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The dissociation constants for acetic acid and $\text{HCN}$ at $\text{25}{}^\circ \text{C}$ are $\text{1}\text{.5 }\times \text{ 1}{{\text{0}}^{-5}}$ and $\text{4}\text{.5 }\times \text{ 1}{{\text{0}}^{-10}}$, respectively. The equilibrium constant for the equilibrium will be:$\text{C}{{\text{N}}^{-}}\text{ + C}{{\text{H}}_{3}}\text{COOH }\rightleftharpoons \text{ HCN + C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}$(A) $3.0\text{ }\times \text{ 1}{{\text{0}}^{5}}$(B) $3.0\text{ }\times \text{ 1}{{\text{0}}^{-5}}$(C) $3.0\text{ }\times \text{ 1}{{\text{0}}^{-4}}$(D) $3.0\text{ }\times \text{ 1}{{\text{0}}^{4}}$

Last updated date: 24th Jun 2024
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Hint: For this problem, we have to write the individual equation for the formation of both hydrogen cyanide and acetic acid ions and then we have to write the value of the equilibrium constant. After adding both the equations we will get the equilibrium constant for the given reaction.

Complete step by step solution:
-In the given question, we have to calculate the equilibrium constant of the given reaction by using the given data.
-As we know that the dissociation constant for both acid and base tells us about the strength of acid as well as the base.
-The more is the ability of a molecule to dissociate into its respective ion, more will be the dissociation constant of the molecule.
-Now, firstly we have to write the equation for the preparation of hydrogen cyanide that is:
$\text{HCN }\rightleftharpoons \text{ }{{\text{H}}^{+}}\text{ + C}{{\text{N}}^{-}}$ …..(1)
-Now, it is given in the question that the value of the dissociation or equilibrium constant (${{\text{K}}_{1}}$)of the hydrogen cyanide is $\text{4}\text{.5 }\times \text{ 1}{{\text{0}}^{-10}}$.
-Now, we will write the reaction of the formation of the acetic acid that is:
$\text{C}{{\text{H}}_{3}}\text{COOH }\rightleftharpoons \text{ }{{\text{H}}^{+}}\text{ + C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}$ ……. (2)
-It is given that the value of dissociation or equilibrium constant (${{\text{K}}_{2}}$) of the acetic acid will be $\text{1}\text{.5 }\times \text{ 1}{{\text{0}}^{-5}}$.
-Now, if we will reverse the equation (1), then we will get the equilibrium constant (${{\text{K}}_{3}}$) as:
${{\text{H}}^{+}}\text{ + C}{{\text{N}}^{-}}\text{ }\rightleftharpoons \text{ HCN}$
- So, ${{\text{K}}_{3}}\text{ = }\dfrac{1}{{{\text{K}}_{1}}}$
${{\text{K}}_{3}}\text{ = }\dfrac{1}{4.5\text{ }\times \text{ 1}{{\text{0}}^{-10}}}\text{ }....\text{ (3)}$
-Now, by adding equation (1) and (3) we will get,
$\text{C}{{\text{N}}^{-}}\text{ + C}{{\text{H}}_{3}}\text{COOH }\rightleftharpoons \text{ HCN + C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}$
- And the value of equilibrium constant (K) for this equation will be equal to:
$\text{K = }{{\text{K}}_{1}}\text{ }\times \text{ }{{\text{K}}_{3}}$
$\text{K = }\dfrac{1}{4.5\ \times \text{ 1}{{\text{0}}^{-10}}}\text{ }\times \text{ 1}\text{.5 }\times \text{ 1}{{\text{0}}^{-5}}\text{ = 3 }\times \text{ 1}{{\text{0}}^{4}}$

Therefore, option (D) is the correct answer.

Note: In the above problem, the equilibrium constant is defined as the ratio of the concentration of products to the concentration of the reactant. The equilibrium constant depends on the temperature due to which if the temperature is changed the K will also change but in the above reaction, the temperature is constant.