# pKa and Ka

Background informationThe Kavalue is a value used to describe the tendency of compounds or ions to dissociate. The Kavalue is also called the dissociationconstant, the ionisation constant, and the acid constant.Topic:EquilibriumsThe pH scaleThe pKavalue is related to the Kavalue in a logic way. pKavalues are easier to remember than Kavalues andpKavalues are in many cases easier to use than Kavalue...

### Background information

The Kavalue is a value used to describe the tendency of compounds or ions to dissociate. The Kavalue is also called the dissociationconstant, the ionisation constant, and the acid constant.

 Topic:Equilibriums The pH scale

The pKavalue is related to the Kavalue in a logic way. pKavalues are easier to remember than Kavalues andpKavalues are in many cases easier to use than Kavalues for fast approximations of concentrations of compounds and ions inequilibriums.

## Definition of pKaand Ka

The definition of Kais: [H+].[B] / [HB], where B is the conjugate base of the acid HB.

The pKavalue is defined from Ka, and can be calculated from the Kavalue from the equation pKa =-Log10(Ka)

 Topic:Ideal Gas Law Ionic Strength

## Example on how pKaand Kavalues are used

An example of ammonium and ammonia and how Kaand pKavalues are used is given below. The Ka value of NH4+is5.75.10-10under ideal conditions at 25 degrees Celsius.

This Kavalue is used to determine how much of the NH4+is dissociated into its conjugate base NH3by thereaction NH4+=> NH3+ H+.

Do also notice that the reaction NH4+NH3+ H+can go either way, depending on conditions.

By introducing the parameter TAN (Total Ammonium Nitrogen) which is ([NH4+] + [NH3]) it can be calculated how much of theTAN is on the form NH4+and NH3at any given pH.

The calculations allowing this is a bit complicated, but when you have been through them once, it is really simple:

TAN = [NH4+] + [NH3] = [NH3].(1+[NH4+] / [NH3])

Notice that normal rules such as multiplication before addition applies.

By removing the second part of the above equation only:

 Topic:pKa& Ka The basics

TAN = [NH3].(1+[NH4+] / [NH3])

remains. This equation can be rearranged to:

[NH3] = TAN / (1+[NH4+] / [NH3])
By multiplying the denominator in the last part of the above equation with [H+] one gets:
[NH3] = TAN / (1+[H+].[NH4+] / [NH3].[H+])(*)

The definition of Ka said that Ka = [H+].[B] / [HB]. Written in the context of the above example, Ka of ammonium orNH4+is: [H+].[NH3] / [NH4+]. This is not enough for a smooth substitutionin (*) so we calculate 1 / Ka to [NH4+] / [H+].[NH3]
which can be substituted into(*):
[NH3] = TAN / (1+([H+].[NH4+] / [NH3].)[H+])[NH3] = TAN / (1+ [H+] / Ka)(#)
From this equation [NH3] can be calculated when TAN, [H+] and Ka are known.

Let’s say that TAN is 0.1 M, pH is 8.24 and the Kavalue is 5.75.10-10, equivalent of a pKavalue of 9.24.If weplace these values into (#) we get that:
[NH3] = 0.1 / (1 + 5.75.10-9/ 5.75.10-10) or 0.1/11 = 0.009 M

The trick with using pKa values is, that in equilibriums like the ammonium/ammonia equilibrium you can always tellthat if the pH value is 1 unit lowerthan the pKavalue, the concentration of ammonia [NH3] is 1/11 of the total TAN concentration because of the base 10 log relationshipbetween Kaand pKa. Had the pH value been 7.24, or 2 units less than the pKa value, 1/101 of the TAN had been [NH3].

With a little experience one can give rough estimates of ions and compounds in equilibrium without a calculator just by looking at the pKavalueand the type of equilibrium.