Weak acids/bases partially
ionised in aq solutions
Ethanoic acid
Aqueous ammonia
Equilibrium to the left
Monoprotic = donates 1 H⁺ per molecule
Diprotic = donates 2 H⁺ per molecule
pH = -log₁₀[H⁺]
[H⁺] = 10⁻ᵖᴴ
pH of strong acids & alkalis
mixtures can be calculated
Annotations:
You need to:
1) Work out moles for each reactants
2) Find the moles of the excess reactant
3) Use c = 1000n ÷ v where N is the moles of excess and V is the total volume
4) If the acid is in excess, use c to find pH. If the alkali is in excess, use [H⁺] = Kw ÷ [OH⁻] and then find pH.
298K (25℃) Kw = 1x10⁻¹⁴ mol²dm⁻⁶
H₂O can be a
acid or base
Temp increases,
Kw increases
Shifts equilibrium
to right
H₂O ⇌ H⁺ + OH⁻ ΔH = +ve
Kw = [H⁺] [OH⁻]
[H⁺] = √Kw
Annotations:
Use this if you need:
1) To find [H⁺] in water at 298K
2) To find the pH of water at 298K
[H⁺] = Kw
[OH⁻]
Annotations:
Use this if you need:
1) To find pH of strong bases at 298K
2) To find conc of a strong base, rearrange to [OH⁻] = Kw ÷ [H⁺]
Weak acids
HA ⇌ H⁺ +A⁻ ΔH = +ve
Temp changes rate of
dissociation
Equilibrium constant is Ka
Ka = [H⁺] [A⁻]
[HA]
mol dm ⁻³
Larger Ka = Stronger acid
Temp increases, Ka increases
If acid is 50% ionised,
Ka = [H⁺]
[H⁺] = √Ka x [HA]
Annotations:
Use when you have a concentration of a weak acid and a Ka constant
[H⁺] = Ka x [HA] ÷ [A⁻]
Annotations:
Use when there is a mixture of excess weak acid and strong alkali
p Ka = -log₁₀Ka
Annotations:
Use this when you are not given a concentration for the weak acid and so can't use [H⁺] = √Ka x [HA]
pH of weak acids & strong alkalis
mixtures can be calculated
Annotations:
You need to (if acid is excess):
1) Work out moles for each reactants
2) Find the moles of the excess reactant
3) Find the amount of [A⁻] formed. This is just the same as the moles of alkali added
4) Use [H⁺] = Ka x [HA] ÷ [A⁻] and then find pH.
You need to (if alkali is excess):
1) Work out moles for each reactants
2) Find the moles of the excess reactant
3) Use c = 1000n ÷ v where N is the moles of excess and V is the total volume
4) If the alkali is in excess, use [H⁺] = Kw ÷ [OH⁻] and then find pH.