Bronstead-Lowry Acid -> Proton donor
Bronstead-Lowry Acid -> Proton acceptor
Strong acid -> fully dissociates (Ionised) in aqueous solution
Weak acid -> partially dissociates in aqueous solution
Conjugate Acid -> formed when a base accepts a proton
Conjugate Base -> formed when an acid donates a proton to another species
pH:
The pH of a solution is defined mathematically as;
We can again rearrange this to find the concentration of H+
. pH should always be quoted to 2 decimal
places
Ionic Product of Water (Kw):
Water dissociates very slightly according to the following equilibrium:
From this equilibrium we can set up an equation for
Kc:
However, as the concentration of water is so much larger than the [H+]
or [OH-], we can rearrange this equation to get the equation for Kw;
At 298k, Kw is given the value 1.0×10^-14
We can then use Kw to find the concentration of H+ ions from the
concentration of OH- ions
Acid Dissociation Constant (Ka):
Weak acids partially dissociate when in solution forming H+
ions and A-
(the conjugate base).
We can therefore establish the following equilibrium equation;
Note, [H+
] = [A-
] in WEAK ACIDS only so, we can replace [A-
] by [H+
], and then square
root to find [H+
]
We can rearrange this to find the [H+
] when we have the concentrations of HA and A-
. Ka will be given in
the question
As an acid increases in strength, Ka will increase and pKa (-logKa) will decrease
Buffer Solutions:
A buffer solution is a solution which resists changes in pH when small amounts of acid or base are added to
it. This is due to very large concentrations of the weak acid [HA] and the conjugate base [A-
].
When an acid is added, the concentration of H+
ions increases, so the equilibria shifts to the left, producing
more HA from A-
. However, as the relative ratios are the same, the pH remains relatively constant.
When a base is added, it reacts with H+
ions decreasing their concentration, so the equilibria shifts to the
right, producing more H+
and A-
from HA. However, as the relative ratios are the same, the pH remains
relatively constant.
To form a Buffer solution, you use a weak acid and a sodium carboxylate salt (e.g. Ethanoic acid & sodium
ethanoate)
Alternatively you can use excess weak acid and a strong base (Ethanoic acid & NaOH)
To calculate a buffer solution you first have to find the equilibrium moles
If weak acid & Sodium carboxylate salt are used, find the moles of each and these are the
equilibrium moles
If weak acid & strong base are used, find moles of each, subtract moles of strong base away from
weak acid, giving the equilibrium moles of HA & then the equilibrium moles of A-
are the moles of
HA used up
Once equilibrium moles have been found use them in the Ka expression to find [H+
]
There is an exception to this though, when EXACTLY half the weak acid has been used up, this is known as
the half neutralisation point. [HX] and [X-
] cancel out in the equation so you get Ka = [H+
]. To find the pH
you just then –logKa
Titrations:
There are 4 types of titrations we need to know at A-level; Strong acid v Strong base, Strong acid v Weak
base, Weak acid v Strong base & Weak acid v Weak base;
The point where the pH flips from positive to negative is known
as the point of equivalence or the end point. This is usually
centred around pH 7, but strong acids pull it down, strong bases
pull it up.
To find the equivalence point we use indicators (weak acids).
The weak acid is a different colour to the conjugate base, hence
when the pH flips at the point of equivalence the colour
changes.
Phenolphthalein -> pKin occurs pH 8-10 (colourless at low pH to
pink at high pH)
Methyl Orange -> pKin occurs pH 3-5 (red at low pH, yellow at
high pH)
You cannot use an indicator with weak acid v weak base as there is not vertical section for the point of
equivalence, instead you must use a pH meter.
Find moles of both to find the volume where point of equivalence occurs at.