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# 21.3: Buffers

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define and give an example of a buffer.
• explain the effect of a strong acid or base on a buffer system.
• explain the mechanism by which a buffer solution resists changes in pH.
• given appropriate information, calculate the pH of a buffer.
• describe how to make a buffer solution.

• buffer

## Introduction

There are many situations in which it is desirable to keep the pH of a solution close to a particular value even though quantities of acids and/or bases are added to the solution. Many organic and biochemical reactions require acids or bases, but if the pH goes too high or too low, the products will be destroyed. For these reactions, it is necessary to keep the pH within a very small range even while acids or bases are added to the reaction. Chemists use mixtures called buffers to keep the reaction solutions within the necessary pH range. Buffers are mixtures of chemicals that cause a solution to resist changes in pH.

Buffers are very important to many biological reactions. Human blood is a substance whose function is very dependent on the function of buffers. Human blood must maintain a nearly constant pH between 7.3\begin{align*}7.3\end{align*} and 7.5\begin{align*}7.5\end{align*}. A change of just 0.2-0.4 pH units outside this range could cause loss of consciousness or even death. The pH of human blood can change slightly depending on foods we eat and the rate at which we inhale and exhale CO2\begin{align*}\mathrm{CO}_2\end{align*}. Fortunately, the human blood stream has buffers that are able to resist these pH changes.

## Buffers

A buffer is a solution that maintains the pH level when small amounts of acid or base are added to the system. Buffer solutions contain either a weak acid and its conjugate base or a weak base and its conjugate acid. A common procedure for producing a buffer in the lab is to make a solution containing both a weak acid and a salt of its conjugate base.

For example, you could make a buffer by preparing a solution containing acetic acid and sodium acetate. This common buffer would contain acetic acid, CH3COOH\begin{align*}\mathrm{CH}_3\mathrm{COOH}\end{align*}, and the acetate ion, CH3COO\begin{align*}\mathrm{CH}_3\mathrm{COO}^-\end{align*}. The solution will remain in the pH range of 3.7 – 5.8 even if small amounts of acids or bases are added. Another example is a solution of hydrogen phosphate, HPO24\begin{align*}\mathrm{HPO}_4^{2-}\end{align*}, plus the phosphate, PO34\begin{align*}\mathrm{PO}_4^{3-}\end{align*}. This will buffer a solution in the pH range of 11.3 - 13.3.

How is it possible that a solution will not change its pH when an acid or base is added? Let’s examine the acetic acid/acetate ion buffer. The ionization equation for acetic acid is shown below.

HC2H3O2(aq)H++C2H3O2\begin{align*}\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_{2(aq)} \rightarrow \mathrm{H}^+ + \mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-\end{align*}

If acid (H+\begin{align*}\mathrm{H}^+\end{align*} ions) is added to this solution, the equilibrium will shift toward the reactants to use up some of the added hydrogen ions. Equilibrium will be re-established in the solution with different concentrations of the three species in the reaction. Although the H+\begin{align*}\mathrm{H}^+\end{align*} concentration will be higher than its initial value, the change will be partially corrected due to the presence of C2H3O2\begin{align*}\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-\end{align*} ions. Similarly, if a base is added to this solution, the base will remove hydrogen ions and the equilibrium will shift to the right to partially counteract the stress. Again, equilibrium will be re-established with new concentrations. The existence of large quantities of both undissociated acid molecules and acetate ions in the solution is what allows the buffer to consume quite a large amount of added acid or base without the pH changing significantly.

Examine what happens to 1.00\begin{align*}1.00\end{align*} liter of pure water to which 0.100\begin{align*}0.100\end{align*} mole of gaseous HCl\begin{align*}\mathrm{HCl}\end{align*} is added. The original concentration of hydrogen ion in the pure water is 1.00/times107 M\begin{align*}1.00 /times 10^{-7} \ \mathrm{M}\end{align*} and therefore, the pH is 7. After the 0.100\begin{align*}0.100\end{align*} mole of HCl\begin{align*}\mathrm{HCl}\end{align*} is added, the concentration of hydrogen ions will be 0.100 M\begin{align*}0.100 \ \mathrm{M}\end{align*} (plus the original 1.00×107 M\begin{align*}1.00 \times 10^{-7} \ \mathrm{M}\end{align*}, which can be neglected as insignificant). This new concentration of hydrogen ions will produce a pH=1\begin{align*}\mathrm{pH} = 1\end{align*}. So, the addition of the 0.100\begin{align*}0.100\end{align*} mole of gaseous HCl\begin{align*}\mathrm{HCl}\end{align*} caused the pH of the pure water to change from 7\begin{align*}7\end{align*} to 1\begin{align*}1\end{align*}.

Let's now see what happens if this same amount of gaseous HCl\begin{align*}\mathrm{HCl}\end{align*} is added to an acetic acid-acetate ion buffer. Let's say we made this solution to contain 0.50 M\begin{align*}0.50 \ \mathrm{M}\end{align*} acetic acid and 0.50 M\begin{align*}0.50 \ \mathrm{M}\end{align*} acetate ion (0.50 M\begin{align*}0.50 \ \mathrm{M}\end{align*} sodium acetate which fully dissociated). The acetic acid dissociation equation will reach equilibrium at its Ka\begin{align*}K_a\end{align*} is 1.8×105\begin{align*}1.8 \times 10^{-5}\end{align*}.

Ka=[H+][C2H3O2][HC2H3O2]=1.8×105\begin{align*}K_a = \frac {[\mathrm{H}^+][\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]} {[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]} = 1.8 \times 10^{-5}\end{align*}

We know both the acetic acid and the acetate ion concentrations will be 0.50 M\begin{align*}0.50 \ \mathrm{M}\end{align*}, so we can plug these values into the expression and solve for [H+]\begin{align*}[\mathrm{H}^+]\end{align*}.

[H+]=(1.8×105)[HC2H3O2][C2H3O2]=(1.8×105)(0.50)(0.50)=1.8×105 M\begin{align*}[\mathrm{H}^+] = \frac {(1.8 \times 10^{-5})[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]} {[\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]} = \frac {(1.8 \times 10^{-5})(0.50)} {(0.50)} = 1.8 \times 10^{-5} \ \mathrm{M}\end{align*}

Then we can insert the hydrogen ion concentration into the pH formula and determine the original pH of the buffer solution.

\begin{align*}\mathrm{pH} = -\log (1.8 \times 10^{-5}) = 4.74\end{align*}

Next, we will add the same \begin{align*}0.100\end{align*} mole of gaseous \begin{align*}\mathrm{HCl}\end{align*} to this buffer solution and calculate the pH of the solution after the acid has been added and equilibrium has been re-established.

When we add \begin{align*}0.100\end{align*} mole \begin{align*}\mathrm{HCl}\end{align*} gas to this solution, the added hydrogen ion will combine with acetate ion to produce more undissociated acid. A small amount of the newly formed acetic acid may dissociate again, but the amount is minimal and can be neglected. The new \begin{align*}[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]\end{align*} will equal \begin{align*}0.60 \ \mathrm{M}\end{align*} (the original \begin{align*}0.50 \ \mathrm{M}\end{align*} plus the added \begin{align*}0.10 \ \mathrm{M}\end{align*}) and the new \begin{align*}[\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]\end{align*} will equal \begin{align*}0.40 \ \mathrm{M}\end{align*} (the original \begin{align*}0.50 \ \mathrm{M}\end{align*} minus the \begin{align*}0.10 \ \mathrm{M}\end{align*} that reacted with the added hydrogen ions). We can now plug these values into the \begin{align*}K_a\end{align*} expression, calculate the new \begin{align*}[\mathrm{H}^+]\end{align*}, and find the new pH.

\begin{align*}K_a = \frac {[\mathrm{H}^+][\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]} {[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]} = 1.8 \times 10^{-5}\end{align*}
\begin{align*}[\mathrm{H}^+] = \frac {(1.8 \times 10^{-5})[\mathrm{HC}_2\mathrm{H}_3\mathrm{O}_2]} {[\mathrm{C}_2\mathrm{H}_3\mathrm{O}_2^-]} = \frac {(1.8 \times 10^{-5})(0.60)} {(0.40)} = 2.7 \times 10^{-5} \ \mathrm{M}\end{align*}
\begin{align*}\mathrm{pH} = -\log (2.7 \times 10^{-5}) = 4.57\end{align*}

The same quantity of \begin{align*}\mathrm{HCl}\end{align*} gas that changed the pH of pure water from \begin{align*}7\end{align*} to \begin{align*}1\end{align*} has changed the pH of this buffer from \begin{align*}4.74\end{align*} to \begin{align*}4.57\end{align*}. The change in pH is only \begin{align*}0.17\end{align*}, which is a function of the buffer. Buffers resist change to pH. We could do this same calculation by adding a base instead of an acid and show that the pH increases by this same slight amount. We will leave this as an exercise for you to complete on your own.

Example:

Which of the following combinations would you expect to make a useful buffer solution:

1. \begin{align*}\mathrm{HClO}_4\end{align*}/\begin{align*}\mathrm{ClO}_4^-\end{align*}
2. \begin{align*}\mathrm{CH}_3\mathrm{NH}_2\end{align*}/\begin{align*}\mathrm{CH}_3\mathrm{NH}_3^+\end{align*}

Solution:

1. \begin{align*}\mathrm{HClO}_4\end{align*}/\begin{align*}\mathrm{ClO}_4^-\end{align*}:\begin{align*}\mathrm{HClO}_4\end{align*} is a strong acid and buffers are made from weak acids and their conjugate bases or weak bases and their conjugate acids. Therefore, this cannot be made into a buffer.
2. \begin{align*}\mathrm{CH}_3\mathrm{NH}_2\end{align*}/\begin{align*}\mathrm{CH}_3\mathrm{NH}_3^+\end{align*}:\begin{align*}\mathrm{CH}_3\mathrm{NH}_2\end{align*} is a weak base and \begin{align*}\mathrm{CH}_3\mathrm{NH}_3^+\end{align*} is the conjugate acid of this base. Therefore, this can be made into a buffer solution.

The first video considers the mathematics of buffer solutions and the second video shows a laboratory example of buffering (5g; 1l I&E Stand.): http://www.youtube.com/watch?v=O_QlZe4fv4g (10:18), (5g) http://www.youtube.com/watch?v=g_ZK2ABUjvA (2:39).

## The Buffer in Blood

The primary buffer found in your bloodstream is carbonic acid, \begin{align*}\mathrm{H}_2\mathrm{CO}_3\end{align*}. The carbonic acid is present due to carbon dioxide from your respiratory system dissolving in water.

\begin{align*}\mathrm{CO}_{2(g)} + \mathrm{H}_2\mathrm{O}_{(l)} \rightleftharpoons \mathrm{H}_2\mathrm{CO}_{3(aq)}\end{align*}

The amount of carbonic acid in your bloodstream is affected by the rate of your respiration. If you breathe rapidly, you reduce the amount of \begin{align*}\mathrm{CO}_2\end{align*} in your bloodstream, and the equilibrium shown above shifts toward the reactants, thus lowering the amount of \begin{align*}\mathrm{H}_2\mathrm{CO}_3\end{align*}. If you breathe slowly, the amount of \begin{align*}\mathrm{CO}_2\end{align*} in your bloodstream increases, and the equilibrium shifts toward the products, increasing the amount of \begin{align*}\mathrm{H}_2\mathrm{CO}_3\end{align*} in your system.

Once the \begin{align*}\mathrm{H}_2\mathrm{CO}_3\end{align*} is produced by dissolving carbon dioxide, the carbonic acid dissociates in your blood as shown below:

\begin{align*}\mathrm{H}_2\mathrm{CO}_{3(aq)} \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-\end{align*}

The buffering system in your blood is composed of the weak acid, \begin{align*}\mathrm{H}_2\mathrm{CO}_3\end{align*}, and its conjugate base, \begin{align*}\mathrm{HCO}_3^-\end{align*}.

Any changes in blood pH that could be caused by food intake will be buffered by this equilibrium system. If acid is added to your blood, the equilibrium will shift toward the reactants, using up the hydrogen ions. If base were added to your blood, thus reacting with hydrogen ions, the equilibrium will shift toward the products to generate more hydrogen ions. This buffer in your blood is very efficient at keeping your blood pH in the necessary range.

Some people, when they get nervous, begin breathing very fast or very slow. Breathing too fast is called hyperventilating, and breathing too slow is called hypoventilating. Your respiratory rate is normally controlled by the amount of carbon dioxide in the blood. Your body receives instructions to breathe faster or slower to adjust the amount of carbon dioxide in your blood in order to properly regulate the buffer system. When people breathe too fast or too slow because of other reasons, the \begin{align*}\mathrm{CO}_2\end{align*} content of the blood becomes incorrect and the pH of the blood rises or lowers outside the acceptable range of 7.3 - 7.5. When this happens, the person can pass out. People who hyperventilate when excited or nervous are sometimes advised to carry a lunch sack or something similar to breathe into when they are feeling light-headed. Breathing into a sack returns air with the same concentration of carbon dioxide that was exhaled. This keeps the amount of carbon dioxide in the blood up and prevents a loss of consciousness.

## Lesson Summary

• A buffer is a solution of a weak acid and its conjugate base or a weak base and its conjugate acid that resists changes in pH when an acid or base is added to it.
• Adding a strong acid to a buffer only decreases the pH by a small amount.
• Adding a strong base to a buffer only increases the pH by a small amount.

To see a short animated video showing concentration changes as strong acid or base is added to a buffer, follow the link below.

## Review Questions

1. Define a buffer solution.
2. One of the following statements of buffers is incorrect. Which one?
1. A buffer may be prepared from a weak acid and its conjugate base salt.
2. A buffer may be prepared from a weak base and its conjugate acid salt.
3. A buffer is a solution that can resist changes in pH when any amount of acid or base is added to it.
4. A buffer is a solution that can resist changes in pH when a small amount of acid or base is added to it.
3. Which of the following pairs of aqueous \begin{align*}1.0 \ \mathrm{mol/L}\end{align*} solutions could be chosen to prepare a buffer: i. \begin{align*}\mathrm{NH}_4\mathrm{HSO}_{4(aq)}\end{align*} and \begin{align*}\mathrm{H}_2\mathrm{SO}_{4(aq)}\end{align*}, ii. \begin{align*}\mathrm{HNO}_{2(aq)}\end{align*} and \begin{align*}\mathrm{NaNO}_{2(aq)}\end{align*}, iii. \begin{align*}\mathrm{NH}_4\mathrm{Cl}_{(aq)}\end{align*} and \begin{align*}\mathrm{NH}_{3(aq)}\end{align*}?
1. i and iii only
2. ii and iii only
3. i, ii, and iii
4. None of these solutions is a buffer.
4. Which of the following would form a buffer solution if combined in appropriate amounts?
1. \begin{align*}\mathrm{HCl}\end{align*} and \begin{align*}\mathrm{NaCl}\end{align*}
2. \begin{align*}\mathrm{HCN}\end{align*} and \begin{align*}\mathrm{NaCN}\end{align*}
3. \begin{align*}\mathrm{H}_2\mathrm{S}\end{align*} and \begin{align*}\mathrm{NaSH}\end{align*}
4. \begin{align*}\mathrm{HNO}_3\end{align*} and \begin{align*}\mathrm{NaNO}_3\end{align*}
5. A buffer is made up of a weak acid and a conjugate base. A small amount of acid is added to the buffer. What happens to the resulting solution?
1. The acid dissociation constant goes up.
2. The concentration of the weak acid in the buffer goes down.
3. The pH of the solution goes up.
4. The pH remains almost the same.

All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative Commons license CC-BY-NC-SA.

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