Question

Derive Ostwald's Dilution law for NH4OH.

Answer 1

The degree of dissociation of weak electrolytes is inversely proportional to the square root of concentration. It is called Ostwald's dilution law. α=cKa As the temperature increases, degree of dissociation will increase. α2α1=Ka2Ka1 If concentration is same.

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Answer 2

The formula for the ionization constant is $K_{b} = C\alpha ^{2}$ for the weak base NH4OH as derived from Ostwald's Dilution Law.

NH4OH is a weak base because :

• The iconic constituents of NH4OH when ionized in water are NH4+ i.e. Ammonium and OH-  i.e. Hydroxide ions
• But it only partially ionizes emitting a limiting number of OH ions which makes it a weak base.
• As it is a weak base or weak electrolyte, we can apply Ostwald's Dilution law to it.

Let the initial concentration of NH4OH before ionization be 'C'. Then the concentration of the reaction products at initial and at equilibrium are :

Reaction                      :    $NH_{4}OH ( aq)$    ⇄       $NH_{4}^{+}(aq) + OH^{-} ( aq)$

Initial Concentration :             C                            0                 0

Equilibrium Concentration :   C(1-$\alpha$)                    C$\alpha$                C$\alpha$

Thus, the initial concentration of our weak base NH4OH is C and the product ions are 0. Considering "$\alpha$" as the degree of dissociation and ionization, we have the following concentrations at equilibrium :

• The concentration of reactant NH4OH : $C(1-\alpha )$
• The concentration of products NH4 and OH ions : $C\alpha$

Applying Ostwald's dilution law for calculating ionization constant $K_{b}$ for the weak base NH4OH :

            $K_{b} = [NH_{4}^{+} ] * [OH^{-}] / [NH_{4}OH]$

                 $= (C\alpha ) * (C\alpha ) / C(1-\alpha )$

                 $= C\alpha ^{2} / (1-\alpha )$

In the case of weak electrolytes like NH4OH, the value of the degree of ionization i.e. $\alpha$ is very small and can be taken as 0. Thus,

$1 - \alpha$  can be approximated as 1.

Applying weak electrolyte approximation :

                     $K_{b} = C\alpha ^{2}$

                     $\alpha = \sqrt{K_{b}/C }$

Hence, the formula for the ionization constant is $K_{b} = C\alpha ^{2}$ , and the formula for the degree of ionization is $\alpha = \sqrt{K_{b}/C }$ for the weak base NH4OH.

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