# pKa and Ka

### Background information

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

 Topic: Equilibriums The pH scale

The pKa value is related to the Ka value in a logic way. pKa values are easier to remember than Ka values andpKa values are in many cases easier to use than Ka values for fast approximations of concentrations of compounds and ions inequilibriums.

## Definition of pKa and Ka

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

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

 Topic: Ideal Gas Law Ionic Strength

## Example on how pKa and Ka values are used

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

This Ka value is used to determine how much of the NH4+ is dissociated into its conjugate base NH3 by 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 NH3 at 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 Ka value is 5.75.10-10, equivalent of a pKa value 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 pKa value, the concentration of ammonia [NH3] is 1/11 of the total TAN concentration because of the base 10 log relationshipbetween Ka and 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 pKa valueand the type of equilibrium.

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