Question

The dissociation constant for acetic acid and $\text{ HCN }$ at $\text{ 2}{{\text{5}}^{\text{0}}}\text{C }$ are $\text{ 1}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-5}}\text{ }$ and $\text{ 4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-10}}\text{ }$ respectively. The equilibrium constant for the following equilibrium would be:
$\text{C}{{\text{N}}^{-}}\text{ + C}{{\text{H}}_{\text{3}}}\text{COOH}\rightleftharpoons \text{HCN + C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}}$
(A) $\text{ 3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}$
(B)$\text{ 3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{5}}$
(C) $\text{ 3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-5}}$
(D)$\text{ 3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-4}}$

Answer 1

The equilibrium constant (K) establishes the relation between the concentration of reactant and product .For a reversible reaction, in which the acid generates its conjugate base and base accepts the proton to form a conjugate acid, the equilibrium constant (K) is related to the dissociation constant of acid $\text{ }{{\text{K}}_{\text{a}}}$and the dissociation constant of the base $\text{ }{{\text{K}}_{\text{b}}}$ as:
$\text{ K = }{{\text{K}}_{\text{a}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{K}}_{\text{b}}}$

Complete step by step solution:
We need to consider the following equations for the equilibrium constants.
Here they are:
The dissociation of acetic acid into acetate ion and hydrogen ion is as follows:
$\text{ C}{{\text{H}}_{\text{3}}}\text{COOH }\rightleftharpoons \text{ C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}}\text{ + }{{\text{H}}^{+}}$ (1)
Let’s the dissociation constant for the acetic acid be $\text{ }{{\text{K}}_{\text{1}}}\text{ }$.We have given the value of $\text{ }{{\text{K}}_{\text{1}}}\text{ }$ which is ,
$\text{ }{{\text{K}}_{\text{1}}}=\text{ 1}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-5}}$
Now let us move to the equation of HCN. So here it is:
$\text{HCN }\rightleftharpoons {{\text{H}}^{+}}\text{ + C}{{\text{N}}^{-}}$ (2)
The $\text{ HCN }$dissociates into the hydrogen ion and cyanide ion.
In this case, the $\text{ }{{\text{K}}_{\text{2}}}$ represents the value of the dissociation constant of HCN. The value of the dissociation constant of $\text{ HCN }$ is given as,
$\text{ }{{\text{K}}_{\text{2}}}\text{ = 4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-10}}$
Now on reversing the reaction of dissociation of $\text{ HCN }$.We get the following equation,
$\text{ }{{\text{H}}^{+}}\text{ + C}{{\text{N}}^{-}}\text{ }\rightleftharpoons \text{HCN }$
Let us name this equation as equation 3. In this case, the $\text{ }{{\text{K}}_{\text{3}}}$ represents the dissociation constant. The value of $\text{ }{{\text{K}}_{\text{3}}}$ can also is written as $\text{ }\dfrac{\text{1}}{{{\text{K}}_{\text{2}}}}$ .
Therefore, we can say that the value of $\text{ }{{\text{K}}_{\text{3}}}\text{ }$is
$\text{ }{{\text{K}}_{\text{3}}}\text{=}\dfrac{\text{1}}{{{\text{K}}_{\text{2}}}}=\dfrac{1}{4.5\times {{10}^{-10}}}$
Now among the above equations that we have got, we have to add equation 1 and equation 2, to get the value of K, which is the equilibrium constant.
So adding the equation 1 and equation 3 we get the following equation:

& \text{ C}{{\text{H}}_{\text{3}}}\text{COOH }\rightleftharpoons \text{ C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}}\text{ + }{{\text{H}}^{+}} \\\ & \+ \\\ & \text{ }{{\text{H}}^{+}}\text{ + C}{{\text{N}}^{-}}\text{ }\rightleftharpoons \text{HCN} \\\ & \overline{\text{C}{{\text{H}}_{\text{3}}}\text{COOH + C}{{\text{N}}^{-}}\text{ }\rightleftharpoons \text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{-}}\text{+ HCN }} \\\ \end{aligned}$$ Now finding the value of K or the equilibrium constant. The equilibrium constant K is equal to the product of the dissociation constant of acetic acid and hydrogen cyanide. $\text{K = }{{\text{K}}_{\text{1}}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{K}}_{\text{3}}}\text{ = 1}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-5}}\text{ }\\!\\!\times\\!\\!\text{ }\dfrac{\text{1}}{\text{4}\text{.5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{-10}}}\text{ = 3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}$ Therefore, we can say that the value of the equilibrium constant is $\text{ 3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ }$. **Hence, (A) is the correct option.** **Note:** We should know that the equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. Chemical equilibrium is a state which is approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.