In chemistry and to an extent biochemistry, acid dissociation constants ae essential for understanding many fundamental reactions. The **pKa **values are a convenient way to specify the dissociation constants for weak acid or basic groups and hence are considered as extremely informative.

The **pKa ****values **help in comparing the compounds with their acid strengths, base strengths, Gibbs free energy change and equilibrium constants of ionization reactions. For any acid base equilibrium reaction, the pKa allow an easy prediction of the favoured direction for that equilibrium as well as concentration of individual species at a given pH.

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For acid dissociation reaction:

$HA_{(aq)} \leftrightarrow A^{-} (aq) + H^{+}_{(aq)}$

The pKa of the acid HA is then defined as, pKa = $-log K_{a}$

$K_{a}$ = $[A- (aq)]$ $\frac{[H+ (aq)]}{[HA (aq)]}$

The thermodynamic definition of any equilibrium constant is in terms of activities and activities are the products of activity coefficients and concentration. Moreover, as the concentration of water is constant in all but the most concentrated solutions.

The exact nature of the solvated proton in water is unknown at this time and is the subject of much curiosity but the standard method is to use a nominal explanation for the solvated proton and represent it by [H+ (aq)]. This is acceptable because solvated proton exist in a cluster of many water molecules.

There is another method of finding $pK_a$ and that is to describe the equilibrium constant for protonation of a base $K_{b}$.

It’s simple to show that $pK_a$ and $pK_b$ are related through the $pK_w$.

Here the Kw is the product of the $[H^{+}]$ and $[OH^{-}]$ concentrations. The value of $pK_w$ varies with temperature and is very close to 14.0 for pure water at room temperature.

The pKa values can be readily measured experimentally but often chemists are interested in $pK_a$ as they need to find values of molecules that is yet to be synthesised.

The prediction of $pK_a$ in advance saves time and effort. Although the chemical accuracy is hard to achieve and by calculating acid dissociation constants the same process could be made little easier. In any kind of drug or pharmaceutical industries the knowledge of acid dissociation makes it easy to get a broader idea of the molecule to be synthesised or distinguished from others.

- Absolute $pK_a$ calculations using thermodynamic cycles
- Absolute $pK_a$ for gas phase cycles
- Absolute $pK_a$ for solution phase free energy calculations

(Acid) $pK_a$ + (conjugate base) $pK_b$ = 14

Similarly we have,

(Base) $pK_b$ + $pK_a$ (conjugate acid) = 14

Acids are in equilibrium with their conjugate base forms. One of these species will be charged and equilibrium ratio will determine the extent to which the molecule is ionised in the solution.

The expression for acid dissociation at equilibrium constant is

- $K_a$ = $\frac{[A^{-}] [H^{+}]}{[HA]}$

- $Log K_a$ = $log [A^{-}] + log [H^{+}] – log [HA]$

- - $log K_a$ = $-log [A^{-}] – log [H^{+}] + log [HA]$ … (multiplying both sides of eqn by -1)

- $pK_a$ = $pH + log$ $\frac{[HA]}{[A^{-}]}$ …. (substituting $pK_a$ for –$log K_a$ and $pH$ for – $log [H^{+}]$

The $pK_a$ is a property of the molecule the $pH$ is a property of the solvent.

In an

The $pK_a$ = $pH + log$ $\frac{(undissociated \ acid)}{(conjugate \ base \ of \ the \ acid)}$

Weak acid like acetic acid the amount of ionisation is much less and the value of Ka is rather small.

The acetic acid $CH_{3}COOH$ has Ka = 1.76 $\times$ 10$^{-5}$

To avoid small numbers the $K_a$ is expressed in logarithm form $pK_a$.

Where, $pK_a$ = $-log_{10} K_a$

The $pK_a$ = $-log [1.76 \times 10^{-5}]$ = -(- 4.75) = 4.75

The acidity of phenols arise from greater resonance stabilisation of phenoxide anion compared with phenol itself.

The strength of an acid is derived from the equilibrium between acid and water.

$HX + H_{2}O \leftrightarrow H_{3} O^{+} + X^{-}$

Therefore, $K_a$ = $[H_{3}O^{+}]$$\frac{[X^{-}]}{[HX]}$

This value of $K_a$ is then converted into $pK_a$.

The acidity of phenol is $10^{7}$ times more than cyclo-hexanol.

The $pK_a$ of phenol is calculated as 9.95.

The relationship between $pK_a$ and pH is based on the dissociation of acid at equilibrium constant. The values are considered from the logarithmic values of the same. Therefore, $pK_a$ = $pH + log$ $\frac{[HA]}{[A-]}$. If the pH and the molarity of a weak acid are known then the value of Ka can be calculated.

The pH = $-log_{10} [H^{+} (aq)]$

2.50 = $-log_{10} [H^{+} (aq)]$

$[H^{+} (aq)]$ = 3.16 $\times$ 10$^{-3}$ mol per dm$^{3}$

Ka = $\frac{(3.16 \times 10^{-3})^{2}}{0.1}$ = $1 \times 10^{-4} mol$ $dm^{-3}$

pKa = $-log_{10} Ka$

pKa = $-log_{10} (1 \times 10^{-4})$ =

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