ACIDS AND BASES
Definitions:

An acid is a proton donor.
HCl(g) + H2O(l)

H3O+(aq) + Cl-(aq)
A base is a proton acceptor.
NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)

When we remove a proton from an acid we form its conjugate
base.

When we donate a proton to a base we form its conjugate acid.

The conjugate base of a weak acid is basic.

The conjugate acid of a weak base is acidic.

The conjugates of strong acids and bases are neutral.

Some substances are amphiprotic. This means they can either
donate or accept protons eg water, hydrogen carbonate ion.

Strong acids and bases dissociate completely.

Weak acids and bases only partially dissociate.
pH
pH = -log10[H3O+]
In any aqueous solution the product of the hydronium ions and the hydroxide
ions is constant. It is referred to as Kw.
Kw = [H3O+][OH-] = (10-7)2 = 10-14
pH calculations
Calculate the pH of 0.005 mol L-1 HCl.
1]
HCl is a strong acid, therefore [H3O+] = 0.005 mol L-1
pH = -log10[H3O+] = - log10(0.005)
=
Calculate the pH of 0.02 mol L-1 sodium hydroxide solution.
2]
Sodium hydroxide is a strong base, therefore [OH-] = 0.02 mol L-1.
Kw = [H3O+][OH-] = 1 x 10-14
[H3O+] =
1x10 14
1x10 14
=
=
0.02
[OH  ]
pH =
Ka and pKa

Weak acids are in equilibrium.
CH3COOH(aq) + H2O(l)

 CH3COO-(aq) + H3O+(aq)
We can write an equilibrium expression:
Kc =
[CH 3COO  ][ H 3O  ]
[CH 3COOH ][ H 2 O]
The concentration of water in dilute aqueous solutions is just about constant,
so we can include it in the Kc and write a new expression:
[CH 3COO  ][ H 3O  ]
Kc[H2O] =
[CH 3COOH ]
Ka =
[CH 3COO  ][ H 3O  ]
[CH 3COOH ]

Ka is called the acid dissociation constant.

The stronger the acid, the more it dissociates. Therefore a strong
acid will have a large Ka value and a weak acid will have a small
Ka.
The larger the Ka the stronger the acid.

Ka is only constant for a given temperature (like all equilibrium
constants).

pKa = - log10Ka
The larger the pKa the weaker the acid.
Calculating pH of weak acids:
With a weak acid there is an equilibrium between the acid molecules and the
hydronium ions. The [H3O+] has to be found using the Ka.
Example: Find the pH of 0.10 mol L-1 ethanoic acid solution.
Ka(CH3COOH) = 1.74 x 10-5.
Step 1.
Write the equation for the dissociation of the acid:
CH3COOH(aq) + H2O(l)
Step 2:

Write the expression for Ka.
Ka =
[CH 3COO  ][ H 3O  ]
[CH 3 COOH ]
Step 3:
Make the assumptions: (2 assumptions)
1.
[H3O+] = [ CH3COO-]
2.
[CH3COOH]  starting concentration 0.10 mol L-1.
3.
Step 4:
Put the assumptions into the equation and find [H3O+].
[CH 3COO  ][ H 3O  ]
[ H 3O  ]2
-5
1.74 x 10 =
=
[CH 3COOH ]
0.10
-5
+
2
1.74 x 10 x 0.10 = [H3O ]
[H3O+] =
Step 5:
Calculate the pH.
pH = -log10[H3O+]
pH =
Calculation pH of weak bases:

Weak bases react with water according to the following equilibrium
equation:
B- + H2O  HB + OHeg NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

The expression for the equilibrium constant Kb is:
Kb =

If we multiply Ka by Kb we get Kw.
Acetic acid is a weak acid. The acetate ion, its conjugate base is basic.
CH3COOH + H2O  CH3COO- + H3O+
CH3COO- + H2O  CH3COOH + OH[CH 3COO  ][ H 3O  ]
Ka =
[CH 3COOH ]
Kb =
[CH 3COOH ][OH  ]
CH 3COO  ]
[CH 3COO  ][ H 3O  ] [CH 3COOH ][OH  ]
Ka x K b =
x
[CH 3COOH ]
[CH 3COO  ]
=
[H3O+][OH-]
= Kw
Ka x Kb = Kw
pH of salts of weak acids or bases:
The pH of a solution of an acidic salt (eg ammonium chloride) or a basic salt
(eg Sodium ethanoate) can be calculated in exactly the same way as that of
acids and bases such as ethanoic acid or ammonia.
Example:
Calculate the pH of a 0.20 mol L-1 solution of NH4NO3, if
+
Ka(NH4 ) = 5.75 x 10-10.
Step 1:
Write the hydrolysis equation:
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Step 2:
Write the expression for the dissociation constant:
Ka =
Step 3:
Make the assumptions:
[NH3] = [H3O+]
[NH4+]  0.20 mol L-1
Step 4:
Find [H3O+] and hence calculate the pH.