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Question: Why is \({{K}_{{{a}_{1}}}}<<{{K}_{{{a}_{2}}}}\) for \({{H}_{2}}S{{O}_{4}}\) in water? \(S{{O}_{2}}...

Why is Ka1<<Ka2{{K}_{{{a}_{1}}}}<<{{K}_{{{a}_{2}}}} for H2SO4{{H}_{2}}S{{O}_{4}} in water?
SO2S{{O}_{2}} reacts with chlorine in the presence of charcoal.

Explanation

Solution

The extent to which ionization occurs depends upon the strength of the bond and extent of the salvation of ions produced. The terms dissociation and ionization are used with different meanings. Dissociation refers to the process of separation of ions in the water already existing as such in the solid state of the solute and ionization corresponds to a process in which a neutral molecule splits into charged ions in the solution.

Complete answer:
Arrhenius concept of acids and bases becomes useful in case of ionization of acids and bases in an aqueous medium. Strong acids like sulphuric acid (H2SO4{{H}_{2}}S{{O}_{4}}) dissociates into ions in an aqueous medium as proton donors (H+{{H}^{+}} ).
According to the Arrhenius concept, strong acids and strong bases can completely dissociate and produce H+{{H}^{+}} and OHO{{H}^{-}} ions respectively in the aqueous medium. Consider, the acid-base dissociation equilibrium of a weak acid HA
HA(aq)acid+H2O(l)baseH3O+(aq)+A(aq)conjugatebaseconjugateacid\underset{acid}{\mathop{HA(aq)}}\,+\underset{conjugateacid}{\mathop{\underset{base}{\mathop{{{H}_{2}}O(l)}}\,\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+\underset{conjugatebase}{\mathop{{{A}^{-}}(aq)}}\,}}\,
It follows that strong acid dissociates completely in water, the resulting base formed would be weak, is known as weak conjugate base.
H2SO4{{H}_{2}}S{{O}_{4}} is a strong base, which readily releases proton, and form HSO4HS{{O}_{4}}^{-} is a weak conjugate base. But HSO4HS{{O}_{4}}^{-} is a negatively charged molecule that resists the releasing of a proton. The dissociation constants of both reactions are represented Ka1&Ka2{{K}_{{{a}_{1}}}}\And {{K}_{{{a}_{2}}}} respectively.
The equations of dissociation H2SO4{{H}_{2}}S{{O}_{4}} are,

 $\begin{aligned}  

& {{H}{2}}S{{O}{4}}\rightleftharpoons HS{{O}{4}}^{-}+{{H}^{+}};{{K}{{{a}{1}}}}=First\text{ }dissociation\text{ }constant \\
& HS{{O}
{4}}^{-}\rightleftharpoons S{{O}{4}}^{-2}+{{H}^{+}};{{K}{{{a}{2}}}}=\operatorname{Sec}ond\text{ }dissociation\text{ }constant \\
\end{aligned}Hence, Hence,{{K}
{{{a}{1}}}}<<{{K}{{{a}{2}}}}forfor{{H}{2}}S{{O}{4}}inwater.Whenin water. WhenS{{O}{2}}reactswithchlorineinthepresenceofcharcoaltogivesulphurylchloride.reacts with chlorine in the presence of charcoal to give sulphuryl chloride. S{{O}{2}}+C{{l}{2}}\to S{{O}{2}}C{{l}{2}}$

Note:
Some compounds like phenolphthalein act as conjugated bases in aqueous medium.Such compounds are used as indicators in acid-base titrations and finding out the concentration of H+{{H}^{+}} ion concentration.