Maths and Chemistry for Biologists

Chemistry 3Acids, Bases and pH

This section of the course covers –

• the nature of acids and bases• how their strengths are defined• the hydrogen ion concentrations of solutions of

acids and bases• the pH scale

What are acids and bases?

Acids are molecules that can donate protons to other substances. The substances that accept

the proton are called bases. If we represent the acid by HA and the base by B

then the reaction is

HA + B A- + BH+

The product of this reaction is a salt. For example

HCl + NH3 NH4+Cl-

Here hydrochloric acid is reacting with ammonia to produce ammonium chloride

Acids in water

When an acid is dissolved in water the water acts as a base

HA + H2O A- + H3O+

acid base conjugate base conjugate acid

The ion A- is called the conjugate base of the acid because it is capable of reacting with some other

acid to go back to HA

H3O+, which is called the hydroxonium ion, is the conjugate acid of water because it can donate a proton to some other base and go back to H2O

Bases in waterWhen a base is dissolved in water the water acts

as an acid

B + H2O BH+ + OH-

base acid conjugate acid conjugate base

The reasons why BH+ is called the conjugate acid of B and OH- is called the conjugate base of H2O

are just the same as in the previous slide

OH- is called the hydroxide ion and occurs in substances such as NaOH which react with

acids to form salts plus water

NaOH + HCl NaCl + H2O

Important facts

Water acts as an acid or a base depending on what is dissolved in it

An acid/base reaction is always of the form

Acid + Base Conjugate base + Conjugate acid

of the acid of the base

Acids in biology

These are of two main types - Carboxylic acids where the -O-H is the acidic group. R meansany other chemical group. (NB alcohols, R-OH, are not acidic)

Phosphates. Here one or bothof the Rs can be H so there can betwo or even three acid groups

R CO

OH

PRO

RO

OH

O

Bases in biology

One main type – amines

These can be primary (one R group), secondary (two R groups) or tertiary (three R groups)

(NB amides are not basic)

NR

H

H

NR

R

H

NR

R

RR C

O

NH2

Amino acids

CH2

NH2

COOH

A very important group of biomolecules are the amino acids. They have both

a carboxylic acid and an amine group. So they are both acids and bases. The structure shown is glycine. In water the molecule is ionised as shown in the lower structure.

CH2

NH3

COO

Strengths of acids

Measured by the extent to which they dissociate in water; that is, for the process below (now shown as an equilibrium), what fraction of HA dissociates to A-

HA + H2O A- + H3O+

For strong acids such as HCl, H2SO4 and HNO3 the process is essentially complete

So in a 0.1 M solution of HCl [H3O+] = 0.1 M (note that the [ ] indicate the concentration of H30+)

Strengths of weak acids

Weak acids dissociate only partially. For the process

HA + H2O A- + H3O+

we can write an equilibrium expression

O][HA][H]O][H[A

K 2

3-

where K is the equilibrium constant and the concentrations are those at equilibrium (for more details about equilibria see section Chem 6)

contd

O][HA][H]O][H[A

K 2

3-

We can simplify this by noting that the concentration of water is very high (55.6 M)* and effectively constant so the it can be included in K to give Ka (Ka = 55.6 x K). In addition it is usual to write H+ as a shorthand for H3O+ so the equation becomes

[HA]]][H[A

K-

a

Ka is called the acid dissociation constant

*(1,000 g of H2O is 1000/18 = 55.6 mol)

Meaning of Ka

The bigger is Ka then the stronger is the acid; that is, the more H+ there will be in the solution

What sort of values does Ka have?

For CH3COOH, Ka = 1.74 x 10-5 M

For HCOOH, Ka = 1.78 x 10-4 M

So formic acid is about 10 x stronger than acetic acid

These numbers are inconvenient and there is a better way of doing it

pKa

We DEFINE a quantity called pKa as

pKa = -log10 Ka

Then for acetic acid pKa = 4.76

and for formic acid pKa = 3.75

This gives us a scale of acid strength where the numbers are small and positive and where the value

changes by 1 as the acid strength changes 10x

The SMALLER the pKa the STRONGER the acid

One for you to do

Phosphoric acid, H3PO4, has three dissociable hydrogens with the Ka values shown. Calculate

the pKa values

H3PO4 H2PO4- HPO4

2- PO43-

Ka 7.08 x 10-3 M 6.31 x 10-8 M 4.68 x 10-13 M

Answer

pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.33

Note that the species are progressively weaker acids. HPO4

2- is 1010 x weaker than is H3PO4. In fact PO4

3- is quite a strong base

What about bases?

O][B][H]][OH[BH

K 2

-

We could write an equation

B + H20 BH+ + OH-

and then an equilibrium expression

[B]]][OH[BH

K-

b

or

and then define pKb = -log10 Kb

- but we usually don’t!

The pKa of the conjugate acid

Instead we talk about the pKa of the conjugate acid.

That is, we consider the process

BH+ + H2O B + H3O+

from which ][BH

][B][H Ka

e.g. for ammonia (NH3) we quote the pKa of the conjugate acid (NH4

+, pKa = 9.2) as a measure of base strength

The hydrogen ion concentration

The hydrogen ion concentration (or more properly [H3O+]) is of crucial importance in biology

For a strong acid it is equal to the concentration

0.1 M HCl, [H+] = 0.1 M, 0.1 M H2SO4, [H+] = 0.2 M

For weak acids it depends on both the concentration and the pKa

0.1 M acetic acid, [H+] = 1.32 x 10-3 M or 1.32 mM

The pH

Again, these numbers are not convenient so we

DEFINE the quantity pH = -log10 [H+]

For [H+] = 1.32 x 10-3 M, pH = 2.88

NB the HIGER the [H+] the LOWER is the pH

[H+] = 10-4 M, pH = 4, [H+] = 10-3 M, pH = 3

Calculation of [H+] for weak acids

Suppose we make a 0.1 M solution of acetic acid and that some of it dissociates to give x M of H+

CH3COOH CH3COO- + H+

At equilibrium (0.1 – x) M x M x M

x- 0.1x

COOH][CH

]][HCOO[CH K

2

3

-3

a

We know that Ka = 1.74 x 10-5 M so we can solve for x

Water is both an acid and a base

Water undergoes self-dissociation

2 H2O H3O+ + OH-

Or, in shorthand, H2O H+ + OH-

The ionic product of water Kw is DEFINED as

Kw = [H+][OH-] M2 and pKw = -log10 Kw

At 25 oC, Kw = 10-14 M2 and pKw = 14

The pH of neutral water

In neutral water [H+] = [OH-]

Since [H+][OH-] = 10-14 M2 then [H+] = 10-7 M

so the pH = 7

So a solution with a pH less than 7 is acidic and a solution with a pH greater than 7 is basic

pH of solutions of strong bases

We can calculate these using the value of Kw

For example in 0.1 M NaOH, [OH-] = 0.1 M

But [H+][OH-] = 10-14 M2 or

So [H+] = 10-14/0.1 = 10-13 and pH = 13

][OH

10 ][H -

14-

Relationship between pKa and pKb

[B]]][OH[BH

K-

b

][BH

][B][H Ka

For a base

For its conjugate acid

So[B]

]][OH[BH

][BH

][B][H K K

-

ba

xx

Hence Ka.Kb = [H+][OH-] = Kw or pKa + pKb = pKw

Ones for you to do

What is the pH of a solution with [H+] of a) 5 mM, b) 50 mM?

What is the [H+] of a solution of pH = 6?

What is the pH of a 1mM solution of NaOH?

Answers

5 mM is 5 x 10-3 M so pH = -log (5 x 10-3) = 2.30

50 mM is 5 x 10-2 M so pH = -log (5 x 10-2) = 1.30

[H+] = antilog (-pH) so if pH = 6.0,

[H+] = antilog -6 = 10-6 M or 1 M

[OH-] = 10-3 M so [H+] = 10-14/10-3 M = 10-11 M

pH = -log 10-11 = 11