Summary Notes 3.1.12 - Acids and bases (A-level only)

Taylor Notes Acid-Base Equilibria

Brønsted-Lowry Theory
When hydrogen bromide is dissolved in water, it acts as an acid

HBr + H2O ⇌ H3O+ + Br-

HBr acts as a Brønsted-Lowry acid because it donates protons

H2O acts as a Brønsted-Lowry base because it accepts protons

The proton donor and acceptor on the right-hand side of the equation are called the conjugate acid and conjugate
base

Br- acts as the conjugate base because it accepts a proton

H3O+ acts as the conjugate acid because it donates a proton

Equation Acid Base Conjugate Acid Conjugate Base
H2O + NH3 ⇌ OH− + NH4+ H2O NH3 NH4+ OH−
H2O + HCl ⇌ Cl− + H3O+ HCl H2O H3O+ Cl−
OH- + HCOOH ⇌ HCOO- + H2O HCOOH OH- H2O HCOO-
H2SO4 + NH3 ⇌ HSO4− + NH4+ H2SO4 NH3 NH4+ HSO4−
CH3COOH + HCl ⇌ CH3COOH2+ Cl- HCl CH3COOH CH3COOH2+ Cl-

pH Calculations
The H+ ion is responsible for a substance being acidic The higher the concentration of this ion, the greater the
acidity

pH = -log10[H+] [H+] = 10-pH

Strong Acids
A strong acid is one which fully dissociates HCl (aq)  H+ (aq) + Cl- (aq)

The hydrogen chloride shown above is a strong acid because the reaction is irreversible (goes to completion)

As strong acids fully dissociate, the concentration of H + ions can easily be calculated from the concentration of the
acidic molecule, HX

[HX] = [H+]  1 mol dm-3 of HCl  1 mol dm-3 H+

Calculate the pH of the following solutions:
a) 0.10 mol dm-3 HCl

b) 0.010 mol dm-3 HNO3

Calculate the concentration of the following solution:
a) pH of HCl = 2.51

b) pH of HNO3 = 1.10

The examples above are all monoprotic strong acids as they fully dissociate to produce one H + ion per molecule

Sulphuric acid is an example of a strong diprotic acid H2SO4 (aq)  2H+ (aq) + SO42- (aq)

Calculate the pH of the following solution a) 0.15 mol dm -3 H2SO4

, Taylor Notes Acid-Base Equilibria

Weak Acids
A weak acid is one which partially dissociates in solution HA (aq) ⇌ H+ (aq) + A- (aq)

The chemical equation shows that HA is a weak acid because the reaction is reversible, therefore it partially
dissociates

As weak acids partially dissociate, the concentration of H + ions cannot be assumed to be the same as the
concentration of the acidic molecule, HA

To perform pH calculations with weak acids, we must use the acid dissociation constant, K a

[H+][A−]
Ka =
[HA]

A stronger acid will have a higher value of K a because the equilibrium position will lie further to the right

A ‘weaker’ weak acid will have a lower value of K a because the equilibrium position will lie further to the left

For pure weak acids [H+] = [A-] or [H+] = √ Ka x [ HA ]

[H+]2 or [H+] = √ Ka x [ HA ]
Ka =
[HA]

Calculate the pH of the following solutions of ethanoic acid (K a = 1.75 x 10-5)
a) 2 mol dm-3 of CH3COOH

b) 0.200 mol dm-3 CH3COOH

Calculate the value of Ka for the following weak acids based on their pH and concentrations
a) HNO2 of concentration 0.100 mol dm-3 with a pH = 2.17




b) HBrO of concentration 1.00 mol dm-3 with a pH of 4.35


Ka values are small and inconvenient Sometimes pKa is used instead as it gives a more convenient number in the
same way that pH values are easier than [H +]

pKa = -log10(Ka) or Ka = 10-Ka

Calculate the pKa of the following weak acids
a) CH3COOH (Ka = 1.75 x 10-5)

A stronger weak acid will have a higher value of K a because the equilibrium position will lie further to the right This
means it will have a lower value of pKa

A ‘weaker’ weak acid will have a lower value of K a because the equilibrium will lie further to the left This means it
will have a higher value of pKa