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Question: The equilibrium constant for the reaction given below is \(3.6 \times {10^{ - 7}}\) \({\text{OC}}{...

The equilibrium constant for the reaction given below is 3.6×10−73.6 \times {10^{ - 7}}
OCl−+H2O(l)⇌HOCl(aq.) + OH(aq.){\text{OC}}{{\text{l}}^ - } + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{HOCl(aq}}{\text{.) + OH(aq}}{\text{.)}}
What is Ka{{\text{K}}_a} for HOCl{\text{HOCl}}?
A. 2.8×10−82.8 \times {10^{ - 8}}
B. 3.6×10−73.6 \times {10^{ - 7}}
C. 6×10−46 \times {10^{ - 4}}
D. 2.8×10−62.8 \times {10^{ - 6}}

Explanation

Solution

Hint: This question is related to the concept of hydrolysis of salt and degree of hydrolysis is actually the fraction of total salt that is hydrolyzed at equilibrium.
Formula used:
Reaction: aA+bB⇌cC + dD{\text{aA}} + {\text{bB}} \rightleftharpoons {\text{cC + dD}}
Equilibrium constant i.e. Kc=[C]c[D]d[A]a[B]b{{\text{K}}_c} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}} where [C],[D],[A]\left[ {\text{C}} \right],\left[ {\text{D}} \right],\left[ {\text{A}} \right]and [B]\left[ {\text{B}} \right] are the molar concentrations in mol per litre and a,b,c{\text{a,b,c}} and d{\text{d}} are the stoichiometric coefficients.
Given that equilibrium constant i.e. Ka=3.6×10−7{{\text{K}}_a} = 3.6 \times {10^{ - 7}}
Reaction is: OCl−+H2O(l)⇌HOCl(aq.) + OH(aq.){\text{OC}}{{\text{l}}^ - } + {{\text{H}}_2}{\text{O(l)}} \rightleftharpoons {\text{HOCl(aq}}{\text{.) + OH(aq}}{\text{.)}}
Ka{{\text{K}}_a} for the above reaction :
Keq=[HOCl][OH−][H2O][OCl−] Kb=[HOCl][OH−][OCl−]  {{\text{K}}_{eq}} = \dfrac{{\left[ {{\text{HOCl}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\text{H}}_2}{\text{O}}} \right]\left[ {{\text{OC}}{{\text{l}}^ - }} \right]}} \\\ {{\text{K}}_b} = \dfrac{{\left[ {{\text{HOCl}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{OC}}{{\text{l}}^ - }} \right]}} \\\
Also, Ka×Kb=Kw{{\text{K}}_a} \times {{\text{K}}_b} = {{\text{K}}_w}
∴3.6×10−7×Ka=10−14 Ka≈2.8×10−8  \therefore 3.6 \times {10^{ - 7}} \times {{\text{K}}_a} = {10^{ - 14}} \\\ {{\text{K}}_a} \approx 2.8 \times {10^{ - 8}} \\\
Hence the correct option is A.
Note: As we know that all chemical reactions move towards a chemical equilibrium where the forward rate of reaction is same as the backward rate of reaction therefore on the basis of concentrations of all the different reaction species at equilibrium we defines the term equilibrium constant of a chemical reaction denoted by K{\text{K}}.