# Finding the pH

- Created by: Hattie
- Created on: 03-04-14 12:45

## Other slides in this set

### 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

complete

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

solution.

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.

Assumptions:

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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