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# Dissociation Constant

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 Sub Topics A dissociation constant is a specific type of equilibrium constant that measures the propensity of a larger object to separate reversibly into smaller components, as when a complex falls apart into its components molecules or when a salt splits up into its component ions. The dissociation constant is usually denoted Ka and is the inverse of the affinity constant.The equilibrium constant for the dissociation of a weak acid is called the acid dissociation constant or acidity constant. The equilibrium constant for the dissociation of a weak base is called the base dissociation constant of an acid is linked to the base dissociation constant of its conjugate base.

## Dissociation Constant Definition

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"A dissociation constant is a constant whose numerical value depends on the equilibrium between the dissociated and undissociated forms of a molecule. Higher the dissociation constant greater the dissociation."
In chemistry a dissociation constant is a specific type of equilibrium constant that measures the propensity of a larger object to separate reversibly into smaller components as when a complex falls apart into its component molecules or when a salt splits up into its components ions. The dissociation constants is often also denoted as Kd and is the inverse of the affinity constant. In the special case of salts, the dissociation constant can also be called ionization constant.

## Dissociation Constant Equation

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In an aqueous solution an acid (HA) will dissociate into the carboxylate anion (A-) and hydrogen ion (H+) and may be represented by the general equation.

HA(aq) $\Leftrightarrow$ H+ + A-

At equilibrium the ratio of the products (ions) to the reactant (non ionized electrolyte) is related by the equation.

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

Where Ka is the dissociation constant. This expression shows that Ka increases if there is increased ionization and vice versa.

## Base Dissociation Constant

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Consider a dilute aqueous solution of a weak base BOH. The equilibrium expression and the Kdiss expression are

BOH $\rightleftharpoons$ B+ (aq) + OH- (aq)

$K_{diss} = \frac{[B^{+}][OH^{-}]}{[BOH]}$ = Kb

The constant Kb is called the base dissociation constant. Its magnitude is a measure of the strength of the base. The larger the value of Kb the greater is the OH- ion concentration and the greater is the strength of the base.

## Water Dissociation Constant

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Water dissociates to a very small extent into H+ and OH- ions expressed by the following equilibrium.

H2O $\rightleftharpoons$ H+ + OH-

Applying the law of chemical equilibrium the dissociation constant (Kc) of water is given as

$K_{c} = \frac{[H^{+}][OH^{-}]}{[H_{2}O]}$

Since dissociation of water takes place to a very small extent the concentration of water may be taken to be constant say k.

Hence Kc $\times$ [H2O][OH-]
Kc $\times$ K = [H+][OH-]

The product of two constants Kc and k gives another constant which is designated as Kw.

$K_{w} = [H^{+}][OH^{-}] = 1 \times 10^{-14}$ at 25oC

Kw is known as dissociation constant of water or more commonly ionic product of water.

## Equilibrium Dissociation Constant

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At equilibrium the forward rate of reaction is equal to the reverse rate of reaction. Defining the forward and reverse reaction rate constants as k and k-1 respectively, the equilibrium dissociation constant (KD) can be expressed as

$K_{D} = \frac{[R][L]}{[RL]}$

The equilibrium constant for a dissociation reaction is usually referred to as the dissociation constant and the products of that dissociation are then written in the numerator. For reactions not at equilibrium the value for the product of the concentrations of all products divided by the product of the concentrations of all reactants is frequently referred to as the mass action ratio.

## pKa Dissociation Constant

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The dissociation constant (Ka) is usually expressed in terms of its negative logarithm and the relationship between pKa and pH is derived from the Henderson-Hasselbalch equation.

Consider the ionization of some weak acid HA occurring with an acid dissociation constant Ka. Then

HA $\rightleftharpoons$ H+ + A-

and

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

Rearranging this expressions in terms of the parameter of interest [H+] we have

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

Taking the logarithm of both sides gives

log [H+] = log Ka + log10 $\frac{[HA]}{[A^{-}]}$

If we change the signs and define pKa = -log Ka, we have

pH = pKa - log10 $\frac{[HA]}{[A^{-}]}$

or

pH = pKa + log10 $\frac{[A^{-}]}{[HA]}$

## Dissociation Constant Table

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The most accessible constant is the overall dissociation constant so this constant is often taken as a measure of strength in acetic acid. The table given below gives the overall dissociation constants for some acids and bases in acetic acid.

 Acid pKHX Base pKa Perchloric acid 4.87 Tribenzylamine 5.36 Sulfuric acid 7.24 Diethylaniline 5.78 p-Toluenesulfonic acid 8.46 Potassium acetate 6.10 Sodium acetate 6.58 Lithium acetate 6.79 2,5-Dichloroaniline 9.48 Urea 10.24

## Affinity Constant

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Affinity and rate constants are operationally defined and depend on experimental conditions such as temperature pH and buffer composition. The affinity constant is the ratio between the two rate constants and the affinity constant can therefore be obtained from the combination of an infinite number of rate constants.

The affinity constant provides information on changes in free energy between the free and bound reaction states while kinetic analysis provides additional information related to energy changes between free state the transition state and the bound state.

## Calculating Dissociation Constant

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Solved problems based on dissociation constant is gvien below.

### Solved Examples

Question 1: The equivalent conductance of an aqueous acetic acid solution concentration 0.10M is 5.2 ohm-1equiv-1cm2 at 298K. Calculate the dissociation constant of acetic acid.
Solution:

$\Lambda$ = 390.71 ohm-1 equiv-1cm2

Ka = $\frac{[0.10\ equiv L^{-1}][5.2\ ohm^{-1}equiv^{-1}cm^{2}]^{2}}{[390.71ohm^{-1}equiv^{-1}cm^{2}][390.71-5.2]ohm^{-1}equiv^{-1}cm^{2}}$

= 1.8 $\times$ 10-5 equiv L-1

= 1.8 $\times$ 10-5 mol L-1

Question 2: At 25oC a 0.100M solution of acetic acid is 1.34% ionized. Calculate the degree of dissociation.
Solution:

Since acetic acid is 1.34% ionized the degree of dissociation
$\alpha = \frac{1.34}{100}$
= 0.0134