# 6.8: Buffer Solutions

Adding as little as 0.1 mL of concentrated $$HCl$$ to a liter of $$H_2O$$ shifts the pH from 7.0 to 3.0. Adding the same amount of HCl to a liter of a solution that is 0.1 M in acetic acid and 0.1 M in sodium acetate, however, results in a negligible change in pH. Why do these two solutions respond so differently to the addition of HCl?

A mixture of acetic acid and sodium acetate is one example of an acid–base buffer. To understand how this buffer works to limit the change in pH, we need to consider its acid dissociation reaction

$\mathrm{CH_3COOH}(aq)+\mathrm{H_2O}(l)\rightleftharpoons\mathrm{H_3O^+}(aq)+\mathrm{CH_3COO^-}(aq)$

and its corresponding acid dissociation constant

$K_\mathrm a=\mathrm{\dfrac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}=1.75\times10^{-5}}\label{6.52}$

Taking the negative log of the terms in Equation $$\ref{6.52}$$ and solving for pH leaves us with the result shown here.

$\mathrm{pH=p\mathit K_a+\log\dfrac{[CH_3COO^-]}{[CH_3COOH]}=4.76+\log\dfrac{[CH_3COO^-]}{[CH_3COOH]}}\label{6.53}$

Buffering occurs because of the logarithmic relationship between pH and the ratio of the concentrations of acetate and acetic acid. Here is an example to illustrate this point. If the concentrations of acetic acid and acetate are equal, the buffer’s pH is 4.76. If we convert 10% of the acetate to acetic acid, by adding a strong acid, the ratio [CH3COO] / [CH3COOH] changes from 1.00 to 0.818, and the pH decreases from 4.76 to 4.67—a decrease of only 0.09 pH units.

## 6.8.1 Systematic Solution to Buffer Problems

Equation $$\ref{6.53}$$ is written in terms of the equilibrium concentrations of CH3COOH and CH3COO . A more useful relationship relates a buffer’s pH to the initial concentrations of the weak acid and the weak base. We can derive a general buffer equation by considering the following reactions for a weak acid, HA, and the salt of its conjugate weak base, NaA.

$\mathrm{NaA}(aq)\rightarrow\mathrm{Na^+}(aq)+\mathrm{A^-}(aq)$

$\mathrm{HA}(aq)+\mathrm{H_2O}(l)\rightleftharpoons\mathrm{H_3O^+}(aq)+\mathrm{A^-}(aq)$

$\mathrm{2H_2O}(l)\rightleftharpoons\mathrm{H_3O^+}(aq)+\mathrm{OH^-}(aq)$

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

$\mathrm{\mathit K_a=\dfrac{[H_3O^+][A^-]}{[HA]}}\label{6.54}$

$\mathrm{\mathit K_w=[H_3O^+][OH^-]}$

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

$C_\mathrm{HA}+C_\mathrm{NaA}=\mathrm{[HA]+[A^-]}\label{6.55}$

$C_\mathrm{NaA}=\mathrm{[Na^+]}\label{6.56}$

and a charge balance equation

$\mathrm{[H_3O^+]+[Na^+]=[OH^-]+[A^-]}\label{6.57}$

Substituting Equation $$\ref{6.56}$$ into Equation $$\ref{6.57}$$ and solving for [A] gives

$\mathrm{[A^-]=\mathit C_{NaA}-[OH^-]+[H_3O^+]}\label{6.58}$

Next, we substitute Equation $$\ref{6.58}$$ into Equation $$\ref{6.55}$$, which gives the concentration of HA as

$\mathrm{[HA]=\mathit C_{HA}+[OH^-]-[H_3O^+]}\label{6.59}$

Finally, substituting Equation $$\ref{6.58}$$ and $$\ref{6.59}$$ into Equation $$\ref{6.54}$$ and solving for pH gives a general equation for a buffer’s pH.

$\mathrm{pH=p\mathit K_a+\log\dfrac{\mathit C_{NaA}-[OH^-]+[H_3O^+]}{\mathit C_{HA}+[OH^-]-[H_3O^+]}}$

If the initial concentrations of the weak acid, CHA, and the weak base, CNaA are greater than [H3O+] and [OH], then we can simplify the general equation to the Henderson–Hasselbalch approximation.

$\mathrm{pH=p\mathit K_a+\log\dfrac{\mathit C_{NaA}}{\mathit C_{HA}}}\label{6.60}$

Henderson and Hasselbalch

Lawrence Henderson (1878-1942) first developed a relationship between $$[H_3O^+]$$, $$[HA]$$, and $$[A^−]$$ while studying the buffering of blood. Kurt Hasselbalch (1874-1962) modified Henderson’s equation by transforming it to the logarithmic form shown in Equation $$\ref{6.60}$$.

The assumptions leading to Equation $$\ref{6.60}$$ produce a minimal error in pH (<±5%) for larger concentrations of HA and A–, for concentrations of HA and A– that are similar in magnitude, and for weak acid’s with pKa values closer to 7. For most problems in this textbook, Equation $$\ref{6.60}$$ provides acceptable results. Be sure, however, to test your assumptions.

For a discussion of the Henderson–Hasselbalch approximation, including the error inherent in Equation $$\ref{6.60}$$, see Po, H. N.; Senozan, N. M. “The Henderson–Hasselbalch approximation: Its History and Limitations,” J. Chem. Educ. 2001, 78, 1499–1503.

As outlined below, the Henderson–Hasselbalch approximation provides a simple way to calculate the pH of a buffer, and to determine the change in pH upon adding a strong acid or strong base.

Example 6.12

Calculate the pH of a buffer that is 0.020 M in NH3 and 0.030 M in NH4Cl. What is the pH after adding 1.0 mL of 0.10 M NaOH to 0.10 L of this buffer?

Solution

The acid dissociation constant for NH4+ is 5.70 × 10–10, or a pKa of 9.24. Substituting the initial concentrations of NH3 and NH4Cl into Equation $$\ref{6.60}$$ and solving, we find that the buffer’s pH is

$\textrm{pH}=9.24+\log\dfrac{0.020}{0.030}=9.06$

Adding NaOH converts a portion of the NH4+ to NH3 as a result of the following reaction

$\mathrm{NH_4^++OH^-\rightleftharpoons H_2O+NH_3}$

Because this reaction’s equilibrium constant is so large (it is 5.7 × 104), we may treat the reaction as if it goes to completion. The new concentrations of NH4+ and NH3 are

$C_\mathrm{NH_4^+}=\dfrac{\mathrm{mol\;NH_4^+-mol\;OH^-}}{V_\mathrm{total}}$

$\mathrm{=\dfrac{(0.030\;M)(0.10\;L)-(0.10\;M)(1.0\times10^{-3}\;L)}{0.10\;L+1.0\times10^{-3}\;L}=0.029\;M}$

$C_\mathrm{NH_3}=\dfrac{\mathrm{mol\;NH_3+mol\;OH^-}}{V_\mathrm{total}}$

$\mathrm{=\dfrac{(0.020\;M)(0.10\;L)+(0.10\;M)(1.0\times10^{-3}\;L)}{0.10\;L+1.0\times10^{-3}\;L}=0.021\;M}$

Substituting these concentrations into the Equation $$\ref{6.60}$$ gives a pH of

$\textrm{pH}=9.24+\log\dfrac{0.021}{0.029}=9.10$

With a pH of 9.06, the concentration of H3O+ is 8.71×10–10 and the concentration of OH is 1.15×10–5. Because both of these concentrations are much smaller than either CNH3 or CNH4Cl, the approximations leading to Equation $$\ref{6.60}$$ are reasonable.

Tote that adding NaOH increases the pH from 9.06 to 9.10. As we expect, adding a base makes the pH more basic. Checking to see that the pH changes in the right direction is one way to catch a calculation error.

Exercise 6.12

Calculate the pH of a buffer that is 0.10 M in KH2PO4 and 0.050 M in Na2HPO4. What is the pH after adding 5.0 mL of 0.20 M HCl to 0.10 L of this buffer. Use Appendix 11 to find the appropriate Ka value.

We can use a multiprotic weak acid to prepare buffers at as many different pH’s as there are acidic protons, with the Henderson–Hasselbalch approximation applying in each case. For example, using malonic acid (pKa1 = 2.85 and pKa1 = 5.70) we can prepare buffers with pH values of

$\mathrm{pH}=2.85+\log\dfrac{C_\mathrm{HM^-}}{C_\mathrm{H_2M}}$

$\mathrm{pH}=5.70+\log\dfrac{C_\mathrm{M^{2-}}}{C_\mathrm{HM^-}}$

where H2M, HM and M2– are malonic acid’s different acid–base forms.

Although our treatment of buffers relies on acid–base chemistry, we can extend the use of buffers to equilibria involving complexation or redox reactions. For example, the Nernst equation for a solution containing Fe2+ and Fe3+ is similar in form to the Henderson-Hasselbalch approximation.

$E=E^\circ_\mathrm{Fe^{3+}/Fe^{2+}}-0.05916\log\mathrm{\dfrac{[Fe^{2+}]}{[Fe^{3+}]}}$

A solution containing similar concentrations of Fe2+ and Fe3+ is buffered to a potential near the standard state reduction potential for Fe3+. We call such solutions redox buffers. Adding a strong oxidizing agent or a strong reducing agent to a redox buffer results in a small change in potential.

## 6.8.2 Representing Buffer Solutions with Ladder Diagrams

A ladder diagram provides a simple graphical description of a solution’s predominate species as a function of solution conditions. It also provides a convenient way to show the range of solution conditions over which a buffer is effective. For example, an acid–base buffer exists when the concentrations of the weak acid and its conjugate weak base are similar. For convenience, let’s assume that an acid–base buffer exists when

$\mathrm{\dfrac{1}{10}\leq\dfrac{[CH_3COO^-]}{[CH_3COOH]}\leq\dfrac{10}{1}}$

Substituting these ratios into the Henderson–Hasselbalch approximation

$\mathrm{pH=p\mathit K_a+\log\dfrac{1}{10}=p\mathit K_a-1}$

$\mathrm{pH=p\mathit K_a+\log\dfrac{10}{1}=p\mathit K_a+1}$

shows that an acid–base buffer works over a pH range of pKa ± 1.

Using the same approach, it is easy to show that a metal-ligand complexation buffer for MLn exists when

$\textrm{pL}=\log K_n\pm 1\hspace{4mm}\mathrm{or}\hspace{4mm}\mathrm{pL}=\dfrac{1}{n}\log\beta_n\pm \dfrac 1n$

$\color{red} {\mathrm{pL=-\log[L]}}$

where Kn or βn is the relevant stepwise or overall formation constant. For an oxidizing agent and its conjugate reducing agent, a redox buffer exists when

$E=E^\circ\pm\dfrac{1}{n}\times\dfrac{RT}{F}=E^\circ\pm\dfrac{0.05916}{n}\mathrm{(at\>25^\circ\>C)}$

Ladder diagrams showing buffer regions for several equilibria are shown in Figure 6.14.

Figure 6.14: Ladder diagrams showing buffer regions for (a) an acid–base buffer for HF and F; (b) a metal–ligand complexation buffer for Ca2+ and Ca(EDTA)2–; and (c) an oxidation–reduction (redox) buffer for Sn4+ and Sn2+.

## 6.8.3 Preparing Buffers

Buffer capacity is the ability of a buffer to resist a change in pH when adding a strong acid or a strong base. A buffer’s capacity to resist a change in pH is a function of the concentrations of the weak acid and the weak base, as well as their relative proportions. The importance of the weak acid’s concentration and the weak base’s concentration is obvious. The more moles of weak acid and weak base a buffer has, the more strong base or strong acid it can neutralize without significantly changing its pH.

Athough higher concentrations of buffering agents provide greater buffer capacity, there are reasons for using smaller concentrations, including the formation of unwanted precipitates and the tolerance of cells for high concentrations of dissolved salts.

The relative proportions of a weak acid and a weak base also affects how much the pH changes when adding a strong acid or a strong base. Buffers that are equimolar in weak acid and weak base require a greater amount of strong acid or strong base to bring about a one unit change in pH. Consequently, a buffer is most effective against the addition of strong acids or strong bases when its pH is near the weak acid’s pKa value.

Buffer solutions are often prepared using standard “recipes” found in the chemical literature.3 In addition, there are computer programs and on-line calculators to aid in preparing buffers.4 Perhaps the simplest way to make a buffer, however, is to prepare a solution containing an appropriate conjugate weak acid and weak base and measure its pH. You can then adjust the pH to the desired value by adding small portions of either a strong acid or a strong base.

A good “rule of thumb” when choosing a buffer is to select one whose reagents have a pKa value close to your desired pH.