2.13: Buffers

Based on OpenStax Buffers

Introduction

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 (Figure \(\PageIndex{1}\)). 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)).

Figure \(\PageIndex{1}\): (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)

Note

Although the term buffer usually refers to an acid-base buffer, it is possible to buffer the concentration of other species using appropriate equilibria.

 

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):

\[\ce{CH_3 CO_2 H (aq) + H2O(l) <=> H3O^{+}(aq) + CH_3 CO_2^{-}(aq)} \nonumber \]

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). Figure \(\PageIndex{2}\) 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.

Example 14.20: 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.

1. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
2. Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.
3. 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:

Substituting the equilibrium concentration terms into the K_a expression, assuming \(x \ll 0.10\), and solving the simplified equation for \(x\) yields

\[x=1.8 \times 10^{-5}~\text{M} \nonumber \]

\[\left[ \ce{H3O^{+}} \right]=0+x=1.8 \times 10^{-5}~\text{M} \nonumber \]

\[\begin{align*}
pH &=-\log \left[ \ce{H3O^{+}} \right] \\[4pt]
&=-\log \left(1.8 \times 10^{-5}\right) \\[4pt]
&= 4.74
\end{align*} \nonumber \]

(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.

\[0.0010~ \cancel{\text{L}} \times \left(\dfrac{0.10 ~\text{mol} ~\ce{NaOH}}{1 ~ \text{L}}\right)=1.0 \times 10^{-4}~\text{mol} ~ \ce{NaOH} \nonumber \]

The initial molar amount of acetic acid is

\[0.100 ~ \cancel{\text{L}} \times\left(\frac{0.100~ \text{mol} ~ \ce{CH3CO2H} }{1 ~ \cancel{\text{L}} }\right)=1.00 \times 10^{-2} ~\text{mol} ~ \ce{CH3CO2H} \nonumber \]

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

\[\left(1.0 \times 10^{-2}\right)-\left(0.01 \times 10^{-2}\right)=0.99 \times 10^{-2} ~\text{mol} ~ \ce{CH3CO2H} \nonumber \]

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

\[\left(1.0 \times 10^{-2}\right) + \left(0.01 \times 10^{-2}\right)=1.01 \times 10^{-2}~ \text{mol} ~ \ce{NaCH3CO2} \nonumber \]

Compute molar concentrations for the two buffer components:

\[\begin{align*}
\left[ \ce{CH3CO2H} \right] & =\frac{9.9 \times 10^{-3} ~\text{mol} }{0.101~\text{L} }=0.098 ~\text{M} \\[4pt]
\left[ \ce{NaCH3CO2} \right] & =\frac{1.01 \times 10^{-2} ~\text{mol} }{0.101 ~\text{L}}=0.100~\text{M}
\end{align*} \nonumber \]

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

\[\begin{align*}
\left[ \ce{H3O^{+}} \right] &=10^{-4.74}=1.8 \times 10^{-5} ~ \text{M} \\[4pt]
\text{mol} ~\ce{H3O^{+}} & =(0.100~\text{L})\left(1.8 \times 10^{-5}~\text{M} \right)=1.8 \times 10^{-6}~ \text{mol}~\ce{H3O^{+}}
\end{align*} \nonumber \]

The amount of hydroxide ion added to the solution is

\[mol OH^{-}=(0.0010 L)(0.10 M)=1.0 \times 10^{-4} mol \ce{OH^{-}} \nonumber \]

The added hydroxide will neutralize hydronium ion via the reaction

\[\ce{H3O^{+}(aq) + OH^{-}(aq) \leftrightharpoons 2 H2O(l)} \nonumber \]

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

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

corresponding to a hydroxide molarity of

\[\dfrac{9.8 \times 10^{-5}~\text{mol}~\ce{OH^{-}}}{0.101~\text{L}}=9.7 \times 10^{-4} M \nonumber \]

The pH of the solution is then calculated to be

\[pH =14.00- pOH =14.00 - -\log \left(9.7 \times 10^{-4}\right)=10.99 \nonumber \]

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).

Exercise \(\PageIndex{1}\)

Show that adding 1.0 mL of \(0.10~\text{M}~\ce{HCl}\) changes the pH of 100 mL of a \(1.8 \times 10^{−5} ~\text{M}~\ce{HCl}\) solution from 4.74 to 3.00.

Answer

Initial pH of \(1.8 \times 10^{−5} ~\text{M}~\ce{HCl}\);

\[\text{pH} = −\log[\ce{H3O^{+}}] = −\log[1.8 \times 10^{−5}] = 4.74 \nonumber \]

Moles of H₃O⁺ in 100 mL \(1.8 \times 10^{−5} ~\text{M}~\ce{HCl}\); \[(1.8 \times 10^{−5}~\text{moles/L})\times (0.100 ~\text{L} = 1.8 \times 10^{−6}~\text{mol}\nonumber \]

Moles of H₃O⁺ added by addition of 1.0 mL of 0.10 M HCl: \[(0.10~\text{moles/L}) \times (0.0010~ \text{L}) = 1.0 \times 10^{−4}~\text{mol} \nonumber \]

Final pH after addition of 1.0 mL of 0.10 M HCl:

\[pH =-\log \left[ \ce{H3O^{+}} \right]=-\log \left(\frac{\text { total moles } \ce{H3O^{+}}}{\text {total volume }}\right)=-\log \left(\frac{1.0 \times 10^{-4} mol +1.8 \times 10^{-6}~\text{mol}}{101~\text{mL} \left(\frac{1~\text{L} }{1000~\text{mL}}\right)}\right) = 3.00 \nonumber \]

Buffer Capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant. 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.

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.

The Henderson-Hasselbalch Equation

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

\[K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{A^{-}} \right]}{[ \ce{HA} ]} \nonumber \]

Rearranging to solve for [H₃O⁺] yields:

\[\left[ \ce{H3O^{+}} \right]=K_{ a } \times \frac{[ \ce{HA} ]}{\left[ \ce{A^{-}} \right]} \nonumber \]

Taking the negative logarithm of both sides of this equation gives

\[-\log \left[ \ce{H3O^{+}} \right]=-\log K_{ a }-\log \frac{[ \ce{HA} ]}{\left[ \ce{A^{-}} \right]} \nonumber \]

which can be written as

\[pH = p K_{ a }+\log \frac{\left[ \ce{A^{-}} \right]}{[ \ce{HA} ]} \label{EQ:HH} \]

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. This equation is often used in an approximate form \(\ref{EQ:HHappr}\), where the [A-] and [HA] are not solved for using an equilibrium calculation, but are assumed to be equal to the initial concentration of the salt containing the A- ion (\([salt]_i\)) and the initial concentration of HA (\([HA]_i\)) based on what was mixed into the solution. Many write equation \(\ref{EQ:HH}\) but utilize it as the approximate form, which is equivalent to the "small x" approximation in an equilibrium calculation.

\[pH = p K_{ a }+\log \frac{\left[ salt \right]_i}{[ \ce{HA} ]_i} \label{EQ:HHappr} \]

This last form is usually referred to as Henderson-Hasselbalch equation and used to estimate the pH of buffers.

What is ignored by the approximation

Based on "Buffer Solutions" by David Harvey.

We can derive a general buffer equation by considering the following reactions for a weak acid, HA, and the soluble salt of its conjugate weak base, NaA.

\[\begin{array}{c}{\mathrm{NaA}(s) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{A}^{-}(a q)} \\ {\mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q)} \\ {2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)}\end{array} \nonumber\]

Because the concentrations of Na⁺, A^–, HA, H₃O⁺, and OH^– are unknown, we need five equations to define the solution’s composition. Two of these equations are the equilibrium constant expressions for HA and H₂O.

\[K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \label{6.3}\]

\[K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \nonumber\]

The remaining three equations are mass balance equations for HA and Na⁺

\[C_{\mathrm{HA}}+C_{\mathrm{NaA}}=[\mathrm{HA}]+\left[\mathrm{A}^{-}\right] \label{6.4}\]

\[C_{\mathrm{NaA}}=\left[\mathrm{Na}^{+}\right] \label{6.5}\]

and a charge balance equation

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\left[\mathrm{Na}^{+}\right]=\left[\mathrm{OH}^{-}\right]+\left[\mathrm{A}^{-}\right] \label{6.6}\]

Substituting Equation \ref{6.5} into Equation \ref{6.6} and solving for [A^–] gives

\[\left[\mathrm{A}^{-}\right]=C_{\mathrm{NaA} }-\left[\mathrm{OH}^{-}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \label{6.7}\]

Next, we substitute Equation \ref{6.7} into Equation \ref{6.4}, which gives the concentration of HA as

\[[\mathrm{HA}]=C_{\mathrm{HA}}+\left[\mathrm{OH}^{-}\right]-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \label{6.8}\]

Finally, we substitute Equation \ref{6.7} and Equation \ref{6.8} into Equation \ref{6.3} and solve for pH to arrive at a general equation for a buffer’s pH.

\[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{C_{\mathrm{NaA} }-\left[\mathrm{OH}^{-}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{C_{\mathrm{HA}}+\left[\mathrm{OH}^{-}\right]-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\]If the initial concentrations of the weak acid, C_HA, and the weak base, C_NaA, are significantly greater than [H₃O⁺] and [OH^–], then this becomes the usual approximate form of the Henderson-Hasselbach (equation \(\ref{EQ:HHappr}\)), where \(C_{NaA} = [salt]_i\) and \(C_{HA} = [HA]_i\).

Henderson-Hasselbalch and buffer capacity

The buffer capacity is generally understood to be the amount of acid or base that can be added while changing the pH by a maximum of 1 pH unit. This means we want to see what happens to the [A-] to [HA] ratio when the pH changes by that much. An ideal buffer has the pH = pK_a, which implies we are interested in the case:

\[pH \pm 1 = pK_a \pm 1 = pK_a + log \frac{[A^-]}{[HA]} \nonumber\]

Which simplifies to

\[\pm 1 = log \frac{[A-]}{[HA]} \implies 10^{\pm 1} =  \frac{[A^-]}{[HA]} \nonumber\]

So the buffer runs out when the ratio in the logarithm becomes 1/10 or 10/1.

Portrait of a Chemist: 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.