## Preparing a Buffer Solution with a Specific pH

A buffer is a solution of weak acid and conjugate base or weak base and conjugate acid used to resist pH change with added solute.

### Learning Objectives

Describe the properties of a buffer solution.

### Key Takeaways

#### Key Points

• Buffer solutions are resistant to pH change because of the presence of an equilibrium between the acid (HA) and its conjugate base (A-).
• When some strong acid is added to a buffer, the equilibrium is shifted to the left, and the hydrogen ion concentration increases by less than expected for the amount of strong acid added.
• Buffer solutions are necessary in biology for keeping the correct pH for proteins to work.
• Buffers can be prepared in multiple ways by creating a solution of an acid and its conjugate base.

#### Key Terms

• aqueous: Consisting mostly of water.
• equilibrium: The state of a reaction in which the rates of the forward (reactant to product) and reverse (product to reactant) reactions are the same.
• pKa: A quantitative measure of the strength of an acid in solution; a weak acid has a pKa value in the approximate range −2 to 12 in water and a strong acid has a pKa value of less than about −2.

### Buffers

A buffer is an aqueous solution containing a weak acid and its conjugate base or a weak base and its conjugate acid. A buffer’s pH changes very little when a small amount of strong acid or base is added to it. It is used to prevent any change in the pH of a solution, regardless of solute. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. For example, blood in the human body is a buffer solution.

Buffer solutions are resistant to pH change because of the presence of an equilibrium between the acid (HA) and its conjugate base (A). The balanced equation for this reaction is:

${ \text{HA}\rightleftharpoons \text{H} }^{ + }+{ \text{A} }^{ - }$

When some strong acid (more H+) is added to an equilibrium mixture of the weak acid and its conjugate base, the equilibrium is shifted to the left, in accordance with Le Chatelier’s principle. This causes the hydrogen ion (H+) concentration to increase by less than the amount expected for the quantity of strong acid added. Similarly, if a strong base is added to the mixture, the hydrogen ion concentration decreases by less than the amount expected for the quantity of base added. This is because the reaction shifts to the right to accommodate for the loss of H+ in the reaction with the base.

Buffer solutions are necessary in a wide range of applications. In biology, they are necessary for keeping the correct pH for proteins to work; if the pH moves outside of a narrow range, the proteins stop working and can fall apart. A buffer of carbonic acid (H2CO3) and bicarbonate (HCO3) is needed in blood plasma to maintain a pH between 7.35 and 7.45. Industrially, buffer solutions are used in fermentation processes and in setting the correct conditions for dyes used in coloring fabrics.

### Preparing a Buffer Solution

There are a couple of ways to prepare a buffer solution of a specific pH. In the first method, prepare a solution with an acid and its conjugate base by dissolving the acid form of the buffer in about 60% of the volume of water required to obtain the final solution volume. Then, measure the pH of the solution using a pH probe. The pH can be adjusted up to the desired value using a strong base like NaOH. If the buffer is made with a base and its conjugate acid, the pH can be adjusted using a strong acid like HCl. Once the pH is correct, dilute the solution to the final desired volume.

pH probe: The probe can be inserted into a solution to measure the pH (reading 8.61 in this example). Probes need to be regularly calibrated with solutions of known pH to be accurate.

Alternatively, you can prepare solutions of both the acid form and base form of the solution. Both solutions must contain the same buffer concentration as the concentration of the buffer in the final solution. To get the final buffer, add one solution to the other while monitoring the pH.

In a third method, you can determine the exact amount of acid and conjugate base needed to make a buffer of a certain pH, using the Henderson-Hasselbach equation:

$\text{pH}=\text{p}{ \text{K} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )$

where pH is the concentration of [H+], pKa is the acid dissociation constant, and [\text{A}-] and [\text{HA}] are concentrations of the conjugate base and starting acid.

## Calculating the pH of a Buffer Solution

The pH of a buffer solution can be calculated from the equilibrium constant and the initial concentration of the acid.

### Learning Objectives

Calculate the pH of a buffer made only from a weak acid.

### Key Takeaways

#### Key Points

• The strength of a weak acid ( buffer ) is usually represented as an equilibrium constant.
• The acid-dissociation equilibrium constant, which measures the propensity of an acid to dissociate, is described using the equation: ${ \text{K}}_{\text{a} }=\frac { { [\text{H} }^{ + }][{ \text{A} }^{ - }] }{ [\text{HA}] }$.
• Using Ka and the equilibrium equation, you can solve for the concentration of [H+].
• The concentration of [H+] can then be used to calculate the pH of a solution, as part of the equation: pH = -log([H+]).

#### Key Terms

• equilibrium: The state of a reaction in which the rates of the forward (reactant to product) and reverse (product to reactant) reactions are the same.

### What Does pH Mean in a Buffer?

In chemistry, pH is a measure of the hydrogen ion (H+) concentration in a solution. The pH of a buffer can be calculated from the concentrations of the various components of the reaction. The balanced equation for a buffer is:

$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

The strength of a weak acid is usually represented as an equilibrium constant. The acid-dissociation equilibrium constant (Ka), which measures the propensity of an acid to dissociate, for the reaction is:

$\text{K}_{\text{a}} = \frac{\left [\text{H}^{+} \right ]\left [\text{A}^{-} \right ]}{\left [\text{HA} \right ]}$

The greater [H+] x [A] is than [HA], the greater the value of Ka, the more the formation of H+ is favored, and the lower the pH of the solution.

### ICE Tables: A Useful Tool For Solving Equilibrium Problems

ICE (Initial, Change, Equilibrium) tables are very helpful tools for understanding equilibrium and for calculating the pH of a buffer solution. They consist of using the initial concentrations of reactants and products, the change they undergo during the reaction, and their equilibrium concentrations. Consider, for example, the following problem:

Calculate the pH of a buffer solution that initially consists of 0.0500 M NH3 and 0.0350 M NH4+. (Note: Ka for NH4+ is 5.6 x 10-10). The equation for the reaction is as follows:

$\text{NH}_4^+ \rightleftharpoons \text{H}^+ + \text{NH}_3$

We know that initially there is 0.0350 M NH4+ and 0.0500 M NH3. Before the reaction occurs, no H+ is present so it starts at 0.

ICE table – initial: ICE table for the buffer solution of NH4+ and NH3 with the starting concentrations.

During the reaction, the NH4+ will dissociate into H+ and NH3. Because the reaction has a 1:1 stoichiometry, the amount that NH4+ loses is equal to the amounts that H+ and NH3 will gain. This change is represented by the letter x in the following table.

ICE table – change: Describes the change in concentration that occurs during the reaction.

Therefore the equilibrium concentrations will look like this:

ICE table – equilibrium: Describes the final concentration of the reactants and products at equilibrium.

Apply the equilibrium values to the expression for Ka.

${ 5.6 \times 10^{-10}} = \frac { { [\text{H} }^{ + }][{ \text{NH}_3 }] }{ [\text{NH}_4^+] } = \frac { x (0.0500+\text{x})}{ 0.0350-\text{x} }$

Assuming x is negligible compared to 0.0500 and 0.0350 the equation is reduced to:

${ 5.6 \times 10^{-10}} = \frac { { [\text{H} }^{ + }][{ \text{NH}_3 }] }{ [\text{NH}_4^+] } = \frac { \text{x} (0.0500)}{ 0.0350 }$

Solving for x (H+):

x = [H+] = 3.92 x 10-10

pH = -log(3.92 x 10-10)

pH = 9.41

## The Henderson-Hasselbalch Equation

The Henderson–Hasselbalch equation connects the measurable value of the pH of a solution with the theoretical value pKa.

### Learning Objectives

Calculate the pH of a buffer system using the Henderson-Hasselbalch equation.

### Key Takeaways

#### Key Points

• The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid – base reaction.
• The formula for the Henderson–Hasselbalch equation is: $\text{pH}=\text{p}{ \text{K} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )$, where pH is the concentration of [H+], pKa is the acid dissociation constant, and [A] and [HA] are concentrations of the conjugate base and starting acid.
• The equation can be used to determine the amount of acid and conjugate base needed to make a buffer solution of a certain pH.

#### Key Terms

• pKa: A quantitative measure of the strength of an acid in solution; a weak acid has a pKa value in the approximate range -2 to 12 in water and a strong acid has a pKa value of less than about -2.

The Henderson–Hasselbalch equation mathematically connects the measurable pH of a solution with the pKa (which is equal to -log Ka) of the acid. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in an acid-base reaction. The equation can be derived from the formula of pKa for a weak acid or buffer. The balanced equation for an acid dissociation is:

$\text{HA}\rightleftharpoons { \text{H} }^{ + }+{ \text{A} }^{ - }$

The acid dissociation constant is:

${ \text{K} }_{ \text{a} }=\frac { [{ \text{H} }^{ + }][\text{A}^{ - }] }{ [\text{HA}] }$

After taking the log of the entire equation and rearranging it, the result is:

$\text{log}({ \text{K} }_{ \text{a} })=\text{log}[{ \text{H} }^{ + }]+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )$

This equation can be rewritten as:

$-\text{p}{ \text{K} }_{ \text{a} }=-\text{pH}+\text{log}(\frac { [\text{A}^{ - }] }{ [\text{HA}] } )$

Distributing the negative sign gives the final version of the Henderson-Hasselbalch equation:

$\text{pH}=\text{p}{ \text{K} }_{ \text{a} }+\text{log}(\frac { { [\text{A} }^{ - }] }{ [\text{HA}] } )$

In an alternate application, the equation can be used to determine the amount of acid and conjugate base needed to make a buffer of a certain pH. With a given pH and known pKa, the solution of the Henderson-Hasselbalch equation gives the logarithm of a ratio which can be solved by performing the antilogarithm of pH/pK­a:

${ 10 }^{ \text{pH}-\text{p}{ \text{K} }_{ \text{a} } }=\frac { [\text{base}] }{ [\text{acid}] }$

An example of how to use the Henderson-Hasselbalch equation to solve for the pH of a buffer solution is as follows:

What is the pH of a buffer solution consisting of 0.0350 M NH3 and 0.0500 M NH4+ (Ka for NH4+ is 5.6 x 10-10)? The equation for the reaction is:

${\text{NH}_4^+}\rightleftharpoons { \text{H} }^{ + }+{ \text{NH}_3}$

Assuming that the change in concentrations is negligible in order for the system to reach equilibrium, the Henderson-Hasselbalch equation will be:

$\text{pH}=\text{p}{ \text{K} }_{ \text{a} }+\text{log}(\frac { { [\text{NH}_3}] }{ [\text{NH}_4^+] } )$

$\text{pH}=9.25+\text{log}(\frac{0.0350}{0.0500} )$

pH = 9.095

## Calculating Changes in a Buffer Solution

The changed pH of a buffer solution in response to the addition of an acid or a base can be calculated.

### Learning Objectives

Calculate the final pH of a solution generated by the addition of a strong acid or base to a buffer.

### Key Takeaways

#### Key Points

• If the concentrations of the weak acid and its conjugate base in a buffer solution are reasonably high, then the solution is resistant to changes in hydrogen ion concentration, or pH.
• The change in pH of a buffer solution with an added acid or base can be calculated by combining the balanced equation for the reaction and the equilibrium acid dissociation constant (Ka).
• Comparing the final pH of a solution with and without the buffer components shows the effectiveness of the buffer in resisting a change in pH.

#### Key Terms

• pH: The negative of the logarithm to base 10 of the concentration of hydrogen ions, measured in moles per liter; a measure of acidity or alkalinity of a substance, which takes numerical values from 0 (maximum acidity) through 7 (neutral) to 14 (maximum alkalinity).
• acid dissociation constant: Quantitative measure of the strength of an acid in solution; typically written as a ratio of the equilibrium concentrations of products to reactants.

If the concentrations of a solution of a weak acid and its conjugate base are reasonably high, then the solution is resistant to changes in hydrogen ion concentration. These solutions are known as buffers. It is possible to calculate how the pH of the solution will change in response to the addition of an acid or a base to a buffer solution.

### Calculating Changes in a Buffer Solution, Example 1:

A solution is 0.050 M in acetic acid (HC2H3O2) and 0.050 M NaC2H3O2. Calculate the change in pH when 0.001 mole of hydrochloric acid (HCl) is added to a liter of solution, assuming that the volume increase upon adding the HCl is negligible. Compare this to the pH if the same amount of HCl is added to a liter of pure water.

#### Step 1:

${\text{HC}_2\text{H}_3\text{O}_2}(\text{aq})\leftrightharpoons {\text{H}^+}(\text{aq})+{\text{C}_2\text{H}_3\text{O}_2^-}(\text{aq})$

Recall that sodium acetate,
NaC2H3O2, dissociates into its component ions, Na+ and C2H3O2 (the acetate ion) upon dissolution in water. Therefore, the solution will contain both acetic acid and acetate ions.

Before adding HCl, the acetic acid equilibrium constant is:

${ \text{K} }_{ \text{a} }=\frac { { [\text{H} }^{ + }]{ [\text{C}_2{\text{H}}_3{\text{O}}_2 }^{ - }] }{ [\text{H}{\text{C}}_2{\text{H}}_3{\text{O}}_2] } =\frac { \text{x}(0.050) }{ (0.050) }$

(assuming that x is small compared to 0.050 M in the equilibrium concentrations)

Therefore:

$\text{x}=[\text{H}^+]={ \text{K} }_{ \text{a} }=1.76\times 1{ 0 }^{ -5 }\text{M}$

$\text{pH}={ \text{pK} }_{ \text{a} }=4.75$

In this example, ignoring the x in the [C2H3O2] and [HC2H3O2] terms was justified because the value is small compared to 0.050.

#### Step 2:

The added protons from HCl combine with the acetate ions to form more acetic acid:

$\text{C}_2\text{H}_3\text{O}_2^{ - }+{ \text{H} }^{ + }(\text{from HCl})\rightarrow \text{HC}_2\text{H}_3\text{O}_2$

Since all of the H+ will be consumed, the new concentrations will be $[\text{HC}_2\text{H}_3\text{O}_2]=0.051 \text{M}$ and $[\text{C}_2\text{H}_3\text{O}_2^-]=0.049 \text{M}$ before the new equilibrium is to be established. Then, we consider the equilibrium conentrations for the dissociation of acetic acid, as in Step 1:

${\text{HC}_2\text{H}_3\text{O}_2}(\text{aq})\leftrightharpoons {\text{H}^+}(\text{aq})+{\text{C}_2\text{H}_3\text{O}_2^-}(\text{aq})$

we have,

${ \text{K} }_{ \text{a} }=\frac { \text{x}(0.049) }{ (0.051) }$

$x=[\text{H}^+]=(1.76\times 1{ 0 }^{ -5 })\frac { 0.051 }{ 0.049 } =1.83\times 1{ 0 }^{ -5 }\text{M}$

$\text{pH}=-\text{log}([{ \text{H} }^{ + }])=4.74$

In the presence of the acetic acid-acetate buffer system, the pH only drops from 4.75 to 4.74 upon addition of 0.001 mol of strong acid HCl, a difference of only 0.01 pH unit.

#### Step 3:

Adding 0.001 M HCl to pure water, the pH is:

$\text{pH}=-\text{log}([{ \text{H} }^{ + }])=3.00$

In the absence of HC2H3O2 and C2H3O2, the same concentration of HCl would produce a pH of 3.00.

### Calculating Changes in a Buffer Solution, Example 2:

A formic acid buffer is prepared with 0.010 M each of formic acid (HCOOH) and sodium formate (NaCOOH). The Ka for formic acid is 1.8 x 10-4. What is the pH of the solution? What is the pH if 0.0020 M of solid sodium hydroxide (NaOH) is added to a liter of buffer? What would be the pH of the sodium hydroxide solution without the buffer? What would the pH have been after adding sodium hydroxide if the buffer concentrations had been 0.10 M instead of 0.010 M?

#### Step 1:

Solving for the buffer pH:

$\text{HCOOH} \leftrightharpoons {\text{H}^+} + {\text{HCOO}^-}$

Assuming x is negligible, the Ka expression looks like:

${ \text{K} }_{ \text{a} }=\frac { \text{x}(0.010) }{ (0.010) }$

1.8 x 10-4 = x = [H+]

pH = -log [H+] = 3.74

Buffer: pH = 3.74

#### Step 2:

Solving for the buffer pH after 0.0020 M NaOH has been added:

$\text{OH}^- + \text{HCOOH} \rightarrow {\text{H}_2O} + {\text{HCOO}^-}$

The concentration of HCOOH would change from 0.010 M to 0.0080 M and the concentration of HCOO would change from 0.010 M to 0.0120 M.

${ \text{K} }_{ \text{a} }=\frac { \text{x}(0.0120) }{ (0.0080) }$

After adding NaOH, solving for $\text{x}=[\text{H}^+]$ and then calculating the pH = 3.92. The pH went up from 3.74 to 3.92 upon addition of 0.002 M of NaOH.

#### Step 3:

Solving for the pH of a 0.0020 M solution of NaOH:

pOH = -log (0.0020)

pOH = 2.70

pH = 14 – pOH

pH = 11.30

Without buffer: pH = 11.30

#### Step 4:

Solving for the pH of the buffer solution if 0.1000 M solutions of the weak acid and its conjugate base had been used and the same amount of NaOH had been added:

The concentration of HCOOH would change from 0.1000 M to 0.0980 M and the concentration of HCOO would change from 0.1000 M to 0.1020 M.

${ \text{K} }_{ \text{a} }=\frac { \text{x}(0.1020) }{ (0.0980) }$

pH if 0.1000 M concentrations had been used = 3.77

This shows the dramatic effect of the formic acid-formate buffer in keeping the solution acidic in spite of the added base. It also shows the importance of using high buffer component concentrations so that the buffering capacity of the solution is not exceeded.

## Buffers Containing a Base and Conjugate Acid

An alkaline buffer can be made from a mixture of the base and its conjugate acid, but the formulas for determining pH take a different form.

### Learning Objectives

Calculate the pH of an alkaline buffer system consisting of a weak base and its conjugate acid.

### Key Takeaways

#### Key Points

• The pH of bases is usually calculated using the hydroxide ion (OH) concentration to find the pOH first.
• The formula for pOH is pOH=-log[OH-]. A base dissociation constant (Kb) indicates the strength of the base.
• The pH of a basic solution can be calculated by using the equation: pH = 14.00 – pOH.

#### Key Terms

• alkaline: Having a pH greater than 7.
• buffers: A weak acid or base used to maintain the acidity (pH) of a solution near a chosen value and which prevent a rapid change in pH when acids or bases are added to the solution.

A base is a substance that decreases the hydrogen ion (H+) concentration of a solution. In the more generalized Brønsted-Lowry definition, the hydroxide ion (OH) is the base because it is the substance that combines with the proton. Ammonia and some organic nitrogen compounds can combine with protons in solution and act as Brønsted-Lowry bases. These compounds are generally weaker bases than the hydroxide ion because they have less attraction for protons. For example, when ammonia competes with OH for protons in an aqueous solution, it is only partially successful. It can combine with only a portion of the H+ ions, so it will have a measurable equilibrium constant. Reactions with weak bases result in a relatively low pH compared to strong bases. Bases range from a pH of greater than 7 (7 is neutral like pure water) to 14 (though some bases are greater than 14).

An alkaline buffer can be made from a mixture of a base and its conjugate acid, similar to the way in which weak acids and their conjugate bases can be used to make a buffer.

Ammonia to ammonium ion: Two-dimensional image depicting the association of proton (H+) with the weak base ammonia (NH3) to form its conjugate acid, ammonium ion (NH4+).

### Calculating the pH of a Base

The pH of bases is usually calculated using the OH concentration to find the pOH first. This is done because the H+ concentration is not a part of the reaction, while the OH concentration is. The formula for pOH is:

$\text{pOH}=-\text{log}([\text{O}{ \text{H} }^{ - }])$

By multiplying a conjugate acid (such as NH4+) and a conjugate base (such as NH3) the following is given:

${ \text{K} }_{ \text{a} }\times { \text{K} }_{ \text{b} }=\frac { [{ \text{H} }_{ 3 }{ \text{O} }^{ + }][\text{N}{ \text{H} }_{ 3 }] }{ [\text{N}{ \text{H} }_{ 4 }^{ + }] } \times \frac { [\text{N}{ \text{H} }_{ 4 }^{ + }][{ \text{OH} }^{ - }] }{ [\text{N}{ \text{H} }_{ 3 }] }$

${ \text{K} }_{ \text{a} }\times { \text{K} }_{ \text{b} }=[{ \text{H} }_{ 3 }{ \text{O} }^{ + }][{ \text{OH} }^{ - }]={ \text{K} }_{ \text{w} }$

${ \text{log}(\text{K} }_{ \text{a} })+{ \text{log}(\text{K} }_{ \text{b} })=\text{log}({ \text{K} }_{ \text{w} })$

${ \text{pK} }_{ \text{a} }+{ \text{pK} }_{ \text{b} }=\text{p}{ \text{K} }_{ \text{w} }=14.00$

The pH can be calculated using the formula:

${ \text{pH}={14}-\text{pOH} }$

Weak bases exist in chemical equilibrium much in the same way as weak acids do. A base dissociation constant (Kb) indicates the strength of the base. For example, when ammonia is put in water, the following equilibrium is set up:

$\text{N}{ \text{H} }_{ 3 }+{ \text{H}_2\text{O}}\rightleftharpoons \text{N}{ \text{H} }_{ 4 }^{ + } +{\text{OH}^-}$

${ \text{K} }_{ \text{b} }=\frac { [\text{N}{ \text{H} }_{ 4 }^{ + }][{ \text{OH} }^{ - }] }{ [\text{N}{ \text{H} }_{ 3 }] }$

Bases that have a large Kb will ionize more completely, meaning they are stronger bases. NaOH (sodium hydroxide) is a stronger base than (CH3CH2)2NH (diethylamine) which is a stronger base than NH3 (ammonia). As the bases get weaker, the Kb values get smaller.

Example:

Calculate the pH of a buffer solution consisting of 0.051 M NH3 and 0.037 M NH4+. The Kb for NH3 = 1.8 x 10-5.

$\text{N}{ \text{H} }_{ 3 }+{ \text{H}_2\text{O}}\rightleftharpoons \text{N}{ \text{H} }_{ 4 }^{ + } +{\text{OH}^-}$

${ \text{K} }_{ \text{b} }=\frac { [\text{N}{ \text{H} }_{ 4 }^{ + }][{ \text{OH} }^{ - }] }{ [\text{N}{ \text{H} }_{ 3 }] }$

Assuming the change (x) is negligible to 0.051 M and 0.037 M solutions:

${ \text{K} }_{ \text{b} }=\frac { [0.037][\text{x}] }{ [0.051] }$

1.8 x 10-5$=\frac { [0.037][\text{x}] }{ [0.051] }$

x = [OH] = 2.48 x 10-5

pOH = 4.61

pH = 14 – 4.61 = 9.39