Acid-Base Equilibria

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.
CH3COOH(aq)+H2O(l)CH3COO(aq)+H3O+(aq){\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 Ka 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 pKa is the negative log of the Ka value.
pH=pKa+log[A][HA]{\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 pKa 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.
pH=pKa+log[A][HA]1.4=4.1+log[A][HA]2.7=log[A][HA]102.7=[A][HA]0.0020=[A][HA]\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 [A]/[HA]\lbrack \rm{A}^-\rbrack/\lbrack \rm{HA}\rbrack is close to 1, the pH will be close to the pKa of the weak acid. Also, consider that pKa=logKa{\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 Ka 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:
pH=pKa{\rm{pH}}={\rm{p}}K_{\rm{a}}
So the ideal buffer is one that has a Ka value close to the desired proton concentration [H+], or pKa close to the desired pH.
Step-By-Step Example
Calculating the pH of a Buffer Solution
Ascorbic acid (HC6H7O6) dissociates according to the equation:
HC6H7O6(aq)+H2O(l)C6H7O6(aq)+H3O+(aq){\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 pKa of 4.1.

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

Step 1
Use the Henderson-Hasselbalch equation.
pH=pKa+log[A][HA]=4.1+log[C6H7O6][HC6H7O6]=4.1+log0.81.0=4.10.097=4.0\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.
Solution
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 pKa 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.
HA(aq)+H2O(l)A(aq)+H3O+(aq){\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.
Step-By-Step Example
Calculating the pH of a Buffer Solution After the Addition of a Strong Acid
What is the pH of the buffer? Ka of ascorbic acid is 7.9×1057.9\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.
Step 1
Write a balanced equation for the reaction.
C6H7O6(aq)+H3O+(aq)HC6H7O6(aq)+H2O(l){\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)
Step 2
Use the given volume and molarity to calculate the initial number of moles of HCl.
initial moles of HCl=(volume of HCl)(molarity of HCl)=(50.0mL)(1L1,000mL)(0.10M)=0.0050molHCl\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.
Step 3
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).
Step 4

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.

Solution
To solve for pH, use the expression for Ka and the Henderson-Hasselbalch equation with pKa=logKa{\rm{p}}K_{\rm{a}}=-\log{K_{\rm{a}}}.
pH=pKa+log[A][HA]=pKa+log[C6H7O6][HC6H7O6]=(log7.9×105)+log0.0200.035=4.100.243=3.9\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}