General Chemistry

In the previous two posts, we talked about the acid-base strength and their quantitative description using Ka, pKa, Kb ,and pKb.

Now, sometimes, we need to do calculations which can be made easier if we find a way to link these quantities, and the good news is that they are correlated as follows:


Ka · Kb = Kw = 10-14

pKa + pKb = 14


The Ka and Kb Relationship

First, remember that this relationship is only for a conjugate acid-base pair. You cannot randomly pick any acid and a base and use this correlation for your calculations.

So, let’s see how they are derived by using as an example acetic acid (CH3CO2H) and its conjugate base acetate ion (CH3CO2).

The equations and dissociation constants for the acid and the base are:


CH3CO2H ⇆ H+ + CH3CO2


\[{K_{\rm{a}}}\;{\rm{ = }}\,\frac{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{][C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H]}}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{0}}{\rm{.74}}\;{\rm{ – }}\;{\rm{x}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.7  \times  1}}{{\rm{0}}^{{\rm{ – 5}}}}\]




\[{K_{\rm{b}}}\;{\rm{ = }}\,\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{H][O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{O}}_{\rm{2}}}^{\rm{ – }}{\rm{][}}{{\rm{H}}_{\rm{2}}}{\rm{O]}}}}\]


Remember, the concentration of water is constant is included in the expression of Ka and Kb.


Now, it turns out that if we multiply the Ka and Kb of the acid-base pair, the expression for the water ionization constant is obtained:

We know that Kw is constant and equals 10-14 at 25 oC, therefore, as Ka or Kb increases, the other one decreases which proves mathematically that this relationship is only applicable to conjugate acid-base pair.

Now, what does this correlation mean? Where is water ionization come from in the acid-base dissociation?

This may look strange at first, but if we add the two equations together, we can see that the net equation represents the autoionization of water:

CH3CO2H ⇆ H+ + CH3CO2   Ka = 1.8 x 105

CH3CO2+H2O ⇆ CH3CO2H + OH  Kb= 5.6 x 1010


H2O ⇆ H+ + OH  Kw


So, does this mean if we dissolve acetic acid in water, there is really nothing happening, and it is just water autoionization which would have happened even in pure water? In other words, the second reaction is the continuation of the first – the acid dissociates, and the acetate ion formed in this reaction, reacts with water in the next.

The important part here is that the two reactions do not occur to the same extent. The values of Ka and Kb are different, and therefore, there is an imbalance between the [H+] and [OH]. The Ka, in this case, is significantly (105 x ) larger than the Kb and most of the protons formed in the acid-dissociation step stay in the solution thus making it acidic.

One application of this correlation is when we need to determine the pH of a solution with a weak base or an acid.

For example,

Calculate the pH of the buffer solution consisting of 0.75 M ammonia (NH3) and 0.95 NH4Cl. Kb (NH3) = 1.8 x 10-5.


This is a buffer solution, and the problem goes a little beyond the scope of the topic, however, it is a good illustration of where the Ka and Kb relationships are used. You can find examples in buffer solutions here.

The pH of a buffer solution is calculated using the Henderson–Hasselbalch Equation:


\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[}}{{\rm{A}}^{\rm{ – }}}{\rm{]}}}}{{{\rm{[HA]}}}}\]


In this example, the base is NH3, and the acid is the NH4+.


\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}\]


We have the concentration of ammonia and the ammonium ion, so the only missing part is the pKa.

Remember, Ka is the dissociation constant of the acid or the conjugate acid, while Kb is the dissociation constant of the base. So, for these components, the corresponding dissociation equation would be:


NH3(aq) + H2O(aq) ⇆ NH4+(aq) + OH(aq) Kb (NH3) = 1.8 x 10-5

NH4+(aq) + H2O(aq) ⇆ NH3(aq) + H3O+(aq) Ka – to be determined


The pKa can be calculated by either first determining the Ka from Kb since Ka · Kb = Kw,




by first determining the pKb and converting using the relationship: pKa + pKb = 14 (see below).


Using the first equation, we get that:


\[\left[ {{\rm{p}}{K_{\rm{a}}}} \right]\;{\rm{ = }}\,\frac{{{K_{\rm{w}}}}}{{\left[ {{\rm{p}}{K_{\rm{b}}}} \right]}}\; = \,\frac{{1.0\, \times {{10}^{ – 14}}}}{{1.8\, \times \,{{10}^{ – 5}}}}\, = \,5.6\, \times \,{10^{ – 10}}\]


pKa = -log Ka = – log 5.6 x10-10 = 9.25


Once the pKa is known, plug the values in the Henderson–Hasselbalch Equation:


\[{\rm{pH}}\; = \;{\rm{p}}{K_{\rm{a}}}\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}\]

\[{\rm{pH}}\; = \;9.25\;{\rm{ + }}\;{\rm{log}}\frac{{{\rm{0}}{\rm{.75}}\,M}}{{{\rm{0}}{\rm{.95}}\,M}}\, = \,\,9.15\]


The pKa and pKb Relationship

If take a log of both sides of the equation above, we can obtain a relationship of pKa and pKb:


Log (Ka · Kb) = Kw = 10-14

log Ka + log Kb = log 10-14

log Ka + log Kb = -14


Rearranging the equation, we get negative logs, which represent the pKa and pKb:


-log Ka – log Kb = 14

pKa + pKb = 14


This correlation can be used, for example, for determining the conjugate base strength of acids based on their pKa values.

Let’s say we need to determine the stronger base comparing CN and NO2, given the pKa values of HCN (pK2.1) and HNO2 (pK3.2).

To answer this faster, you need to remember that the stronger the acid, the weaker its conjugate base, and vice versa, the weaker the acid, the stronger its conjugate base. Therefore, HCN, being a stronger acid (smaller pKa), will have a weaker conjugate base.

Mathematically, you can calculate the corresponding pKb values keeping in mind that the smaller the pKb, the stronger the base.

pKb = 14 – pKa

pKb (CN)= 14 – 2.1 = 11.9

pKb (NO2) = 14 – 3.2 = 10.8

Therefore, NO2is a stronger base.


Going back to the example of NH3/NH4 buffer, we can also determine the pKa by first calculating the pKb:


pKb = -log Kb = -log 1.8 x 10-5 = 4.75


Therefore, the pKa is:

Ka = 14 – pKb = 14 – 4.75 = 9.25


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