IB Acids And Bases Calculations

Hi I am doing the IB Diploma Programme so I need the IB Acids And Bases Calculations soc an you please help me in the same???

[Rajkumar Agarwal]

The IB Diploma Programme, for students aged 16 to 19, is an academically challenging and balanced programme of education that prepares students for success at university and life beyond.

IB Acids And Bases Calculations

State the expression for the ionic product constant of water (Kw).

Water is in equilibrium with its dissociated ions (hydrogen and hydroxide).

The equilibrium:

H2O H+ + OH-

Can be expressed according to the equilibrium law:
Kc = [H+][OH-]
[H2O]

However, as the concentration of the water effectively remains constant on both sides of the equilibrium then the [H2O] term can be removed to a very close approximation and the equilibrium constant denoted as Kw (sometimes called the ionic product of water).

This gives:

Kw = [H+][OH-]

The constant Kw remains unchanged at constant temperature (as all good constants should!).

At 25ºC the value of Kw = 1 x 10-14 mol2 dm-6

As the concentration of the hydrogen ions equals the concentration of the hydroxide ions (see note 1) then the concentration of hydrogen ions in pure water at 25ºC = the square root of the ionic product of water:

= 1 x 10-7 mol dm-3

All equilibrium constants are temperature dependent (and this one is no exception)

Deduce [H+(aq)] and [OH-(aq)] for water at different temperatures given Kw values.
Variation of Kw with temperature

The equilibrium

H2O H+ + OH-

involves the breaking of bonds and is therefore endothermic - energy must be applied to break one of the the H-O-H bonds to give the ions. Consequently, according to Le Chatelier, an increase in temperature favours the forward reaction - i.e. the position of equilibrium shifts towards the right hand side and Kw becomes larger.

However, as the ratio of hydrogen ions to hydroxide ions in pure water must remain 1:1, then if we know the value of Kw, it is a simple matter to calculate the value of either H+ or/and OH- to obtain the concentrations and hence the values of pH and pOH.

Example: Calculate the pH when Kw = 6,5 x 10-14 mol2 dm-6

As...

Kw = [H+][OH-]

and...

[H+] =[OH-]

Then...

Kw = [H+]2

Therefore...

[H+] = √ Kw

[H+] = √ 6.5 x 10-14

[H+] =2.55 x 10-7

pH = 6.59

The pOH value will also be the same as [H+] =[OH-]