Other slides in this set

Slide 2

Preview of page 2

Here's a taster:

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

Preview of page 3

Here's a taster:

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

Preview of page 4

Here's a taster:

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

Preview of page 5

Here's a taster:

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


No comments have yet been made

Similar Chemistry resources:

See all Chemistry resources »See all resources »