = \frac{[H_{3} O^{+}]}{1} = 7.94 \times 10^{−4} \end{array}

In [link], the acid with the closest K_a to 7.94 $\times$

10⁻⁴ is HF, with a K_a of 7.2 $\times$

10⁻⁴.

Which acid in [link] is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice.

Which base in [link] is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice.

For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio $\frac{[{OH}^{−}]}{K_{b}}$

that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH⁻] is [OH⁻] = 10^−pOH = 10^−3.35 = 4.467 $\times$

10⁻⁴ M.* * *

We can now solve for K_b of the best base as follows:* * *

\frac{[{OH}^{−}]}{K_{b}} = 1

K_b = [OH⁻] = 4.47 $\times$

10⁻⁴* * *

In [link], the base with the closest K_b to 4.47 $\times$

10⁻⁴ is CH₃NH₂, with a K_b = 4.4 $\times$

10⁻⁴.

Which base in [link] is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice.

Saccharin, C₇H₄NSO₃H, is a weak acid (K_a = 2.1 $\times$

10⁻²). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 $\times$

10⁻³ g of sodium saccharide, Na(C₇H₄NSO₃), what are the final concentrations of saccharine and sodium saccharide in the solution?

The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:* * *

9.75 $\times$

10⁻⁶ mol* * *

The pKa for [HA] is 1.68, so [HA] = $6.2 \times 19^{–9}$

M. Thus, [A–] (saccharin ions) is $3.90 \times 10^{–5}$

M.* * *

Thus, A⁻ is 3.90 $\times$

10⁻⁵ M

What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C₅H₉NO₄, a diprotic acid; K₁ = 8.5 $\times$

10⁻⁵, K₂ = 3.39 $\times$

10⁻¹⁰) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?