Buffers

Buffer Action

Proton concentration [H⁺] and pH of a buffer can be determined using the Henderson-Hasselbalch equation.

A buffer is a solution that resists a change in pH when acid or base is added and contains significant quantities of a weak acid and its conjugate base or a weak base and its conjugate acid. Adding small quantities of acids or bases shifts the equilibrium, reducing the number of protons or hydroxide ions in solution. A buffer can be made by adding a weak acid to a salt of its conjugate base, providing the metal cation will not itself affect pH.

A buffer of acetic acid (CH₃COOH), for example, could be made by mixing acetic acid with potassium acetate (CH₃COOK). In aqueous solution, potassium acetate completely dissociates into potassium ions (K⁺) and acetate ions (CH₃COO^–). Potassium ions (K⁺) do not play a role in the buffer because the pH is affected only by the acid and its conjugate base. The acetic acid molecules can donate a proton to the added base, and the acetate ions (CH₃COO^–) can absorb protons (H⁺) from any acid that is added.

{\rm{CH_3COOH}}(aq)+{\rm{H_2O}}(l)\rightleftarrows{\rm{CH_3COO^-}}(aq)+{\rm{H_3O^+}}(aq)

As protons (H⁺) are added to this (or any) aqueous buffer solution, the CH₃COO^– ions absorb them, producing more molecules of CH₃COOH, and the pH does not change dramatically. Similarly, if OH^– is added instead, it bonds with the H⁺ ions, producing more water molecules and maintaining the pH. The pH of a buffer system will depend on the value of the acid equilibrium constant K_a for the weak acid and the concentrations of acid (HA) and base (A^–) that are initially mixed. These quantities can be related to each other using the Henderson-Hasselbalch equation, formulated by American chemist Lawrence Joseph Henderson and Danish chemist Karl Albert Hasselbalch. The pK_a is the negative log of the K_a value.

{\rm{pH}}={\rm{p}}K{\rm_{a}}+\log\frac{[{\rm{A}}^-]}{[{\rm{HA}}]}

For example, the chemical name for vitamin C is ascorbic acid (HC₆H₇O₆). It has a pK_a of 4.1. The stomach has a pH of 1.4. When an individual ingests a vitamin C tablet, it dissolves in their stomach. The Henderson-Hasselbalch equation can be used to determine the ratio of the concentrations by solving for the term inside the logarithm.

\begin{aligned}\rm{pH}&=\rm{p}K_{\rm{a}}+\log\frac{[\rm{A^-}]}{[\rm{HA}]}\\1.4&=4.1+\log\frac{[\rm{A^-}]}{[\rm{HA}]}\\-2.7&=\log\frac{[\rm{A^-}]}{[\rm{HA}]}\\10^{-2.7}&=\frac{[\rm{A^-}]}{[\rm{HA}]}\\0.0020&=\frac{[\rm{A^-}]}{\rm{[HA]}}\end{aligned}

Note that it is not necessary to know the actual concentrations of A^– and HA because only the ratio between them is needed. This ratio will always be the same at equilibrium.

If the ratio $\lbrack \rm{A}^-\rbrack/\lbrack \rm{HA}\rbrack$ is close to 1, the pH will be close to the pK_a of the weak acid. Also, consider that ${\rm{p}}K_{\rm{a}}=-\log{K_{\rm{a}}}$ and the fact that the logarithm of an exponential is approximately equal to the exponent. The exponent of K_a therefore gives a rough indication of the value of pH. As more A^– is added, the ratio becomes greater, and as HA is added, the ratio becomes smaller. Keeping the ratio within an order of magnitude in either direction therefore keeps the pH range within about ±1.

Knowing the pH of a buffer system is important when deciding which buffer to use for a specific reaction. The best choice of buffer is one that will maintain the optimal pH for the reaction taking place. Buffer pH can be estimated by assuming that the initial concentrations of acid and conjugate base will be the same, making the Henderson-Hasselbalch equation:

{\rm{pH}}={\rm{p}}K_{\rm{a}}

So the ideal buffer is one that has a K_a value close to the desired proton concentration [H⁺], or pK_a close to the desired pH.

Calculating the pH of a Buffer Solution

Ascorbic acid (HC₆H₇O₆) dissociates according to the equation:

{\rm{HC}}_{6}{\rm H}_7{\rm O}_6(aq)+{\rm H_2}{\rm O}(l)\;\rightleftarrows\;{\rm C}_6{\rm H}_7{\rm O}_{6}{^-}(aq)+\;{\rm H}_{3}{\rm O^+}(aq)

A scientist would like to make a buffer that will maintain a pH of 5.0. To do this, the scientist mixes a solution using equal parts 0.80 M HC₆H₇O₆ and 1.0 M NaC₆H₇O₆. Ascorbic acid has a pK_a of 4.1.

What is the pH of this solution? Will this be an effective buffer for maintaining pH around 5.0?

Use the Henderson-Hasselbalch equation.

\begin{aligned}{\rm{pH}}&={\rm{p}}K_{\rm{a}}+\log{\frac{\rm{[A}^-\rm{]}}{\rm{[HA]}}}\\&=4.1+\log{\frac{\rm{[C}_6\rm{H}_7{\rm{O}_6}^-\rm{]}}{\rm{[HC}_6\rm{H}_7\rm{O}_6\rm{]}}}\\&=4.1+\log{\frac{0.8}{1.0}}\\&=4.1-0.097\\&=4.0\end{aligned}

The pH of the solution is 4.0.

A buffer is able to absorb acid or base up to about 1 pH unit up or down from its initial pH. With a pH of 4.0, this buffer would not be very effective at absorbing OH^– ions once the solution reached a pH of about 5.0. Therefore, this is not an ideal solution for maintaining a pH around 5.0. A better choice would be to use an acid with a pK_a closer to 5.0 or begin with a greater initial concentration of base, which would increase the starting pH of the buffer.

Buffer Capacity

The capacity for a buffer to absorb ions is limited by the concentrations of acid and conjugate base in the solution.

A buffer solution has a finite number of acid and conjugate base particles in it. It therefore can absorb a finite number of strong acid or base ions. The buffer capacity is the number of moles of strong acid, protons (H⁺), or strong base, hydroxide ions (OH^–), that can be added to one liter of solution before the pH changes by 1 in either direction.

A liter of 1 M buffer, for example, contains one mole of HA and one mole of A^–.

{\rm{HA}}(aq)+{\rm{H_2O}}(l)\rightleftarrows{\rm{A^-}}(aq)+{\rm{H_3O^+}}(aq)

Le Chatelier's principle states that a change in the temperature, pressure, or concentration of a component will cause the equilibrium condition of a chemical system to change in a way that reduces the change. Therefore adding acid to this buffer solution will drive the equilibrium to the left. Eventually, the solution will run out of A^– particles to absorb the protons (H⁺). At that point, the solution will lose its ability to buffer, and the pH of the solution will decrease rapidly. Adding hydroxide ions (OH^–) will have a similar effect. The OH^– ions will remove H₃O⁺ from solution, driving the reaction to the right until so much HA has dissociated that the solution can no longer absorb OH^– ions. At this point the OH^– will remain ionized in solution, and the pH will rapidly rise. The pH range over which a buffer is effective in stabilizing the pH of a solution is called the buffer range. The effective range of a buffer is within one pH unit above and below the pH of the buffer. Outside of those pH values, the buffer is not able to absorb additional protons or hydroxide (OH^–) ions.

A buffer with a pH of 4.75 has a buffer range between 3.75 and 5.75. Within the buffer range, the slope of the curve is relatively small. Outside this range, the slope of the curve increases dramatically.

The pH of a buffer solution after a known amount of strong acid or base has been added can be used to determine the effectiveness of the buffer.

Calculating the pH of a Buffer Solution After the Addition of a Strong Acid

What is the pH of the buffer? K_a of ascorbic acid is

7.9\times10^{-5}

. A volume of 50.0 mL of 0.10 M HCl is added to a buffer consisting of 0.030 moles of ascorbic acid (HC₆H₇O₆) and 0.025 moles of NaC₆H₇O₆.

Write a balanced equation for the reaction.

{\rm{C}}_6{\rm{H}}_7{\rm{O}_6}^-(aq)+{\rm{H}}_3{\rm{O}}^+(aq)\leftrightarrows{\rm{HC}}_6{\rm{H}}_7{\rm{O}}_6(aq)+{\rm{H}}_2{\rm{O}}(l)

Use the given volume and molarity to calculate the initial number of moles of HCl.

\begin{aligned}\text{initial moles of }\rm{HCl}&=(\text{volume of }{\rm{HCl})(\text{molarity of }\rm{HCl}})\\&=(50.0\;{\rm{mL})}\left(\frac{1\;{\rm{L}}}{1{,}000\;{\rm{mL}}}\right)(0.10\;{\rm{M}})\\&=0.0050\;{\rm{mol\;HCl}}\end{aligned}

Because HCl dissociates completely, the initial moles of H₃O⁺ is 0.0050 mol.

Develop an ICE table to determine the equilibrium concentrations.

The balanced equation shows the molar ratios of all species in the reaction. Based on these ratios, the number of moles of hydronium ions (H₃O⁺) added to the solution equals the decrease in moles of ascorbate (C₆H₇O(₆)_-) and the increase in moles of ascorbic acid (HC₆H₇O₆).

The third column shows the concentration of ascorbic acid, which starts as 0.030 moles. Then 0.0050 moles of HCl is added, which completely dissociates, adding an additional 0.0050 moles of H⁺ to the solution. The H⁺ ions will be absorbed by the C₆H₇O₆^– ions, making the [HC₆H₇O₆] in the final solution 0.035 moles.

The equilibrium amount of H₃O⁺ will be zero in this case because it is entirely absorbed by the conjugate base ascorbate (C₆H₇O₆^–). The first column shows the change in concentration of the conjugate base, which decreases as the HCl is added and is used up in producing weak acid molecules.

To solve for pH, use the expression for K_a and the Henderson-Hasselbalch equation with

{\rm{p}}K_{\rm{a}}=-\log{K_{\rm{a}}}

\begin{aligned}{\rm{pH}}&={\rm{p}}K_{\rm{a}}+\log{\frac{\rm{[A}^-\rm{]}}{\rm{[HA]}}}\\&={\rm{p}}K_{\rm{a}}+\log{\frac{\rm{[C}_6{\rm{H}}_7{\rm{O}_6}^-{\rm{]}}}{\rm{[HC}_6{\rm{H}}_7{\rm{O}}_6{\rm{]}}}}\\&=(-\log7.9\times10^{-5})+\log{\frac{0.020}{0.035}}\\&=4.10-0.243\\&=3.9\end{aligned}

<Polyprotic Solutions >Acid-Base Titrations

Acid-Base Equilibria

Contents

Buffers

Buffer Action

Buffer Capacity