# 10B: Common Ion Effect and Buffers (Worksheet)

- Page ID
- 81631

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Name: ______________________________

Section: _____________________________

Student ID#:__________________________

Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

Last week we looked at how to calculate the concentrations of all species and pH or pOH in a solution of a pure acid or base in water, with no additional amounts of the conjugate added. Now, we need to look at the effect of adding extra amounts of the conjugate base or acid to the solution. The shift in the position of the equilibrium, called the *common ion effect*, changes the pH and imbues the solution with certain properties that are the basis for formulating a *buffer*. Buffer solutions are important for regulating pH in many chemical and biological systems.

## Learning Objectives

- Understand the effect of adding the conjugate to a solution of a weak acid or weak base
- Understand the principles of buffer solutions

## Success Criteria

- Be able to predict the effect of adding additional amounts of conjugate base or acid to a solution
- Be able to calculate the pH of a solution subject to the common ion effect
- Be able to calculate the pH of a buffer solution and to calculate the ratio of conjugate pair concentrations needed to achieve a certain pH

## Common Ion Effect

Thus far we have only considered the equilibrium of a pure acid or base in water, with no other source for its conjugate but the dissociation itself. What would happen if we were to add a significant amount of the conjugate? Suppose we consider the dissociation of a 0.100 M solution of acetic acid, HOAc, for which \(K_a = 1.76 \times 10^{–5}\). [\(\ce{HOAc}\) is a common abbreviation many chemists used for acetic acid, and likewise \(\ce{OAc^{–}}\) is used an abbreviation for the acetate ion.] What are the concentrations of \(\ce{H3O^{+}}\) and \(\ce{OAc^–}\) in this solution and what is the pH?

#### Add the Weak Acid

We would have the following requilibrium in solution

\[\ce{HOA + H3O <=> H3O^{+} + OAc^{-}} \nonumber \]

and the following algebraic expressions for the principal species at equilibrium can be formulated via an ICE Table (Table \(\PageIndex{1}\)).

Table \(\PageIndex{1}\): Equilibration of \(\ce{HOAc}\)

ICE Table |
\(\ce{HOAc}\) | \(\ce{H2O}\) | \(\ce{H3O^{+}}\) | \(\ce{OAc^{–}}\) |
---|---|---|---|---|

Initial |
\(0.100\) | - | \(0\) | \(0\) |

Change |
\(-x_1\) | - | \(+x_1\) | \(+x_1\) |

Equilibrium |
\(0.100 - x_1\) | - | \(x_1\) | \(x_1\) |

However, \(C_{\ce{HOAc}} \gg K_a\), so we can assume \(\ce{[HOAc]} \approx 0.100\, M\) and \(\ce{ [H3O^{+}]} = \ce{[OAc^{–}]} = x_1\). As we have seen, this leads to the simple solution

\[ \begin{align} x_1 &= \sqrt{ C_{\ce{HOAc}} K_a^{\ce{HOAc}} } = \sqrt{ (0.100 \,M) (1.76 \times 10^{-5}) } = 1.33 \times 10^{-3}\,M \label{eq1} \\ &= \ce{[H3O^{+}]} = \ce{[OAc^{-}]} \end{align} \]

From the definition of pH at

\[pH =- \log_{10} [\ce{H3O^{+}}] \label{phdef} \]

thus it follows that for this solution

\[pH = 2.877 \label{ph1} \]

#### Add a salt of the Conjugate Base (Exact Calculation)

Suppose we added sodium acetate, \(\ce{NaOAc}\), in an amount equivalent to 0.200 mole per liter of this solution. Being a strong electrolyte, the sodium acetate will dissociate completely to give 0.200 mole per liter of \(\ce{OAc^{-}}\), in addition to any acetate ion provided by acetic acid’s hydrolysis (this will also add 0.2000 mol of Na to the solution; but sodium ion has no acid-base character, so it will have no effect on the pH of the solution and we can just ignore it). Now we have the following algebraic expressions for the principal species at equilibrium (Table \(\PageIndex{2}\)):

Table \(\PageIndex{2}\): Equilibration of \(\ce{HOAc}\) and \(\ce{NaOAc}\)

ICE Table |
\(\ce{HOAc}\) | \(\ce{H2O}\) | \(\ce{H3O^{+}}\) | \(\ce{OAc^{–}}\) |
---|---|---|---|---|

Initial |
\(0.100\) | - | \(0\) | \(0.200\) |

Change |
\(-x_2\) | - | \(+x_2\) | \(+x_2\) |

Equilibrium |
\(0.100 - x_2\) | - | \(x_2\) | \(0.200 + x_2\) |

Notice that in this solution, unlike the pure acid case, the concentration of hydronium ion does not equal the concentration of conjugate base. By Le Chatelier's Principle, the added \(\ce{OAc^{-}}\) will drive the equilibrium to the left, removing some of the \(\ce{H3O^{+}}\) ion and \(\ce{OAc^{-}}\) ions to reform molecular \(\ce{HOAc}\) and \(\ce{H2O}\) (i.e., \(x_2 < x_1\)). Put another way, adding conjugate base to the solution will make the pH go up as the concentration of \(\ce{H3O^{+}}\) goes down. From our previous calculation for 0.100 M HOAc, we saw that \(x = 1.33 \times 10^{-3} M\) (Equation \ref{eq1}), but now this number will be even smaller, due to the left shifting in the equilibrium position.

To solve for \(x_2\) and the pH of this new solution, we plug the equlibrium values of Table \(\PageIndex{2}\) into the expression for \(K_a\)

\[ K_a = \ce{\dfrac{[H3O^{+}][OAc^{-}]}{[HOAc]}} = \dfrac{x_2(0.200 + x_2)}{0.100 - x_2} \label{nonapp} \]

#### Add a salt of the Conjugate Base (Approximate Calculation)

Unfortunately, Equation \ref{nonapp} is a little awkward to solve for \(x_2\) and therefore pH. We can simplify this equation via an approximation. Compared to the analytical concentration of \(\ce{HOAc}\) with \(C_{\ce{HOAc }} = 0.100\, M\)) and the analytical concentration of \(\ce{NaOAc}\) salt with \(C_{\ce{NaOAc }} = 0.200\, M\), this small value of \(x_2\) is now insignificant, actually falling beyond the significant digits of those concentrations. Therefore, we can simplify the calculation of \(\ce{[H3O^{+}]}\) by assuming that \(x_2 \ll 1\), therefore

- \(\ce{[HOAc]} = C_{\ce{HOAc}} \approx 0.100\) and
- \(\ce{[OAc^{-}]} = C_{\ce{HOAc}} \approx 0.200 \,M\).

Under this assumption, Table \(\PageIndex{2}\) then looks like Table \(\PageIndex{3}\).

Table \(\PageIndex{3}\): Approximation of Equilibration of \(\ce{HOAc}\) and \(\ce{NaOAc}\)

ICE Table |
\(\ce{HOAc}\) | \(\ce{H2O}\) | \(\ce{H3O^{+}}\) | \(\ce{OAc^{–}}\) |
---|---|---|---|---|

Initial |
\(0.100\) | - | \(0\) | \(0.200\) |

Change |
\(-x_3\) | - | \(+x_3\) | \(+x_3\) |

Equilibrium |
\(0.100 - x_3 \approx 0.100\) | - | \(x_3\) | \(0.200 + x_3 \approx 0.200\) |

All we have to do is plug these values into the \(K_a\) expression and solve for \([\ce{H3O^{+}}]\).

\[ K_a = \ce{\dfrac{[H3O^{+}][OAc^{-}]}{[HOAc]}} \approx \dfrac{x_3(0.200)}{0.100} = 1.76 \times 10^{-5} \nonumber \]

Notice that this is a far simplier simple equation to solve for \(x_3\) than Equation \ref{nonapp} as it does not involve handling a quadratic equation, nor even even a square root!.

\[x_3 = \ce{[H3O^{+}]} = 8.80 \times 10^{–6} M \nonumber \]

and via Equation \ref{phdef}:

\[pH = 5.056 \nonumber \]

As expected, the pH has gone up from the weak acid alone in solution (Equation \ref{ph1}). Also, note that the value of \(x_3 = \ce{[H3O^{+}]}\) that we obtain is, indeed, beyond the significant digits of the analytical concentration values of \(\ce{HOAc}\) and \(\ce{OAc^{-}}\). In other words, we were justified in ignoring \(x_3\) in our algebraic expressions for the concentrations of those species (Table \(\PageIndex{3}\)).

#### The Henderson-Hasselbalch Equations

In general, for a solution of an acid with added amounts of conjugate base, except at extreme dilution, we can assume the following relationships:

\[\ce{[HA]} \approx C_{\ce{HA}} \nonumber \]

\[ \ce{[A^{-}]} \approx C_{\ce{A^{-}}} \nonumber \]

\[ K_a = \ce{\dfrac{[H3O^{+}][OAc^{-}]}{[HOAc]}} \approx \dfrac{\ce{[H3O^{+}]} C_{\ce{A^{-}}}}{C_{HA}} \nonumber \]

\[ \ce{[H3O^{+}]} \approx K_a \left( \dfrac{C_{\ce{HA}}}{C_{\ce{A^{-}}}} \right) \label{HH1} \]

By a similar reasoning, for a solution of a base \(\ce{B}\) with added amounts of its conjugate acid, \(\ce{BH^{+}}\), except at extreme dilution, we can assume

\[\ce{[B]} \approx C_{\ce{B}} \nonumber \]

\[ \ce{[BH^{+}]} \approx C_{\ce{BH^{+}}} \nonumber \]

\[ K_b = \ce{\dfrac{[OH^{-}][BH^{+}]}{[B]}} \approx \dfrac{\ce{[OH^{-}]} C_{\ce{BH^{+}}}}{C_{B}} \nonumber \]

\[ \ce{[OH^{-}]} \approx K_b \left( \dfrac{C_{\ce{B}}}{C_{\ce{BH^{+}}}} \right) \label{HH2} \]

If we take base-10 logarithms of both Equation \ref{HH1} and \ref{HH2}, we can obtain equations for the pH and pOH for these kinds of solutions in terms of \(K_a\) and \(K_b\). These are called the Henderson-Hasselbalch equations (or Buffer equations).

For \(\ce{HA/A^{-}}\) solutions:

\[ \log_{10} \ce{[H3O^{+}]} \approx \log_{10} K_a + \log_{10} \left( \dfrac{C_{\ce{HA}}}{C_{\ce{A^{-}}}} \right) \label{HH3a} \]

\[ pH \approx pK_a - \log_{10} \left( \dfrac{C_{\ce{HA}}}{C_{\ce{A^{-}}}} \right) \label{HH3b} \]

\[ \boxed{pH \approx pK_a + \log_{10} \left( \dfrac{C_{\ce{A^{-}}}}{C_{\ce{HA}}} \right)} \label{HH3c} \]

For \(\ce{B/BH^{+}}\) solutions:

\[ \log_{10} \ce{[OH^{-}]} \approx \log_{10} K_b + \log_{10} \left( \dfrac{C_{\ce{B}}}{C_{\ce{BH^{+}}}} \right) \label{HH4a} \]

\[ pOH \approx pK_a - \log_{10} \left( \dfrac{C_{\ce{B}}}{C_{\ce{BH^{+}}}} \right) \label{HH4b} \]

\[ \boxed{ pOH \approx pK_b + \log_{10} \left( \dfrac{C_{\ce{BH^{+}}}}{C_{\ce{B}}} \right)} \label{HH4c} \]

These equations are useful if caring out a series of calculations for the same conjugate acid-base pair, but for a single calculation it is just as quick to plug the numbers into the appropriate \(K_a\) or \(K_b\) expression and take the base-10 logarithm of the result, as shown in the example above.

### Q1

What is the pH of a solution prepared by adding 0.20 mole of formic acid, \(\ce{HCO2H}\), and 0.25 mole 2 of sodium formate, \(\ce{NaHCO2}\), in enough water to make a liter of solution? The \(K_a\) of formic acid is \(1.8 \times 10^{–4}\).

### Q2

What is the pH of a solution prepared by adding 0.60 mole of formic acid and 0.75 mole of sodium formate in enough water to make a liter of solution? How does your answer compare to the answer you obtained in Q1? Explain.

### Q3

The solution prepared in Q2 was diluted with enough water to make ten liters of solution. What is the pH of the dilute solution? How does it compare to the previous two solutions? Explain.

## Buffers

The solutions we have just examined, in which significant amounts of both weak acid and its conjugate base or weak base and its conjugate acid have been mixed together, are buffer solutions. Buffer solutions have two key properties that make them useful in a wide variety of applications, including biological systems:

- The pH of a buffer does not change with moderate dilution.
- The pH of a buffer changes only slightly with addition of small amounts of acid or base.

We can understand the first property from having worked Q1 through Q3. As those problems show, it is not really the actual concentrations of the acid-base conjugate pair that fix the hydronium ion concentration, but rather the ratio between those concentrations. Diluting a buffer solution does not alter the ratio of the numbers of moles of each component, and so the concentration ratio remains constant. (This breaks down, however, at extreme dilution.)

If we want to make a buffer with a certain \(pH\) from a given acid-base conjugate pair, we simply add the a b ingredients in the proper ratio, as required by the acid’s \(K_a\) or bases’s \(K_b\), to achieve the hydronium ion that corresponds to the desired pH. The ability of a buffer to resist changing pH with small amounts of acid or base depends on how little this concentration ratio is altered by the additions. Adding small amounts of a strong acid (e.g., \(\ce{HCl}\)) will neutralize an equivalent amount of the conjugate base and form an equivalent amount of the weak acid. Conversely, adding small amounts of a strong base (e.g., \(\ce{NaOH}\)) will neutralize an equivalent amount of the weak acid and form an equivalent amount of the conjugate base. In both scenarios the ratio between the conjugate pair will be altered, but if the change is small the pH will be virtually unaffected. The **buffer capacity** is the amount of added acid or base the buffer can tolerate before the pH changes significantly. This depends on minimizing the change in the conjugate pair ratio. For this reason, it is a good idea to make the concentrations of the two components fairly strong, so that numerically small changes in either component from added acid or base will cause relatively insignificant changes in the ratio.

To illustrate, suppose we wish to make up a buffer that has a pH of 5.00, using acetic acid and sodium acetate. A pH of 5.00 means

\[\ce{[H3O^{+}]} = 1.0 \times 10^{-5}. \nonumber \]

We can plug this value into the \(K_a\) expression and solve for the ratio \(\ce{[OAc ]/[HOAc]}\).

\[ K_a = \ce{\dfrac{[H3O^{+}][OAc^{-}]}{[HOAc]}} = \dfrac{(1.0 \times 10^{-5}) \ce{[OAc^{-}]}}{\ce{[HOAc]}} = 1.76 \times 10^{-5} \nonumber \]

\[ \ce{\dfrac{[OAc^{-}]}{[HOAc]}} = \dfrac{1.76 \times 10^{-5}}{1.0 \times 10^{-5}} = 1.76 \nonumber \]

Any moderate concentrations that have the ratio (\ce{[OAc ]/[HOAc]} = 1.76\) will achieve a pH of 5.00. Furthermore, notice that this is really a ratio of moles of sodium acetate to acetic acid, because the volume is the same for both. Any specific concentrations in this ratio will achieve the desired pH, but in order to achieve a good buffer capacity we should use relative high concentrations. Suppose we chose to make a buffer solution that has \(\ce{[OAc ]} = 1.76\, M\) and \(\ce{ [HOAc]} = 1.00\, M\). At equilibrium we would have the following concentrations.

Table \(\PageIndex{4}\): Equilibration of \(\ce{HOAc}\) and \(\ce{NaOAc}\) Buffer

ICE Table |
\(\ce{HOAc}\) | \(\ce{H2O}\) | \(\ce{H3O^{+}}\) | \(\ce{OAc^{–}}\) |
---|---|---|---|---|

Equilibrium |
\(1.00\,M\) | - | \(x_2\) | \(1.76\,M\) |

Now suppose we add a small amount of strong acid (e.g., \(\ce{HCl}\)) to supply an additional 0.01 mol/L of \(\ce{H3O^{+}}\). This additional hydronium ion, which is much greater than that provided from the acetic acid's hydrolysis, will shift the \(\ce{HOAc/OAc^{-}}\) equilibrium to the left to consume the extra \(\ce{H3O^{+}}\), thereby forming an equivalent amount of \(\ce{HOAc}\). This is essentially the same as neutralizing an equivalent amount of the base \(\ce{OAc^{-}}\). The changes and new equilibrium are summarized below:

Table \(\PageIndex{5}\): Equilibration of \(\ce{HOAc}\) and \(\ce{NaOAc}\) Buffer

\(\ce{HOAc}\) | \(\ce{H2O}\) | \(\ce{H3O^{+}}\) | \(\ce{OAc^{–}}\) | |
---|---|---|---|---|

Original Equilibrium | \(1.00\,M\) | - | \(10^{-5}\) | \(1.76\,M\) |

Add | 0 | - | \(0.01\,M\) | 0 |

Change | \(+ 0.01\,M\) | - | \(- 0.01\,M\) | \(- 0.01\,M\) |

New Equilibrium | \(1.01\,M\) | - | ? | \(1.75\,M\) |

To find the new \(\ce{H3O^{+}}\) substitute into the \(K_a\) expression:

\[ K_a = \ce{\dfrac{[H3O^{+}][OAc^{-}]}{[HOAc]}} = 1.76 \times 10^{-5} = \dfrac{\ce{[H3O^{+}]} (1.75\,M) }{1.01\,M } \nonumber \]

\[ [\ce{H3O^{+}}] = \dfrac{ (1.76 \times 10^{-5}) (1.01\,M) }{1.75\,M } = 1.02 \times 10^{-5} \,M \nonumber \]

and via Equation \ref{phdef}

\[pH = 4.99 \nonumber \]

As this shows, the small addition of \(\ce{H3O^{+}}\) has negligible effect on the pH because the shift in the equilibrium makes only a minor change in the ratio \(\ce{[OAc^{-}]/[HOAc]}\). This would not be the case if the concentrations of \(\ce{HOAc}\) and \(\ce{OAc^{-}}\) were small numbers. Therefore, the buffer's ability to resist a pH change from small addition of acid depends upon having high concentrations of the acid-base pair. In other words, high concentrations of acid and conjugate base result in a high buffer capacity.

Suppose, instead, that we add a small amount of a strong base (e.g., \(\ce{NaOH}\)) to our pH 5.00 buffer to supply an additional 0.01 mol/L of \(\ce{OH^{-}}\). The addition of hydroxide ion will neutralize an equivalent amount of \(\ce{HOAc}\), driving the acid hydrolysis equilibrium to the right. In this case, \(\ce{[OAc^{-}]}\) will increase by 0.01 mol/L and [HOAc] will decrease by 0.01 mol/L. Substituting into the \(K_a\) expression, we can obtain the new pH as follows:

\[ K_a = \ce{\dfrac{[H3O^{+}][OAc^{-}]}{[HOAc]}} = 1.76 \times 10^{-5} = \dfrac{\ce{[H3O^{+}]} (1.77\,M) }{0.99\,M } \nonumber \]

\[ [\ce{H3O^{+}}] = \dfrac{ (1.76 \times 10^{-5}) (0.99\,M) }{1.77\,M } = 9.8 \times 10^{-6} \,M \nonumber \]

and via Equation \ref{phdef}

\[pH = 5.01 \nonumber \]

In both cases, the small shift in \(pH\) occurs because the amounts of added acid or base are *insignificant *relative to the large amounts of the conjugate acid-base pair.

If the concentration of the conjugate base is small, a buffer will not have good resistance to pH change with added \(\ce{H3O^{+}}\). Likewise, if the concentration of the acid of the conjugate pair is small, a buffer will not have good resistance to pH change with added \(\ce{OH^{-}}\). If we want to make an ideal buffer, with equal resistance to pH changes from added \(\ce{H3O^{+}}\) or \(\ce{OH^{-}}\), we should have the concentrations of both the acid and conjugate base be high and equal. By substitution into either \(K_a\) or the Henderson-Hasselbalch equation, we can see that a buffer for which \(C_{\ce{HA}} = C_{\ce{A^{-}}}\) would have \(\ce{[H3O^{+}]} = K_a\) and \(pH = pK_a\).

It is not always necessary to have equal buffering capacity against both acid or base challenges. For example, the goal may be simply to make a solution with a particular pH to use as a standard for instrument calibration. If a particular pH is desired for a buffer from a given acid-base pair, the concentrations of the two components may need to be unequal. But not all ratios are possible for a given pair. The possible limits are set by the ratio achieved when the pure acid or the pure base hydrolyzes to give the amount of conjugate dictated by its \(K_a\) or \(K_b\). When trying to make a buffer with a certain pH, the following guidelines should be followed in selecting the acid-base pair and choosing the concentration ratios:

- Try a to choose a conjugate pair whose \(pK_a \approx pH\).
- Adjust the ratio \(\ce{[A^{-}]/[HA]}\) or \(\ce{[BH^{+}]/[B]}\) to achieve the desired pH.
- The practical limit on the choice of conjugate acid-base pair is \[pH \approx pK_a \pm 1 \nonumber \] Otherwise, one member of the pair is present in
**too small**an amount to make an effective buffer.

### Q4

A student wishes to prepare a buffer with a pH of 4.52. She starts by preparing one liter of 0.10 M formic acid, \(HCO2H\). How many grams of sodium formate, \(\ce{NaHCO2}\), must she add to the solution to achieve the desired pH? \(K_a = 1.8 \times 10^{-4}\) for formic acid. The formula weight of sodium formate is 68.00 u.

### Q5

A buffer solution is made by adding an equal number of moles of the base pyradine, \(\ce{C5H5N}\), and the chloride salt of it conjugate base, \(\ce{C5H5NHCl}\). What is the pH of the buffer solution? The \(K_b\) of \(\ce{C5H5N}\) is \(1.7 \times 10^{-9}\).