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Question: In the preceding problem, if \(\left[ {{A^ + }} \right]\) and \(\left[ {A{B_2}^ - } \right]\) are y ...

In the preceding problem, if [A+]\left[ {{A^ + }} \right] and [AB2−]\left[ {A{B_2}^ - } \right] are y and x respectively, under equilibrium produce by adding the substance AB to the solvents, then K1/K2{K_1}/{K_2} is equal to.
A) yx(y−x)2\dfrac{y}{x}{\left( {y - x} \right)^2}
B) y2(x−y)x\dfrac{{{y^2}\left( {x - y} \right)}}{x}
C) y2(x+y)x\dfrac{{{y^2}\left( {x + y} \right)}}{x}
D) yx(x−y)\dfrac{y}{x}\left( {x - y} \right)

Explanation

Solution

We know that the equilibrium constant Keq{{\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,
Keq = [C]c[D]d[A]a[B]b{{\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,
Keq{K_{eq}} is the equilibrium constant.
The concentration of the reactants and products are denoted as [A],[B]& [C][D]\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, 2SO2(g)+O2(g)⇄2SO3(g)2S{O_{2\left( g \right)}} + {O_{2\left( g \right)}} \rightleftarrows 2S{O_{3\left( g \right)}}
We can write equilibrium constant as,
Kc=[SO3]2[SO2]2[O2]{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,
[AB]→A++AB2−\left[ {AB} \right] \to {A^ + } + A{B_2}^ -
The concentration of [A+]\left[ {{A^ + }} \right] is y.
The concentration of [AB2−]\left[ {A{B_2}^ - } \right] is x.
Then the concentration of [B−]\left[ {{B^ - }} \right] is (y-x).
The equilibrium constant K1=[A+][B−][AB]{K_1} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}}
The equilibrium constant K2=[AB2−][AB][B−]{K_2} = \dfrac{{\left[ {A{B_2}^ - } \right]}}{{\left[ {AB} \right]\left[ {{B^ - }} \right]}}
K1K2=[A+][(B−)2][AB2−]\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,
K1K2=[y][(y−x)2][x]\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 pHpH of the reaction,
Example:
Write the dissociation equation of the reaction.
HA+H2O→H3O++A−HA + {H_2}O\xrightarrow{{}}{H_3}{O^ + } + {A^ - }
The constant Ka{K_a} of the solution is 4×10−24 \times {10^{ - 2}}.
The dissociation constant of the reaction Ka{K_a} is written as,
Ka=[H3O+][A−][HA]{K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}
Let us imagine the concentration of [H3O+][A−]\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right] as x.
4×10−7=x20.08−x4 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.08 - x}}
⇒x2=4×10−7×0.08\Rightarrow {x^2} = 4 \times {10^{ - 7}} \times 0.08
On multiplying the above values we get,
⇒x=1.78×10−4\Rightarrow x = 1.78 \times {10^{ - 4}}
The concentration of Hydrogen is 1.78×10−41.78 \times {10^{ - 4}}
We can calculate the pHpH of the solution is,
pH=−log⁡[H+]=3.75pH = - \log \left[ {{H^ + }} \right] = 3.75
The pHpH of the solution is 3.753.75.