Question

In the preceding problem, if $\left[ {{A^ + }} \right]$ and $\left[ {A{B_2}^ - } \right]$ are y and x respectively, under equilibrium produce by adding the substance AB to the solvents, then ${K_1}/{K_2}$ is equal to.
A) $\dfrac{y}{x}{\left( {y - x} \right)^2}$
B) $\dfrac{{{y^2}\left( {x - y} \right)}}{x}$
C) $\dfrac{{{y^2}\left( {x + y} \right)}}{x}$
D) $\dfrac{y}{x}\left( {x - y} \right)$

Answer 1

Hint: We know that the equilibrium constant ${{\text{K}}_{{\text{eq}}}}$ gives the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.
We can calculate the equilibrium constant for a reaction by using the formula which is given as below,
${{\text{K}}_{{\text{eq}}}}{\text{ = }}\dfrac{{{{\left[ {\text{C}} \right]}^{\text{c}}}{{\left[ {\text{D}} \right]}^{\text{d}}}}}{{{{\left[ {\text{A}} \right]}^{\text{a}}}{{\left[ {\text{B}} \right]}^{\text{b}}}}}$
Where,
${K_{eq}}$ is the equilibrium constant.
The concentration of the reactants and products are denoted as $\left[ {\text{A}} \right]{\text{,}}\left[ {\text{B}} \right]{\text{\& }}\left[ {\text{C}} \right]\left[ {\text{D}} \right]$ respectively.
a, b, c and d are stoichiometric coefficients
For a reaction, $2S{O_{2\left( g \right)}} + {O_{2\left( g \right)}} \rightleftarrows 2S{O_{3\left( g \right)}}$
We can write equilibrium constant as,
${K_c} = \dfrac{{{{\left[ {S{O_3}} \right]}^2}}}{{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}$

Complete step by step answer:
We can write the chemical equation for this reaction as,
$\left[ {AB} \right] \to {A^ + } + A{B_2}^ - $
The concentration of $\left[ {{A^ + }} \right]$ is y.
The concentration of $\left[ {A{B_2}^ - } \right]$ is x.
Then the concentration of $\left[ {{B^ - }} \right]$ is (y-x).
The equilibrium constant ${K_1} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}}$
The equilibrium constant ${K_2} = \dfrac{{\left[ {A{B_2}^ - } \right]}}{{\left[ {AB} \right]\left[ {{B^ - }} \right]}}$
$\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{{\left( {{B^ - }} \right)}^2}} \right]}}{{\left[ {A{B_2}^ - } \right]}}$
Substitute the values of concentration in the above equation,
$\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{\left[ y \right]\left[ {{{\left( {y - x} \right)}^2}} \right]}}{{\left[ x \right]}}$

So, the correct answer is Option A.

Note: The concept of the equilibrium constant is applied in various fields of chemistry and pharmacology. In protein-ligand binding the equilibrium constant describes the affinity between a protein and a ligand. A little equilibrium constant indicates a more tightly bound the ligand. Within the case of antibody-antigen binding the inverted equilibrium constant is employed and is named affinity constant.
Using the equilibrium constant we can calculate the $pH$ of the reaction,
Example:
Write the dissociation equation of the reaction.
$HA + {H_2}O\xrightarrow{{}}{H_3}{O^ + } + {A^ - }$
The constant ${K_a}$ of the solution is $4 \times {10^{ - 2}}$.
The dissociation constant of the reaction ${K_a}$ is written as,
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
Let us imagine the concentration of \[\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]\] as x.
$4 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.08 - x}}$
$ \Rightarrow {x^2} = 4 \times {10^{ - 7}} \times 0.08$
On multiplying the above values we get,
$ \Rightarrow x = 1.78 \times {10^{ - 4}}$
The concentration of Hydrogen is $1.78 \times {10^{ - 4}}$
We can calculate the $pH$ of the solution is,
$pH = - \log \left[ {{H^ + }} \right] = 3.75$
The $pH$ of the solution is $3.75$.