You should know what acid and base in A-Level Chemistry, let's review their definitions:
Arrhenius definition – when acids/bases dissolve in water then completely/partially dissociate into charged particles (ions)
Brønsted–Lowry definition – an acid is a proton donor and a base is a proton acceptor; acids (proton donors) will never release a H+on its own, it is always combined with H2O to form Hydroxium ions– H3O+
Strong & weak acid/base
strong acid/base
examples: HCl, H2SO4, NaOH, KOH
HCl(g) —> H+(aq) + Cl-(aq)
NaOH(s) + H2O(l) —> Na+(aq)+ OH-(aq)
Strong acids and bases ionise almost completely in water.
*HCl has a pH of 0 = completely ionised
weak acid/base
examples: CH3COOH, HCN, NH3
CH3COOH(aq) <-->CH3COO-(aq)+ H+(aq)
NH3(aq)+ H2O(l) <--> NH4+(aq)+ OH-(aq)
Weak acids and bases only slightly ionise.
Equilibrium is set up with mostly reactants (to the left)
WATER is special – it can behave as a base and an acid.
You can work out the equilibrium constant in the same manner as
i.e. Kc= [H+] [OH-] / [H2O]
However, the equilibrium is very far left and so the equilibrium constant for this reaction is said to have a constant value;
At 298K/1atm, the Kc of water is 1.0 x 10-14 mol2 dm-6
pH value
pH– “power of hydrogen” - is a measure of the hydrogen ion concentration
pH= - log [H+]
CALCULATION- finding the pH of a strong acid
1) Calculate the pH of 0.05 mol dm-3 of nitric acid.
pH = - log[H+]
pH = - log[0.05] =1.3 (strong and very acidic)
2) An acid has a pH of 2.45, what is the hydrogenion concentration?
pH = - log[H+]
[H+] = 10-pH
=3.55 x 10-3 mol dm-3
*NOTE: H2SO4 dissociates to give 2[H+] and you will have to divide the final answer by 2 to find your hydrogen ion concentration
CALCULATION: finding the pH of a weak acid
Weak acids do not fully dissociate so it isn’t as straightforward as above. Another constant called Ka is introduced. There are some assumptions to make first:
a) Only a tiny amount of product dissociates so initial concentration of reactant = equilibrium concentration of reactant
b) All H+ ions come from the acid i.e. concentration of product 1 = concentration of product 2
1) Calculate the hydrogen ion concentration and the pH of a 0.02 mol dm-3 solution of propanoic acid (CH3CH2COOH). The Ka of propanoic acid is1.3 x 10-5 moldm-3.
Ka = [H+]2/[CH3CH2COOH]
[H+] = 5.09 x 10-4
pH = -log[5.09 x 10-4]
pH = 3.29
CALCULATION: finding the pH of a strong base
One OH- ion fully dissociates per mole of base so the concentration of OH- ions and concentration of the base is the same. However to work out pH from the formula, we need [H+]. Therefore, we use our knowledge of (@ 298K), Kw= 1.0 x 10-14 mol2dm-6
1) Find the pH of 0.1 moldm-3 of NaOH at298K.
[H+] = Kw / [OH-]
= 1.0 x 10-14 / 0.1
= 1.0 x 10-13 mol dm-3
Therefore,
pH = -log [1.0x 10-13]
= 13.0 (ph value is large – expectedfor a strong alkali)
CALCULATION: finding the pKa
pKa= - log [Ka]
1) Calculate the pH of 0.05 mol dm-3 ofmethanoic acid (HCOOH). Methanoic acid has a pKa of 3.75.
3.75 = -log[Ka]
Ka= 1.78 x 10-4
1.78 x10-4 = [H+]2/[0.05]
[H+] = 2.98 x 10-3
pH = -log[2.98 x10-3]
pH = 2.53
Lastly, you should be aware that;
· diluting a strong acid (e.g. HCl) by a factor of 10 increases the pH by 1
· diluting a weak acid (e.g. CH3COOH) by a factor of 10 increases the pH by 0.5
Drafted by Eunice Wong (Chemistry)