In the previous two posts, we talked about the acid-base strength and their quantitative description using K_a, pK_a, K_b ,and pK_b.

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:

 

K_a · K_b = K_w = 10^-14

pK_a + pK_b = 14

 

The K_a and K_b 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^–

 

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

 

CH₃CO₂^–+H₂O ⇆ CH₃CO₂H + OH^–

  

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

 

Remember, the concentration of water is constant is included in the expression of K_a and K_b.

 

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

We know that K_w is constant and equals 10^-14 at 25 ^oC, therefore, as Ka or K_b 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^–   K_a = 1.8 x 10^–5

CH₃CO₂^–+H₂O ⇆ CH₃CO₂H + OH^–  K_b= 5.6 x 10^–10

^{____________________________________________________}

H₂O ⇆ H⁺ + OH^–  K_w

 

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 K_a and K_b are different, and therefore, there is an imbalance between the [H⁺] and [OH^–]. The K_a, in this case, is significantly (10⁵ x ) larger than the K_b 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 M NH₄Cl. K_b (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 K_a and K_b 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 NH_3, 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 pK_a.

Remember, K_a is the dissociation constant of the acid or the conjugate acid, while K_b 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) K_b (NH₃) = 1.8 x 10^-5

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

 

The pK_a can be calculated by either first determining the K_a from K_b since K_a · K_b = K_w,

 

or

 

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

 

Using the first equation, we get that:

 

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

 

pK_a = -log Ka = – log 5.6 x10^-10 = 9.25

 

Once the pK_a 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 pK_a and pK_b Relationship

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

 

Log (K_a · K_b) = K_w = 10^-14

log K_a + log K_b = log 10^-14

log K_a + log K_b = -14

 

Rearranging the equation, we get negative logs, which represent the pK_a and pK_b:

 

-log K_a – log K_b = 14

pK_a + pK_b = 14

 

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

Let’s say we need to determine the stronger base comparing CN^– and NO₂^–, given the pK_a values of HCN (pK_a 2.1) and HNO₂ (pK_a 3.2).

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 pK_a), will have a weaker conjugate base.

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

pK_b = 14 – pK_a

pK_b (CN^–)= 14 – 2.1 = 11.9

pK_b (NO₂^–) = 14 – 3.2 = 10.8

Therefore, NO₂^– is a stronger base.

 

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

 

pK_b = -log K_b = -log 1.8 x 10^-5 = 4.75

 

Therefore, the pK_a is:

K_a = 14 – pK_b = 14 – 4.75 = 9.25

 

Check Also

• Definitions of Acids and Bases
• Acid-Base Reactions
• Acid-Base Titrations
• Conjugate Acid and Conjugate Base
• Autoionization of Water and K_w
• The pH and Acidity
• Acid Strength, K_a, and pK_a
• Base Strength, K_b and pK_b
• The pH of a Strong Acid and Base
• pH + pOH = 14
• The pH of a Weak Acid
• The pH of a Weak Base
• ThepH of Polyprotic Acids
• The acidity of a Salt Solution
• The pH of a Salt Solution
• The pH of Salts With Acidic Cations and Basic Anions
• Acids and Bases Practice Problems