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Pka Value

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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How to Calculate pka Value?

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For any nucleophilic substitution reactions the pKa, the values allow prediction of the relevant strength of the nucleophile and the best leaving groups.

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 $K_a$ is the equilibrium constant for the dissociation.

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 $pK_w$.

The relation looks like: $pK_{w}$ = $pK_{b} + pK_{a}$

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.

There are three areas of development for $pK_a$.
  • Absolute $pK_a$ calculations using thermodynamic cycles
  • Absolute $pK_a$ for gas phase cycles
  • Absolute $pK_a$ for solution phase free energy calculations

pka Calculation Formula

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The equilibrium constant for acid is $K_a$ and for base its $K_b$. In water the $pK_b$ scale extends from 0 to 14 with low values representing the strongest bases and hence the relation between $pK_a$ and $pK_b$ is as follows.

(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 relation is known as the Henderson – Hasselbalch equation and gives the relationship between the $pK_a$ of an acid and the ratio of its acid form to conjugate base form at a given $pH$.

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

pka Value of Acetic Acid

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In an aqueous media the pKa value can be derived from the pH value by using an expression commonly known as Henderson – Hasselbalch equation.

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

pka Value of Phenol

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Phenols are converted into salts with strong alkalis such as sodium hydroxide but not with sodium carbonate solution. These are stronger acids than but lower than carboxylic acids.

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.

How to Calculate pKa Value From pH?

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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.

Example: A 0.1 M solution of a weak acid, HA has a pH value of 2.50. Calculate the value of acid dissociation constant Ka.
Solution:

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 = $-log_{10} (1 \times 10^{-4})$ =  4.0
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