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Question: Derive Ostwald’s dilution law for weak acid. Derive \(pH + pOH = 14\) at \({25^ \circ }C\)....

Derive Ostwald’s dilution law for weak acid. Derive pH+pOH=14pH + pOH = 14 at 25∘C{25^ \circ }C.

Explanation

Solution

This question gives the knowledge about the Ostwald’s dilution law. Ostwald’s dilution law is the application of law of mass action. The degree of dissociation of any electrolyte increases as the concentration decreases.

Complete step by step answer:
According to the Ostwald’s dilution law, the degree of dissociation of any electrolyte increases as the concentration decreases.
Consider a reaction of a weak acid as follows:
CH3COOH+H2O⇌CH3COO−+H3O+C{H_3}COOH + {H_2}O \rightleftharpoons C{H_3}CO{O^ - } + {H_3}{O^ + }
In the above reaction, we observe that the initial concentration of acetic acid is cc and after time tt the initial concentration becomes c−cαc - c\alpha , the concentration of acetate ion is cαc\alpha and the hydronium ion concentration is cαc\alpha .
The dissociation constant of the above reaction is as follows:
⇒K=[CH3COO−][H3O+][CH3COOH][H2O]\Rightarrow K = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]\left[ {{H_2}O} \right]}}
On rearranging we get,
⇒K×[H2O]=[CH3COO−][H3O+][CH3COOH]\Rightarrow K \times \left[ {{H_2}O} \right] = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}
The dissociation constant of the acid is as follows:
⇒Ka=[CH3COO−][H3O+][CH3COOH]\Rightarrow {K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}
Substitute the values in above equation as follows:
⇒Ka=(cα)(cα)c−cα\Rightarrow {K_a} = \dfrac{{\left( {c\alpha } \right)\left( {c\alpha } \right)}}{{c - c\alpha }}
On simplifying, we get
⇒Ka=cα21−α\Rightarrow {K_a} = \dfrac{{c{\alpha ^2}}}{{1 - \alpha }}
If the degree of dissociation is less than or equal to 5%5\% then consider 1−α≈11 - \alpha \approx 1, the above equation will be expressed as:
⇒Ka=cα2\Rightarrow {K_a} = c{\alpha ^2}
On rearranging, we get
α=Kac\alpha = \sqrt {\dfrac{{{K_a}}}{c}}
Consider a reaction to derive the relation pH+pOH=14pH + pOH = 14 as follows:
H2O⇌HO−+H+{H_2}O \rightleftharpoons H{O^ - } + {H^ + }
In the above reaction, we observe that the initial concentration of water is cc and after time tt the concentration of hydroxide ion is cαc\alpha and the hydronium ion concentration is cαc\alpha .
The dissociation constant of the above reaction is as follows:
⇒Kd=[HO−][H+][H2O]\Rightarrow {K_d} = \dfrac{{\left[ {H{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {{H_2}O} \right]}}
On rearranging we get,
⇒Kd×[H2O]=[HO−][H+]\Rightarrow {K_d} \times \left[ {{H_2}O} \right] = \left[ {H{O^ - }} \right]\left[ {{H^ + }} \right]
The dissociation constant of the water is as follows:
⇒Kw=[HO−][H+]\Rightarrow {K_w} = \left[ {H{O^ - }} \right]\left[ {{H^ + }} \right]
The dissociation constant of water is also known as ionic product of water or ionic constant of water or autoprotolysis of water or equilibrium constant of water.
Multiply both sides with –log on both sides as follows:
⇒−log⁡Kw=−log⁡[HO−]−log⁡[H+]\Rightarrow - \log {K_w} = - \log \left[ {H{O^ - }} \right] - \log \left[ {{H^ + }} \right]
Using pHpH formulas we conclude that the above equation is equal to,
⇒pKw=pOH+pH\Rightarrow p{K_w} = pOH + pH
As we know, the pKwp{K_w} of water at 25∘C{25^ \circ }C is 1414. Substitute the value of pKwp{K_w} as 1414 in the above equation as follows:
⇒14=pOH+pH\Rightarrow 14 = pOH + pH

Note: Always remember that the degree of dissociation of any electrolyte increases as the concentration decreases. Ostwald’s dilution law is the application of law of mass action. It is valid for weak acids only.