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 dissociation constant, 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 and pK_{a} values are in many cases easier to use than K_{a} values for fast approximations of concentrations of compounds and ions in equilibriums.
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 |
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}^{+} is 5.75^{.}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 the reaction 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 the TAN 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 or
NH_{4}^{+} is: [H^{+}]^{.} [NH_{3}] / [NH_{4}^{+}]. This is not enough for a smooth substitution
in (*) 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 we place 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 tell that if the pH value is 1 unit lower than the pK_{a} value, the concentration of ammonia [NH_{3}] is 1/11 of the total TAN concentration because of the base 10 log relationship between 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} value and the type of equilibrium.
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