The reason that we work with negative logarithms (p-values) can be found in the fact that in chemistry the concentrations of substances often is very small.
Are you used to is, you will find that this is a very easy way of writing data:
pH = 6 instead of [H3O+] = 10-6 mol/l
or:
pOH = 8 instead of [OH-]= 10-8 mol/l
Always when you write concentrations, you must add the unit 'mol/l'. Working with p-values, this is not needed.
Kw
In module 9 was spoken of water as an ampholyte, so a very weak base ánd a very weak acid.
If no other substances are dissolved in water, than the pH of that water, at 25 degrees C is exactly 7 as also the pOH = 7.
With the (water)equilibrium (H2O + H2O H3O+ + OH-) comes of course also an equilibrium condition K:
The waterconcentration is constant, does not change.
Kw is called the water constant and is, at 25°C : 10-14
As in all equilibria, also the water constant depends only on the temperature.
The above formulas can also be written with p-values:
pKw = pH + pOH
or in numbers:
pKw = 14 = 7 + 7
If the pH and the pOH equal each other, so the [H3O+] and the [OH-] are equal,
than the solution is neutral.
Because Kw is a constant (so pKw too) pH and pOH toghether must be 14.
example: pH = 5 and pOH = 9. [att: only at room temperature!]
If the pH becomes bigger, the pOH becomes smaller and v.v.
adding acid makes the [H3O+] larger, so the pH <.
adding base makes the [OH-] larger, zo the pOH <.
KA and KB
If an acid (HA) or a base (A-) is dissolved in water, than an equilibrium will be reached.
We can elaborate an equilibrium constant K for the acid as well as for the base in water:
Because the [H2O] is constant (about 55,5 mol/l), this concentration is included in de constant K.
Ka is the acid constant
Kb is the base constant
the weaker an acid, the smaller the acid constant Ka.
The weaker the base, the smaller the base constant Kb.
A conjugated base belongs to an acid dat donated one proton.
In table I the conjugated acids and bases stand toghether. The difference between left and right is always one proton.
At 25°C the mathematical product of the acid constant and the base constant always will be 10-14
Badly soluble salts (built up of ions with a very strong ionic lattice) where a base is present, will only act as a base if a strong acid is added:
Cu(OH)2(s) + 2HCl(aq) CuCl2(aq) + 2H2O
Cu(OH)2(s) does not give any reaction in water to an indicator, becaus it is badly dissolved; the base (the hydroxyde ions)do not come free automatically.
If you must do a pH-calculation, do not start immediatly the calculation without looking well at the kind of substance you are dealing with; which particles ar present? Are there weak of strong acids or bases?
If you are dealing with an acid solution, always calculate with pH and/or with [H3O+].
If you are dealing with a basic solution, always calculate with pOH and/or [OH-].
As soon as you must fill in a concentration [...] in any formula (for example in Ka), than the unit must be mol/l. If needed, you must first do some conversions.
The way to resolve the problem is as follows:
Take care that you know exactly how much acid was used
Take care that you know exactly how much base was used
Investigate with a certained reaction equation which of the two will be consumed and which reactant will not completely disappear
Calculate how much will remain and convert this to mol/l
Continuously watch the volumes you work with.
Solubility product KS and solubility S
In table XI you can find of a large number of substances if they dissolve in water and, if yes, how much can dissolve befor saturating the solution.
For all substances, also for salts, there is a moment of saturarion.
Badly soluble salts reach that point much faster than well soluble salts.
The solubility of a salt mostly is indicated in the unit mol/liter, sometimes in mol/100 gr.
If a salt in water (partly) dissolves, than the ionic lattice (also partly) falls apart; the number of free particles increases.
Dissolving depends on the strenghth of the ionic lattice, but also of the temperature. At high temperatures the ionic lattice easier fals apart.
Badly soluble salts have ionic lattices that only dissociate in water for a (very) small part.
In that case there is a heterogenious and onesided equilibrium.
The small amount of silver carbonate that dissolves in water (in mol/l) is called the Solubility S at a certain temperature.
Att: in this example S mol silver carbonate will form: 2S mol silver ions and S ml carbonate ions.
The ions spread out homogeniously throughout the whole solution.
Because the solid and undissolved silvercarbonte is present in a heterogious way, the concentration of that solid in the equilibrium condition K may have the value 1.
The ionic product (I.P.) in a saturated solution of a badly soluble salt is called the SOLUBILITY PRODUCT KS
As long as there is not dissolved substance on the bottom, the solution saturated (I.P.= Ks).
If not, than you still can add and the solution is unsaturated (I.P.< Ks).
In special cases you can have oversaturated solutions (I.P. > Ks).
You must be capable to calculate S from Ks and v.v.: Ks from S.
H3O+ + OH-) comes of course also an equilibrium condition K:
K·55,5 = Kw = 10-7·10-7 = 10-14
