The Henderson-Hasselbalch Equation

1. The Henderson-Hasselbalch Equationthe significance of pH

the predominant solution species

2. The significance of pH in solutions which contain many different acid/base pairs

3. Diprotic Acids

4. Triprotic AcidsExample - calculating

concentrations of all phosphate containing species in a solution of KH2PO4 at known pH

5. pH buffer solutions- choosing the conjugate

acid/base pair - calculating the pH

Henderson-Hasselbalch Equation

HA = H+ + A- Ka = [H+] [A-][HA]

An especially convenient form of the equilibrium equation is obtained by re-writing the equilibrium expression using logs -

log10 Ka = log10 [H+] + log10

[A-

][HA]

- pKa = - pH + log10

[A-

][HA]

pH = pKa + log10

[A-

][HA]

Henderson-Hasselbalch

Equation

Use of the Henderson-Hasselbalch Equation

pH = pKa + log10

[A-

][HA]

Henderson-Hasselbalch

Equation

In most practical cases, the pH of the solution is known either from

(1) direct measurement with a pH meter

(2) use of a pH buffer in the solution

When the pH is known, the H.-H. Equation is much more convenient to use than the equilibrium constant expression. It immediately gives the ratio of concentrations of all conjugate acid/base pairs in the solution.

measured or set by buffer

known (from tabulations)

calculate ratio of (base/acid) concentrations

pKa and the Dissociation of Weak Acids

For any conjugate acid/base pair,

HA = H+ + A-

pH = pKa + log10

[A-]

[HA]

Note that pH = pKa when

[base]

[acid]

[A-]

[HA]=1=

because log10 (1) = 0

The Use of pH plots

pH

0 7 14

pKa=4.7

5

Mostly CH3COO-

mostly CH3COOH

CH3COOH = H+ + CH3COO- pKa = 4.75When pH < pKa, acetic acid is mostly protonated.

When pH > pKa, acetic acid is mostly deprotonated.When pH = pKa, concentrations of the protonated and deprotonated forms are equal.

pH, pKa and % Dissociation

When the pH = pKa, half the conjugate acid/base pair

is in the protonated form, half is de-

protonated.

At pH 4.7 [CH3COO-] / [CH3COOH] = 1 (equal)

pH 5.7 [CH3COO-] / [CH3COOH] = 10 (mostly de-protonated)

pH 3.7 [CH3COO-] / [CH3COOH] = 0.1 (mostly protonated)

If the pKa = 4.7 (as for acetic acid):

An Exercise in % Dissociation

A 0.050 M acetic acid solution is made pH 7.00 with added NaOH.

Find [CH3COOH], [CH3COO-], and [H+] in this solution.

pH = pKa + log10

[CH3COO-]

[CH3COOH]

4.75

7.00

ratio = 180

Thus, [CH3COO-] ≈ 0.050 M

[CH3COOH] ≈ 2.8 x 10-4 M

most of the acetic acid is dissociated (%

undissociated is 0.56%)

How to Use the [base] / [acid] Ratio

[base]

[acid]pH = pKa + log10

[acetic acid] + [acetate] = 0.050 M

[acetate] / [acetic acid] = r

[base]

[acid]r =

pH = pKa + log10

(r)

where

[acetic acid] = r * [acetate]

[acetic acid] (1 + r) = 0.050 M

[acetic acid] = 0.050 M / (1 + r) = 2.8 x 10-4 M [acetate] = 0.050 M - 2.8 x 10-4 M ≈ 0.050 M

1. The solution may contain many conjugate acid/base pairs (biological solutions usually do). In order to reproduce a sample, you need to reproduce the pH. This guarantees that all conjugate acid/base pairs will have the same ratio of protonated/deprotonated concentrations as in the original sample.

2. When the pH is known, you can readily calculate the ratio of (protonated/deprotonated) forms of any acid for which you know the pKa.

Summary: Significance of the pH

Ratio of protonated and deprotonated species

Knowledge of the pH completely determines the state of protonation / deprotonation of every Bronsted conjugate acid/base pair in solution.

An example:

A solution at pH 6.9 contains lactic acid (pKa = 3.9). Is lactic acid predominantly in the protonated form or the deprotonated form (lactate ion)?

Ans: Predominantly deprotonated. The ratio

[Lac-]/[HLac] = 103

Diprotic Weak Acid - H2CO3

H2CO3 = H+ + HCO3- pKa1

= 6.4

HCO3- = H+ + CO3

2- pKa2 = 10.3

pH

0 7 14

pKa1=6.

4 mostly CO3

2-

mostly H2CO3

pKa2=10.

3mostly HCO3

-

Triprotic Weak Acid - H3PO4

H3PO4 = H+ + H2PO4- pKa1

= 2.12

H2PO4- = H+ + HPO4

2- pKa2 = 7.21

HPO42- = H+ + PO4

3- pKa3 = 12.67

pH

0 7 14

pKa1=2.1

2mostly H3PO4

pKa2=7.2

1mostly H2PO4

-

mostly HPO4

2-mostly PO4

3-

pKa3=12.67

Effect of pH on Solution Composition: H3PO4

pH

0 7 14

pKa1=2.1

2mostly H3PO4

pKa2=7.2

1mostly H2PO4

-

mostly HPO4

2-

mostly PO4

3-

pKa3=12.6

7

Problem:

A solution is prepared by dissolving 0.100 mole of NaH2PO4 in water to produce 1.00 L of solution. The pH

is then adjusted to pH 8.50 with NaOH.

What are the concentrations of H3PO4, H2PO4-, HPO4

2-, PO4

3-, and H+?

Calculating the Concentrations

pH = pKa + log

[base]

[acid]

Apply H-H eqn to [H2PO4-] and [HPO4

2-] using pKa2 :

8.50 = 7.21 + log10 ( [HPO42-] /

[H2PO4-] )

This gives [HPO42-] / [H2PO4

-] = 20Apply H-H eqn to [H3PO4] and [H2PO4

-] using pKa1

:8.50 = 2.12 + log10 ( [H2PO4

-] / [H3PO4] )

This gives [H2PO4-] / [H3PO4] =

2.4 x 106Apply H-H eqn to [PO43-] and [HPO4

2-] using pKa3 :

8.50 = 12.67 + log10 ( [PO42-] /

[HPO42-] )

This gives [PO43-] / [HPO4

2-] = 6.8 x 10-5

(1)

(2)

(3)

Answers to the Exercise

pH

0 7 14

pKa1=2.12mostl

y H3PO4

pKa2=7.21mostl

y H2PO4

-

mostly HPO4

2-

mostly PO4

3-

pKa3=12.67

solution

Ans: [HPO42-] = 0.95 x10-1 M

[H2PO4-] = 0.047 x10-1

M

[PO43-] = 1.6 x10-4 *

[HPO42-]

[H3PO4] = 2.0 x106 * [H2PO4

-][H3O+] = 3.2 x10-8 M

Problem

Calculate the pH of a 500 mL solution prepared from:

0.050 mol of acetic acid and 0.020 mol sodium acetate.

(a)

HAc = H+ + Ac- pKa = 4.75

?

pH = pKa + log10

[Ac-]

[HAc]

4.75

?

?

Problem

Suppose that 0.010 mol NaOH is added to the buffer of

part (a). What is the pH?

(b)

HAc + OH- = H2O + Ac-pKa = 4.75

pH = pKa + log10

[Ac-][HAc]

4.75

(0.020 + 0.010) / 0.50

(0.050 - 0.010) / 0.50

4.63

As long as the buffer capacity is not exceeded, the change of pH is small, in

this case, 4.35 to 4.63

pH Buffers

1. pH buffers resist a change in pH upon addition of small amounts of either acid or base.

2. Buffer solutions should contain roughly equal concentrations of a conjugate acid and its conjugate base.

3. The conjugate acid/base pair of the buffer should have a pKa that approximately equals the pH.

H+ + C2H3O2- = HC2H3O2OH- + HC2H3O2 = H2O + C2H3O2

-

Added H+ is neutralized by the conjugate base

Added OH- is neutralized by the conjugate acid

For example: a buffer will result from mixing 0.1 M acetic acid and 0.1 M sodium acetate.

pH of Buffer Solutions

Problem:

Prepare a pH 5.00 buffer using sodium acetate and acetic acid

Any solution with this composition (i.e., this ratio of base / acid), will form a

buffer, but higher concentrations provide higher buffering capacity. For example, one

could use

0.178 M sodium acetate + 0.100 M acetic acid

pH = pKa + log10

[base]

[acid]

4.75

5.00 ratio

[base]/[acid] = 1.78

Exercises with Buffers

Use the data in Table 10.2 to design buffers at:

pH 6.9

pH 9.3

pH 3.6

Find the weight of solid compounds you would use to produce 100 mL each buffer.

Choosing a Weak Acid/Base Pair for a Buffer

6.9