AQA A2 Acids, Bases and pH

Acids:

Bronsted-Lowry acid releases protons.

HA(aq) + H2O(l) --> H3O+(aq) + A-(aq)

Bases:

Bronsted-Lowry bases accept protons.

B(aq) + H2O(l) --> BH+(aq) + OH-(aq)

Strong and Weak:

Strong acids and bases ionise almost completely in water:

HCl is a strong acid:
HCl(g) + water --> H+(aq) + Cl-(aq)

NaOH is a strong base:
NaOH(s) + water --> Na+(aq) + OH-(aq)

Weak acids and bases dissociate only slightly in water, so the equilibrium lies far to the left.

Ethanoic acid is a weak acid:
CH3COOH(aq) <==> CH3COO-(aq) + H+(aq)

Ammonia is a weak base:
NH3(aq) + H2O <==> NH4+ + OH-(aq) 

Protons are transferred when acids and bases react.

Acids can't just throw away their protons, there needs to be a base to accept them. Here, acid HA donates a proton to base B:

HA(aq) + B(aq) <==> BH+(aq) + A-(aq)

It's an equilibrium reaction, which means that is anything is added, the equilibrium shifts to minimise the change to the system (Le Chatelier's Principle)

When an acid is added to water, water acts as a base and accepts a proton:

HA(aq) + H2O(l) <==> H3O+(aq) + A-(aq)

Water dissociates slightly in water, but only a tiny amount - the equilibrium lies far to the left.

The dissociation is so small, that the concentration of water can be considered a constant value.

By multiplying Kc by H2O, another constant, the ionic product of water is made. It has the symbol Kw

Kw = [H+][OH-] (Units: mol2dm-6)

The value of Kw under standard conditions is 1 x 10^-14 mol2dm-6.

In pure water, there is one H+ for every OH-, so, with pure water...

Kw = [H+]^2

The pH scale measures hydrogen ion concentration, it usually goes from 0 to 14.

pH = -log[H+] or [H+] = 10^-pH

For strong monoprotic acids, H+ conc. = acid conc.

For strong diprotic acids, H+ conc. = twice acid conc.

Kw is used to find the pH of a base:

1) Find [H+] by using formula for Kw:

[H+] = Kw/[OH-] (remember, Kw is 1x10^-14 unless specified)

2) Use pH formula:

pH = -log[H+]

Ka, the acid dissociation constant is used to work out the pH of weak acids:

Ka = [H+][A-]/[HA]

In weak acids, little dissociation occurs, so we can assume:

[HA]start = [HA]equilibrium

Also, we assume all [H+] comes from the acid, so:

Ka = [H+]^2/[HA]

Process:

1) Use Ka formula to find [H+]:

Ka = [H+]^2/[HA] ===> [H+]^2 = Ka[HA]

2) Square root to find [H+]:

√[H+]^2 = [H+]

3) Use pH formula:

pH = -log[H+]

pKa is calculated in a similar way to pH is with [H+]:

pKa = -log(Ka) or Ka = 10^-pKa