# Buffers and Titration Curves (Worksheet)

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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.

An important property of blood and other physiological components is that they resist change in pH. A buffer system occurs when a weak acid and its conjugate base are present in the same solution. For instance, blood has a pH of about 7.4, and complex chemical systems work to maintain that pH. The most important component of those systems is the carbonic acid buffer system. Not by coincidence, this happens to be the same weak acid found in soft drinks.

Buffer systems are an important application of acid–base equilibria. The study of acid–base equilibria is very useful because many other chemical systems can be understood through the same mathematical approach. The most common experimental method used to study acid–base systems is titration analysis, through which we can determine the pK_{a} of a weak acid and the pK_{b} of its conjugate base, the two essential components of a buffer.

## The Buffer Equation (Henderson-Hasselbach Approximation)

Let’s consider a weak acid equilibrium system and its corresponding equilibrium constant:

\[HA(aq) \rightleftharpoons H^+(aq) + A^–(aq)\]

with \[K_a = \dfrac{[H^+] [A^–]}{[HA]} \label{2}\]

where \(HA\) represents a weak monoprotic acid, and \(A^–\) is its conjugate base. Rearranging Equation \(\ref{2}\) to express the equilibrium constant expression for hydrogen ion concentration,

\[[H^+] = K_a \dfrac{[HA]}{[A^–]}\]

Taking the log of each side and multiplying by –1,

\[– \log_{10} [H^+] = – \log_{10} \left( K_a \dfrac{[HA]}{[A^–]} \right )\]

Algebraically rearranging,

\[– \log_{10} [H^+] = – \log_{10} K_a – \log_{10} \dfrac{[HA]}{[A^–]}\]

\[– \log_{10} [H^+] = –\log_{10} K_a + \log_{10} \dfrac{[A^–]}{[HA]}\]

Using the fact that \(pX = –\log X\) for a generic \(X\) variable, we arrive at the buffer equation:

\[pH = pK_a + \log_{10} \dfrac{[A^–]}{[HA]}\]

Note that since this equation is derived from the equilibrium constant expression, all concentrations must be equilibrium concentrations. However, we often find it useful and accurate to make the approximation that the weak acid is only slightly dissociated. Thus the equilibrium concentration of HA is approximately equal to the initial concentration, or

\[[HA]_{eq} \approx [HA]_o\]

A similar assumption is also valid for weak bases:

\[[A^–]_{eq} \approx [A^–]_o\]

## The Effectiveness of a Buffer

Consider a 100.0 mL solution containing 0.010 mol acetic acid, \(HC_2H_3O_2\), and 0.010 mol sodium acetate, \(NaC_2H_3O_2\). We have

\[[A^–][HA] = 1\]

and

\[\log 1 = 0\]

therefore, \(pH = pK_a\). Looking up \(pK_a\) for acetic acid, we find \(pH = pK_a = 4.75\). Now let’s consider what will happen if we add 0.005 mol of HCl to this solution. The strong acid will react with the acetate ion.

ICE Table | \(H^+(aq)\) | \(C_2H_3O^{2–}(aq)\) | \(\rightleftharpoons\) | \(HC_2H_3O_2(aq)\) |
---|---|---|---|---|

Initial |
0.005 | 0.010 | 0.010 | |

Change |
– 0.005 | – 0.005 | + 0.005 | |

Equilibrium |
0 | 0.005 | 0.015 |

The buffer equation can now be applied to determine the new solution pH:

\[pH = pKa + \log_{10} \dfrac{[A^–]}{[HA]} = 4.75 + \log_{10} \left( \dfrac{0.005 \,mol/0.1000 \,L}{0.015 \,mol/0.1000\, L} \right) = 4.27\]

The pH of the solution changes from 4.75 to 4.27 upon addition of the acid.

## The Ineffectiveness of a Non-buffer

Let’s compare this to what will happen if we add the same amount of \(HCl\) to a **nonbuffered **solution that begins at pH = 4.75. A \(1.8 \times 10^{–5} M\) \(HCl\) solution has a pH of 4.75 (full dissociated). The number of moles of \(H^+(aq)\) in this solution is

\[(0.1000 L )(1.8 \times 10^{–5} mol L ) = 1.8 \times 10^{–6} mol\]

The amount of \(HCl\) added was 0.005 mol, so after the acid is added, the number of moles is

\[0.005\,mol + 0.0000018\, mol = 0.005 mol\]

and the new hydrogen ion concentration is

\[\dfrac{0.005\, mol}{0.1000\, L} = 0.05\, M\]

and the solution pH is

\[pH = – \log_{10} [H^+] = – \log_{10} (0.05) = 1.3\]

In this unbuffered solution, the pH changes from 4.75 to 1.3, which is a **much larger change** than in the buffered solution.

A consideration that must be made when preparing a buffer is to have sufficient quantities of both the weak acid and its conjugate base to completely react with any base or acid that may be added to the system. The **buffer capacity** of a system is defined in terms of the concentrations of the acid–base conjugate pair. Greater concentrations will withstand greater additions of base or acid while still resisting a significant pH change. If we were to add so much acid so that it reacted with all of the base in a buffer system, the buffering capacity of the system would be exceeded, and further additions of acid would result in large changes in pH.

A guideline for preparing a buffer system is to choose an acid with a pK_{a} within one pH unit of the desired buffer. This ensures that the ratio of base to acid will range between 1 to 10 and 10 to 1, and thus sufficient quantities of both acid and base will be present in the buffering system.

## pH Titration

A pH titration is performed by adding small amounts of a titrant to a solution and simultaneously monitoring the solution pH. A typical titration would be to add small amounts of sodium hydroxide solution to a weak acid solution. In this case, the pH during the titration is related to the pKa of the weak acid.

During the titration, acid is being converted to its conjugate base, and a buffer solution is formed. Eventually, the quantity of base added is such that all of the acid has been converted to its conjugate base, and the equivalence point of the titration has been reached. The solution is no longer a buffer at the equivalence point.

## Outside Links

- Barnum, Dennis W. "Predicting Acid-Base Titration Curves without Calculations."
*J. Chem. Educ.***1999***76*938. - de Levie, Robert. "A General Simulator for Acid-Base Titrations."
*J. Chem. Educ.***1999***76*987.