Question

Explain the differing extents of dissociation of ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$in $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{,}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{,}$and $\mathit{N}{\mathit{H}}_{\mathbf{3}}$.

Short answer

Answer

The differing extents of dissociation of ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$in $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{,}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathit{O}$and $\mathit{N}{\mathit{H}}_{\mathbf{3}}$are shown by the order of acidity as – $\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{>}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{>}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}$.

Step-by-step solution

Concept Introduction

The equilibrium law, commonly known as Le Chatelier's principles, is used to forecast the effect of changes on a system in chemical equilibrium (such as the change in temperature or pressure).

Dissociation of H2SO4

When it is said that an acid is a strong acid it means that, that specific acid dissociates and completes completely in water.

Levelling with respect to water is done because water is easily accessible and has the highest value of the Dielectric constant and so it is easily able to break a number of polar bonds.

Being an acid ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$will not dissociate completely in Acetic Acid due to the ${\mathit{H}}^{\mathbf{+}}$contribution by the Acetic Acid (Common Ion Effect) and so it will show relatively less acidic behavior in Acetic Acid.

But when $\mathit{N}{\mathit{H}}_{\mathbf{3}}$is used as the solvent for dissolution of ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$it shows more acidic behavior than in the water as the ${\mathit{H}}^{\mathbf{+}}$furnished by ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$gets used in the reaction with $\mathit{N}{\mathit{H}}_{\mathbf{3}}$and according to the Le Chatelier’s Principle if the concentration of the product is 0 then the reaction will go on and this will go on until the ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$gets used up completely.

Even the ${\mathit{K}}_{\mathbf{a}\mathbf{2}}$of ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$is having a value of $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$will also release more fastly in and so the order of acidity of ${\mathit{H}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{4}}$in water, Acetic Acid and $\mathit{N}{\mathit{H}}_{\mathbf{3}}$are –

$\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{>}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{>}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}$. Therefore, the order is $\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{>}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{>}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}$.