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

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Last updated date: 20th Jun 2024
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Answer
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Hint: The above question is based on chemical kinetics also known as reaction kinetics, is the branch of physical chemistry that is concerned with understanding the rates of chemical reactions. For multiple reaction rates, we should compare the reaction rates of each equation and try to find out the relation between each rate of reaction.

Complete Solution :
A reaction may include one or more than one step. The vast amount of work done in chemical kinetics has led to the conclusion that some chemical reactions go in a single step known as elementary reactions. Other reactions go in more than one step and are said to be stepwise or complex. Measurements of the rates of chemical reactions over a range of conditions can show whether a reaction proceeds in one or more steps.
The rate of reaction is affected by several factors. Some of them are listed below:
- Concentration: We know, the reactions are due to collisions of reactant species. The frequency with which the molecules or ions collide depend on their concentrations. The more crowded molecules are, more likely they are to collide and react with one another. Therefore, an increase in the concentrations of the reactants will usually result in the increase in the reaction rate, while a decrease in the concentrations will usually have a reverse effect.

- Temperature: Usually temperature has a major effect on the rate of a chemical reaction. Molecules at a higher temperature will have more thermal energy. Although collision frequency is greater at higher temperatures, this alone contributes only a very small proportion to the increase in the rate of reaction.

- Catalysts: A catalyst is a substance that alters the rate of a chemical reaction, but it remains chemically unchanged after the completion of reaction. The catalyst increases the rate of the reaction by providing a new reaction mechanism or pathway to occur in a lower activation energy.
In the case of dibasic acid, reaction occurs at multiple rates so first of all we have to find the relation of each rate constant.
For reaction 1
$  {H_2}A \Leftrightarrow H{A^ - } + {H^ + }({K_1})$
$ {K_1} = \dfrac{{[H{A^ - }][{H^ + }]}}{{[{H_2}A]}}$

For reaction 2
$  H{A^ - }\underset {} \leftrightarrows {A^{ - 2}} + {H^ + }({K_2})$
$  {K_2} = \dfrac{{[{A^{ - 2}}][{H^ + }]}}{{[H{A^ - }]}}$

For reaction 3
$ {H_2}A \Leftrightarrow 2{H^ + } + {A^{2 - }}(K)$
$  K = \dfrac{{{{[{H^ + }]}^2}[{A^{2 - }}]}}{{[{H_2}A]}}$

Now, by multiplying $K_1$ and $K_2$, we will get
$  {K_1} \times {K_2} = \dfrac{{[H{A^ - }][{H^ + }]}}{{[{H_2}A]}} \times \dfrac{{[{A^{2 - }}][{H^ + }]}}{{[H{A^ - }]}}$
$  {K_1} \times {K_2} = \dfrac{{{{[{H^ + }]}^2}[{A^{2 - }}]}}{{[{H_2}A]}} = K$
So, the correct answer is “Option D”.

Note: Remember that in case of multi-step reaction. The rate is dependent on the slow step i.e. the step which requires the maximum amount of energy. All other steps which have less energy than the slow step are called fast steps.