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Question: If the equilibrium constant \[C{H_3}COOH + {H_2}O \rightleftharpoons C{H_3}CO{O^ - } + {H_3}{O^ + }\...

If the equilibrium constant CH3COOH+H2O⇌CH3COO−+H3O+C{H_3}COOH + {H_2}O \rightleftharpoons C{H_3}CO{O^ - } + {H_3}{O^ + } is 1.8×10−51.8 \times {10^{ - 5}}, the equilibrium constant for CH3COOH+OH−⇌CH3COO−+H2OC{H_3}COOH + O{H^ - } \rightleftharpoons C{H_3}CO{O^ - } + {H_2}O is?
A. 1.8×10−91.8 \times {10^{ - 9}}
B.1.8×1091.8 \times {10^9}
C.5.55×10−95.55 \times {10^{ - 9}}
D.5.55×10105.55 \times {10^{10}}

Explanation

Solution

To answer this question, you should recall the concept of the equilibrium constant. We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. We shall calculate the equilibrium constant using the relation of equilibrium constant of reactions and equilibrium constant of water.

Complete step by step answer:
The equilibrium constant value of a reaction is independent of the initial values of the concentration of reactants. It is defined as the product of molar concentrations of the products divided by the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical equation at equilibrium. It is only dependent on the temperature at which the reaction is taking place.
The reaction given is: CH3COOH+H2O→CH3COO−+H3O+C{H_3}COOH + {H_2}O \to C{H_3}CO{O^ - } + {H_3}{O^ + }.
The equilibrium constant can be written as:
K1=[CH3COOH][CH3COO−][H3O+] (i){K_{1}} = [C{H_3}COOH][C{H_3}CO{O^ - }][H3{O^ + }]{\text{ }}\left( i \right).
The required reaction is:
CH3COOH+OH−→CH3COO−+H2OC{H_3}COOH + O{H^ - } \to C{H_3}CO{O^ - } + {H_2}O.
The equilibrium constant can be written as
K2=[CH3COOH][OH−][CH3COO−] (ii){K_2} = [C{H_3}COOH][O{H^ - }][C{H_3}CO{O^ - }]{\text{ }}\left( {ii} \right).
The reaction of self ionization of water can be written as:
H2O+H2O⇌H3O++OH−{H_2}O + {H_2}O \rightleftharpoons {H_3}{O^ + } + O{H^ - }.
The equilibrium constant for this reaction can be written as:
Kw=[H3O+][OH−] (iii){K_w} = [{H_3}{O^ + }][O{H^ - }]{\text{ }}\left( {iii} \right).
Comparing all three equation we get:
K1K2=Kw\dfrac{{{K_1}}}{{{K_2}}} = {K_w}.
Rearranging the values we will get:
K2=K1Kw=1.8×10−51×10−19=1.8×10−9{K_2} = \dfrac{{{K_1}}}{{{K_w}}} = \dfrac{{1.8 \times {{10}^{ - 5}}}}{{1 \times {{10}^{ - 19}}}} = 1.8 \times {10^{ - 9}}.

Hence the correct answer is i.e. option A.

Note:
You should know the change in values of concentration, pressure, catalyst, inert gas addition, etc. not affect equilibrium constant. Le Chatelier's principle states that the temperature, concentration, pressure, catalyst, inert gas addition to a chemical reaction can lead to a shift in equilibrium position only. We know that activation energy is the minimum energy required to start a chemical reaction.