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Question: The dissociation constants for acetic acid and \( HCN \) at \( {25^0}C \) are \( 1.5 \times {10^{ - ...

The dissociation constants for acetic acid and HCNHCN at 250C{25^0}C are 1.5×10−51.5 \times {10^{ - 5}} and 4.5×10−104.5 \times {10^{ - 10}} respectively. The equilibrium constant for the following equilibrium would be:
CN−+CH3COOH  ⇌  HCN+CH3COO−C{N^ - } + C{H_3}COOH\,\, \rightleftharpoons \,\,HCN + C{H_3}COO -
(A) 3.0×1043.0 \times {10^4}
(B) 3.0×1053.0 \times {10^5}
(C) 3.0×10−53.0 \times {10^{ - 5}}
(D) 3.0×10−43.0 \times {10^{ - 4}}

Explanation

Solution

Hint : Ethanoic acid is commonly called acetic acid. It is a weak acid that dissociates into acetate ion when it is reacted with hydrogen cyanide. The equilibrium constant is the value of the reaction quotient which is calculated at the equilibrium stage.

Complete step by step solution:
The value of equilibrium tells us about the relationship between the reactants and products in the chemical reaction when the equilibrium is attained. It is denoted by Kc{K_c} . The chemical equilibrium can also be expressed as the ratio of the concentration of products to the ratio of concentration of reactants, each of them raised to the value of their stoichiometric coefficients. In a chemical reaction, many types of equilibrium constants describe the relationship between reactants and products. At the equilibrium stage, the rate of forwarding reaction becomes equal to the rate of backward reaction. It is given as:
Kc=[C]c[D]d[A]a[B]b{K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}
Now, in the given question, we are given equilibrium constants for the dissociation of acetic acid and HCNHCN . The reaction for dissociation of HCNHCN will be:
HCN⇌H++CN−HCN \rightleftharpoons {H^ + } + C{N^ - }
The value of the dissociation constant for this reaction is K1=4.5×10−10{K_1} = 4.5 \times {10^{ - 10}} . The reaction for dissociation of acetic acid will be:
CH3COOH⇌CHCOO−+H+.......(i)C{H_3}COOH \rightleftharpoons CHCO{O^ - } + {H^ + }.......(i)
Value of dissociation constant for this reaction is K2=1.5×10−5{K_2} = 1.5 \times {10^{ - 5}} . Now, we will calculate the equilibrium constant for the given reaction:
CN−+CH3COOH  ⇌  HCN+CH3COO−C{N^ - } + C{H_3}COOH\,\, \rightleftharpoons \,\,HCN + C{H_3}COO -
The reverse equation of dissociation of HCNHCN will be:
H++CN−⇌HCN........(ii){H^ + } + C{N^ - } \rightleftharpoons HCN........(ii)
Its dissociation constant value will become K3=1K1=14.5×10−10{K_3} = \dfrac{1}{{{K_1}}} = \dfrac{1}{{4.5 \times {{10}^{ - 10}}}}
On adding eq. (i)(i) and (ii)(ii) we get:
CN−+CH3COOH  ⇌  HCN+CH3COO−C{N^ - } + C{H_3}COOH\,\, \rightleftharpoons \,\,HCN + C{H_3}COO -
Its equilibrium constant will be given as:
Ka=K1×K3{K_a} = {K_1} \times {K_3}
Ka=K1×K3=1.5×10−5×14.5×10−10=3×104{K_a} = {K_1} \times {K_3} = 1.5 \times {10^{ - 5}} \times \dfrac{1}{{4.5 \times {{10}^{ - 10}}}} = 3 \times {10^4}
Hence, the value of the equilibrium constant for the given reaction will be 3.0×1043.0 \times {10^4} .
Therefore, option (A) is correct.

Note:
The value of equilibrium constant does not get changed when we introduce a catalyst in the reaction. It only changes the rate of reaction. It is only affected by the change in concentration of any reactant or product.