pKa and Ka

Background information

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

Topic:Equilibriums

The pH scale

The pK_avalue is related to the K_avalue in a logic way. pK_avalues are easier to remember than K_avalues andpK_avalues are in many cases easier to use than K_avalues for fast approximations of concentrations of compounds and ions inequilibriums.

Definition of pK_aand K_a

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

The pK_avalue is defined from K_a, and can be calculated from the K_avalue from the equation pKa =-Log₁₀(K_a)

Topic:Ideal Gas Law

Ionic Strength

Example on how pK_aand K_avalues are used

An example of ammonium and ammonia and how K_aand pK_avalues are used is given below. The Ka value of NH₄⁺is5.75^.10^-10under ideal conditions at 25 degrees Celsius.

This K_avalue 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⁺] / K_a)(#)
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 K_avalue is 5.75^.10^-10, equivalent of a pK_avalue 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 pK_avalue, the concentration of ammonia [NH₃] is 1/11 of the total TAN concentration because of the base 10 log relationshipbetween K_aand pK_a. 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 pK_avalueand the type of equilibrium.