Question

For a dibasic acid ${H_2}A \rightleftharpoons H{A^ - } + {H^ + }({K_1})$ $H{A^ - } \rightleftharpoons {A^2}^ - + {H^ + }({K_2})$ ${H_2}A \rightleftharpoons 2{H^ + } + {A^2}^ - (K)$, then
A. $K = {K_1} + {K_2}$
B. $K = {K_1} - {K_2}$
C. $K = {K_1}/{K_2}$
D. $K = {K_1}.{K_2}$

Answer 1

To find out the relationship between dissociation constant ${K_1}$ and dissociation constant ${K_2}$ first from the given reaction form an equation for dissociation constant by dividing the concentration of constituents ion by the concentration of dibasic acid.

Complete step by step answer:
The dibasic acids are defined as those acids which contain two acidic hydrogens in its molecule. The two acidic hydrogens are donated to other groups. The dibasic acid are also known as diprotic acid.
It is given three dibasic acids which dissociates in water to give ions and the dissociation constant is denoted by K.
The dissociation constant is calculated by dividing the concentration of the individual constituent ions by the total concentration of the acid. The general dissociation of acid is shown below.
$HA \to {H^ + } + {A^ - }$
The dissociation constant of acid HA is given as shown below.
${K_c} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$
Reaction 1:
${H_2}A \rightleftharpoons H{A^ - } + {H^ + }$
Here, ${H_2}A$ dissociates into $H{A^ - }$ anion and ${H^ + }$ cation.
The dissociation constant of this reaction is given as shown below.
${K_1} = \dfrac{{[H{A^ - }][{H^ + }]}}{{[{H_2}A]}}$
Reaction 2:
$H{A^ - } \rightleftharpoons {A^2}^ - + {H^ + }$
Here, $H{A^ - }$ dissociates into ${A^{2 - }}$ and ${H^ + }$.
The dissociation constant of this reaction is given as shown below.
${K_2} = \dfrac{{[{A^{2 - }}][{H^ + }]}}{{[H{A^ - }]}}$
Reaction 3:
${H_2}A \rightleftharpoons 2{H^ + } + {A^2}^ -$
Here, ${H_2}A$ dissociates into $2{H^ + }$ and ${A^{2 - }}$.
The dissociation constant of this reaction is given as shown below.
$K = \dfrac{{{{[{H^ + }]}^2}[{A^{2 - }}]}}{{[{H_2}A]}}$
By multiplying ${K_1}$ by ${K_2}$.
$\Rightarrow {K_1} \times {K_2}$
$\Rightarrow \dfrac{{[H{A^ - }][{H^ + }]}}{{[{H_2}A]}} \times \dfrac{{[{A^{2 - }}][{H^ + }]}}{{[H{A^ - }]}}$
Cancel $H{A^ - }$ on both sides.
$\Rightarrow \dfrac{{{{[{H^ + }]}^2}[{A^{2 - }}]}}{{[{H_2}A]}}$
Hence, ${K_1} \times {K_2} = K$
Thus, the dissociation constant K for reaction 3 is obtained by multiplying the dissociation constant ${K_1}$ of reaction 1 with dissociation constant ${K_2}$ of reaction 2.
Therefore, the correct option is D.

Note:
Make sure to apply the relation whose combination results in the dissociation constant K. In the dissociation constant the recipient before the ion in the reaction is substituted above in the concentration of ion.