Dissociation constant

The dissociation constant quantifies the tendency of a compound or an ion to dissociate. Dissociation is the process by which the compound or ion is split into two components that are also ions and or compounds. A general example is the dissociation of the compound AB into A+B which can be written as (1).

       (1) AB ⇔ A + B


   Fig 1

If looking into a solution (fig 1) that has been added some AB, some of the AB has dissociated into A and B and some are still on the form AB.

It can be calculated from the dissociation how much in the solution is on the form AB, A and B constant. However, before the calculation it is important to stress that the dissociation of AB into A and B is a reversible process. A and B can assemble back into AB. In this very simple example, the dissociation constant can be written as Ka = ([A] . [B]) / [AB].

If the dissociation constant of this process (Ka) is e.g. 10-7, 1 mol of AB dissolved into a solution of water will result in a concentration of both A and B of approx. 0.000316 moles and a concentration of AB of 0.999684. So in all, AB is not very soluble.

The calculation is not completely trivial. It requires that a second degree equation is set up. The process is as follows:

1. the equation is set up: 10-7 = ([A] . [B]) / [AB]
2. when AB dissolves, x concentration is lost from AB and x concentration added to A and B so the equation can be rephrased to: 10-7 = (x . x) / 1-x
3. This equation is rewritten to: 10-7 . 1-x = (x . x)
4. Further rewriting : 10-7 - 10-7.x = x 2
5. Further rewriting : 10-7 = x 2 + 10-7.x
6. Further rewriting : 0 = x 2 + 10-7.x - 10-7

This is a second order equation that can be calculated. The result is approx. 0.000362 which is x or [A] and [B]. The concentration of [AB] is 1-x or 1-0.000362. The equation can be solved numerically as well by trial and error.

More complicated cases are calculated using the same method and in most cases the problem is more complicated. Equation (1) is actually a simplification of a general equation. The general equation is written as (2):

       (2) AxBy ⇔ xA + yB

In this case the dissociation constant is written as: Ka = ([A]x . [B]y) / [AxBy]

Acid dissociation constant

Dissociation constants are commonly used to calculate what happens to weak and strong acids when dissociated. For this reason, dissociation constants are often denoted Ka-values. An simple example on how to use Ka-values is given here:
Consider ammonium (NH4+) dissociating reversibly into ammonia (NH3) (3).

       (3) NH4+ ⇔ NH3 + H+

What we are up to is to be able to calculate the concentration of NH3 from the dissociation constant of NH4+. Unfortunately we cannot measure neither NH3 and NH4+ directly. Instead, we are capable of measuring the total concentration of NH3 and NH4+ or [NH3] and [NH4+]. This is also referred to as total ammonia nitrogen or TAN.

From TAN, the concentration of NH3 can be calculated:
TAN = [NH3] + [NH4+] ⇔ TAN = [NH3] . (1 + ([NH4+] / [NH3]) ) ⇔ [NH3] = TAN / (1 + ([NH4+] / [NH3]) )

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

The last equation above can be rewritten by substituting the Ka-value of NH4+ into the equation. The Ka-value of NH4+ is defined as (4):

       (4) Ka(NH4+) = [H+]. [NH3] / [NH4+]

This is easily substituted into [NH3] = TAN / (1 + ([NH4+] . [H+]/ [NH3] .[H+] ) ) and [NH3] becomes (5):

       (5) [NH3] = TAN / (1 + ([H+] / Ka(NH4+) )

Now that we got this equation it is fairly easy to calculate [NH3] when we got TAN and pH. The Ka-value of ammonium can be found in a table. At standard temperature and pressure (STP) the dissociation constant of ammonium is 10-9.24. So when pH is 9.24 ([H+] is 10-9.24) the concentration of ammonia is half the TAN concentration. We can use this observation to figure out the [NH3] to TAN concentration at different pH levels.

When pH is 8.24, [H+] is 10-8.24 and the denominator in (5) is 11. This means that at pH = 8.24 the concentration of NH3 is 10/11 of the TAN concentration. At this pH the concentration of NH4+ is 10/11 of the TAN concentration.

pKa and Ka

The example above illustrates why pKa values are used rather than Ka-values when dissociation constants are mentioned. It is much easier to say that the pKa of ammonium is 9.24 than the Ka-value of ammonium is 10-9.24 or written in full 5.574E-10 or 0.000000000575. Also, the pKa value immediately tells that at a pH value of 9.24 the ion (in this case ammonium) is 50% dissociated.

Same reasoning as with pH-values

It's almost the same with pH values. There's not much information in telling the [H+] is 0.0000001. It's much easier to say that the pH-value is 7.

Table with dissociation values

Actually, instead of making a table inside a text the dissociation constants of all compounds I decided to put them all up in the left margin. However, these are only the chemical equations not the names. So the names of compounds included are: ammonium, methyl-amine, hydrogen-sulphide, hydrogen sulphate, bicarbonate, carbonic acid or carbon dioxide, methyl-mercaptane, nitrous acid, sulfurous acid, formic acid, phosphoric acid, hydroxyl-amine, water, perchloric acid, hydrofluoric acid, and hydrazoid acid.

Temperature and dissociation constants

Temperature has a very important effect on dissociation constant. When looking in tables with dissociation constants what is seen is the dissociation constant at 25o Celsius. A list of such constants can be seen in the literature constants section.

The influence of temperature can be calculated using the van 't Hoff equation. The van 't Hoff equation states amongst others that dissociation constants depend on temperature and the difference in the 'heat of formation' between the ion or compound that dissociates and the ion or compound it dissociates into.

The accuracy of the van 't Hoff equation depend on the accuracy of the heat of formation values available.

If we stick to the ammonium ⇔ ammonia example from the start page it is easy to illustrate how the acid dissociation constant or ionization constant changes with temperature.

To find the temperature dependency we start by finding the difference in heat of formation ΔHo between product and substrate: Σ (ΔHoproducts - ΔHosubstrates)

ΔHo of NH4+: -133.26 kj/mol
ΔHo of NH3: -80.83 kj/mol
ΔHo of H+: 0

The difference is 52.43 kJ.

This is used in the equation (6):

       (6) Loge(Ka' / Ka) = ΔHo / R . (1/T-1/T')

where Ka' is the value we are looking for from our Ka at 25oC. This tells that at 37oC (10-9.24) the Ka of NH4+ is 10-8.88. Please note that the temperature is calculated in Kelvin and not Celcius. In equation (1) this transforms to a value of T' at 310 oK whereas T is 298oK.

Unfortunately, the temperature dependency on all dissociation constants are not as cleary established as one could imagine. In the section named literature constants both Ka-values and how temperature affects them can be found for the most common ions.

Other resources

Dissociation constant definition at Avogradro - chemistry resource
Chemcases


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