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Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base. - Chemistry

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प्रश्न

Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base.

संक्षेप में उत्तर

उत्तर

Let us consider the reactions between a strong acid, HCl, and a weak base, NH4OH, to produce salt, NH4Cl, and water.

\[\ce{HCl_{(aq)} + NH4OH_{(aq)} ⇌ NH4Cl_{(aq)} + H2O_{(l)}}\]

In aqueous solution NH4Cl is completely dissociated as follows:

\[\ce{NH4Cl_{(aq)} -> NH^+_{4(aq)} + Cl^-_{( aq)}}\]

NH4+ is a strong conjugate acid of the weak base NH4OH and it has a tendency to react with OH from water to produce unionised NH4OH shown below.

\[\ce{NH^+_{4(aq)} + H2O_{(l)} ⇌ NH4OH_{(aq)} + H^+_{( aq)}}\]

There is no such tendency shown by Cl and therefore [H+] > [OH] the solution is acidic and the pH is less than 7.

As discussed in the salt hydrolysis of a strong base and weak acid. In this case, also, we can establish a relationship between the Ka and Kb as Kh.Kb = Kw

Let us calculate the Kb value in terms of the degree of hydrolysis (h) and the concentration of salt Kh = h2 C

[H+] = `sqrt("K"_"h"."C")`

[H+] = `sqrt("K"_"w"/"K"_"b"."C")`

pH = –log [H+]

= `-log ("K"_"w"/"K"_"b"."C")^(1/2)`

= `-1/2 log "K"_"w" - 1/2 log "C" + 1/2 log "K"_"b"`

pH = `7- 1/2 "pK"_"b" - 1/2 log "C"`

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Ionic Equilibria - Salt Hydrolysis
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अध्याय 8: Ionic Equilibrium - Evaluation [पृष्ठ ३१]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 8 Ionic Equilibrium
Evaluation | Q 18. | पृष्ठ ३१
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