In the previous two posts, we talked about the acid-base strength and their quantitative description using *K*_{a}, p*K*_{a}, *K*_{b} ,and p*K*_{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}**

**p K_{a }+ pK_{b }= 14**

**The ***K*_{a }and *K*_{b }Relationship

*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 (CH_{3}CO_{2}H) and its conjugate base acetate ion (CH_{3}CO_{2}^{–}).

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

CH_{3}CO_{2}H ⇆ H^{+} + CH_{3}CO_{2}^{–}

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

CH_{3}CO_{2}^{–}+H_{2}O ⇆ CH_{3}CO_{2}H + 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 ^{o}C, therefore, as *K*a 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:

CH_{3}CO_{2}H ⇆ H^{+} + CH_{3}CO_{2}^{–}* K*_{a} = 1.8 x 10^{–}^{5}

CH_{3}CO_{2}^{–}+H_{2}O ⇆ CH_{3}CO_{2}H + OH^{–}* K*_{b}= 5.6 x 10^{–}^{10}

^{____________________________________________________}

H_{2}O ⇆ 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 (10^{5} 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 **(*NH_{3}) and 0.95 *M *NH_{4}Cl. *K*_{b} (NH_{3}) = 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 NH_{4}^{+}.

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

We have the concentration of ammonia and the ammonium ion, so the only **missing part is the p K_{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:

NH_{3}(*aq*) + H_{2}O(*aq*) ⇆ NH_{4}^{+}(*aq*) + OH^{–}(*aq*) *K*_{b} (NH_{3}) = 1.8 x 10^{-5}

NH_{4}^{+}(*aq*) + H_{2}O(*aq*) ⇆ NH_{3}(*aq*) + H_{3}O^{+}(*aq*) *K*_{a }**– to be determined**

The p*K*_{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 p*K*_{b} and converting using the relationship: **p K_{a} + pK_{b} = 14 **(see below).

Using the first equation, we get that:

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

p*K*_{a} = -log Ka = – log 5.6 x10^{-10} = 9.25

Once the p*K*_{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 p***K*_{a }and p*K*_{b }Relationship

*K*

_{a }and p

*K*

_{b }Relationship

If take a log of both sides of the equation above, we can obtain a relationship of p*K*_{a} and p*K*_{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 p*K*_{a} and p*K*_{b}:

-log* K*_{a }– log *K*_{b }= 14

**p K_{a }+ pK_{b }= 14**

** **

This correlation can be used, for example, for determining the conjugate base strength of acids based on their p*K*_{a} values.

Let’s say we need to **determine the stronger base comparing CN ^{–} and NO_{2}^{–}**, given the p

*K*

_{a}values of HCN (p

*K*

_{a }2.1) and HNO

_{2}(p

*K*

_{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 p*K*_{a}), will have a weaker conjugate base.

Mathematically, you can calculate the corresponding p*K*_{b} values keeping in mind that the smaller the p*K*_{b}, the stronger the base.

p*K*_{b }= 14 – p*K*_{a }

p*K*_{b }(CN^{–})= 14 – 2.1 = 11.9

p*K*_{b }(NO_{2}^{–}) = 14 – 3.2 = 10.8

Therefore, NO_{2}^{– }is a stronger base.

** **

Going back to the example of NH_{3}/NH_{4} buffer, we can also determine the pKa by first calculating the p*K*_{b}:

p*K*_{b }= -log *K*_{b }= -log 1.8 x 10^{-5 }= 4.75

Therefore, the p*K*_{a} is:

*K*_{a} = 14 – p*K*_{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 p*K*_{a} - Base Strength,
*K*_{b}and p*K*_{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**