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Slide 1

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Acid Dissociation Constants and Calculating pH
Acids are compounds that have the ability to transfer H+ ions to something
Bases are things that accept these H+ ions.
Acids and bases are said to dissociate when in solution to form their
conjugate base and the proton:
HA H+ (aq) + A- (aq)
An acid is deemed stronger when the position of equilibrium lies further to
the right, i.e the acid has dissociated more.
How strong an acid is depends on the stability of its conjugate base. A strong
acid will have a weak conjugate base.
An acid is called strong when the dissociation of the acid is so far to the right
that it could be considered complete.…read more

Slide 2

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Strong Acids
Strong acids dissociate almost completely in solution.
In solution, the H+ ions will react with water to form oxonium ions.
Therefore, the dissociation in water can be written as: Assumption:
HA (aq) + H2O (l) H3O+ (aq) + A- (aq) That there is a
Which can be simplified to: dissociation of the
acid, so that the
number of moles of
HA (aq) H+ (aq) + A- (aq) by leaving out the water (which is in excess) H+ = the number of
moles of HA
To find the pH of a strong acid, the method is fairly simple.
It can be said, due to the assumption that the dissociation has gone to completion, that
the moles of H+ in solution is equal to the number of moles of HA (the acid) put into
pH = -log[H+ ]
So if 0.01 moles of HCl was put into solution, 0.01 moles of H+ would dissociate, and so
pH = -log[0.01] = 2
If 0.01 moles of H2SO4 was put into solution, 0.02 moles of H+ would dissociate, as there
are two H+ ions to lose in the molecule. Therefore
pH = -log[0.02] = 1.69…read more

Slide 3

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Equilibrium constants are written by doing the
Weak Acids concentration of products divided by the
concentrations of reactants. So in the case of a weak
acid where HA H+ + A- the constant is written as
Strong acids only slightly dissociate in solution. Ka = [H+ ][A- ]/ [HA] This equilibrium constant is called
To find the concentration of H+ in the solution, the acid dissociation constant.
we cannot use the method for strong acids. We must instead start with
finding the equilibrium constant for the acid.
Ka = [H+ ][A- ]/ [HA]
Due to assumption 1, this can then be written as Ka = [H+ ]2 / [HA] 1) [H+ (aq)]=[A- (aq)]
It is assumed this is the
case because while
You will be provided with the Ka value for the acid if pH is going to be there will be H+ ions
calculated. produced from the
natural dissociation of
water, water produces
So to find the pH you need to find the [H+ ]. To do this rearrange the far fewer H+ ions than
equation: the acid, so we do not
introduce a significant
[H+ ]2 = Ka x [HA] inaccuracy by ignoring
Square root this to find [H+ ] the effect of water
2) The amount of HA at
From here it is a simple matter of plugging in this value into the pH equilibrium is the same
formula. as the amount put into
the solution initially. As
pH = -log[H+ ] the fraction of the acid
that has dissociated is
Strong Acid Weak Acid very small so the
concentration of HA has
acid is diluted by a factor of [H+ (aq)] becomes 100x [H+ (aq)] becomes 10x changed only slightly
100 smaller smaller
pH increases by 2 pH increases by 1…read more

Slide 4

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Ionisation of Water
Water is able to dissociate slightly:
H2O(l) H+ (aq) + OH- (aq)
It is acting like a weak acid and so it can be written:
Ka = [H+ ][OH- ] / [H2O]
However we can leave out the H2O term as it is effectively constant because water is such
a weak acid, and water is in excess.
So, Kw = [H+ ][OH- ] Kw is used to denote the ionic product of water.
Water will change pH dependant on temperature because as the forward reaction is
endothermic, a change in temperature will result in a shift in the position of equilibrium.
At 298K, Kw = 1x10-14 mol2dm-6
In pure water, [H+ ] = [OH- ] and so 1x10-14 = [H+ ] 2, so [H+ ] = 1x10-7 mol dm-3
Therefore pH = 7…read more

Slide 5

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Basic Solutions
Kw allows us to calculate the pH of a strong base.
An example of a strong base is sodium hydroxide.
It is assumed that as the base is strong, so we can say that in 0.1M NaOH
solution, the concentration of OH- ions is 0.1M
Therefore Ka = 1x10-14 = [H+ ] x 0.1
Thererfore [H+ ] = 1x10-14 / 0.1
So [H+ ] = 1x10-13
Hence we can say that the pH of the solution is 13.…read more


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