### Background information

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

Topic:Equilibriums | The pH scale |

The pK_{a}value is related to the K_{a}value in a logic way. pK_{a}values are easier to remember than K_{a}values andpK_{a}values are in many cases easier to use than K_{a}values for fast approximations of concentrations of compounds and ions inequilibriums.

## Definition of pK_{a}and K_{a}

The definition of K_{a}is: [H^{+}]^{.}[B] / [HB], where B is the conjugate base of the acid HB.

The pK_{a}value is defined from K_{a}, and can be calculated from the K_{a}value from the equation pKa =-Log_{10}(K_{a})

Topic:Ideal Gas Law | Ionic Strength |

## Example on how pK_{a}and K_{a}values are used

An example of ammonium and ammonia and how K_{a}and pK_{a}values are used is given below. The Ka value of NH_{4}^{+}is5.75^{.}10^{-10}under ideal conditions at 25 degrees Celsius.

This K_{a}value is used to determine how much of the NH_{4}^{+}is dissociated into its conjugate base NH_{3}by thereaction NH_{4}^{+}=> NH_{3}+ H^{+}.

Do also notice that the reaction NH_{4}^{+}NH_{3}+ H^{+}can go either way, depending on conditions.

By introducing the parameter TAN (Total Ammonium Nitrogen) which is ([NH_{4}^{+}] + [NH_{3}]) it can be calculated how much of theTAN is on the form NH_{4}^{+}and NH_{3}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 = [NH_{4}^{+}] + [NH_{3}] = [NH_{3}]^{.}(1+[NH_{4}^{+}] / [NH_{3}])

Notice that normal rules such as multiplication before addition applies.

By removing the second part of the above equation only:

Topic:pK_{a}& K_{a} | The basics |

TAN = [NH_{3}]^{.}(1+[NH_{4}^{+}] / [NH_{3}])

remains. This equation can be rearranged to:

[NH_{3}] = TAN / (1+[NH_{4}^{+}] / [NH_{3}])

By multiplying the denominator in the last part of the above equation with [H^{+}] one gets:

[NH_{3}] = TAN / (1+[H^{+}]^{.}[NH_{4}^{+}] / [NH_{3}]^{.}[H^{+}])**(*)**

The definition of Ka said that Ka = [H^{+}]^{.}[B] / [HB]. Written in the context of the above example, Ka of ammonium orNH_{4}^{+}is: [H^{+}]^{.}[NH_{3}] / [NH_{4}^{+}]. This is not enough for a smooth substitutionin (*) so we calculate 1 / Ka to [NH_{4}^{+}] / [H^{+}]^{.}[NH_{3}]

which can be substituted into**(*)**:

[NH_{3}] = TAN / (1+([H^{+}]^{.}*[NH _{4}^{+}] / [NH_{3}]^{.})[H^{+}]*)[NH

_{3}] = TAN / (1+ [H

^{+}] / K

_{a})

**(#)**

From this equation [NH

_{3}] 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 K_{a}value is 5.75^{.}10^{-10}, equivalent of a pK_{a}value of 9.24.If weplace these values into (#) we get that:

[NH_{3}] = 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 pK_{a}value, the concentration of ammonia [NH_{3}] is 1/11 of the total TAN concentration because of the base 10 log relationshipbetween K_{a}and pK_{a}. Had the pH value been 7.24, or 2 units less than the pKa value, 1/101 of the TAN had been [NH_{3}].

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