Question

The dissociation constants for two acids $H{A_1}$ and $H{A_2}$ are $4.5 \times {10^{ - 4}}$ and $1.8 \times {10^{ - 5}}$ respectively if both are having equal concentrations, then the relative strength of the acid is:
A. $5:2$
B. $2:5$
C. $5:1$
D. $1:5$

Answer 1

Hint: We can calculate the relative strength of the acid using the values of dissociations constant of the two acids. We can take the square root of the dissociation constants of both acids to get the relative strength of the acids.

Complete step by step answer:
Given data contains,
Dissociation constant of $H{A_1}$ is $4.5 \times {10^{ - 4}}$.
Dissociation constant of $H{A_2}$ is $1.8 \times {10^{ - 5}}$.
We can give the dissociation of the acid by the equation,
$HA \to {H^ + } + {A^ - }$
We can see that the dissociation constants for the given acids are low. The concentration of $HA$ could be assumed to be unchanged.
We know the concentration of hydronium ion is given by the square root of the dissociation constant to the concentration of the acid.
We know that concentration of hydronium ion ${H^ + }$ is given as,
$\left[ {{H^ + }} \right] = \sqrt {\dfrac{{{K_a}}}{c}} $
Here ${K_a}$ is the dissociation constant and c is the concentration
We can assume that the concentration is constant. So, the relative strengths of the acid is given as,
Relative strength of the acid=${\text{ = }}\dfrac{{{\text{Strength}}\,{\text{of}}\,{\text{acid - 1}}}}{{{\text{Strength}}\,{\text{of}}\,{\text{acid - 2}}}}$$ = \dfrac{{{H^ + }_{H{A_1}}}}{{{H^ + }_{H{A_2}}}}$
We have to know that concentration of hydronium ion is written as,
$\dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \sqrt {\dfrac{{{K_{a1}}}}{{{K_{a2}}}}} $
We can substitute the values of dissociation constant in the above expression and get the relative strengths of the acid.
$\dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \sqrt {\dfrac{{{K_{a1}}}}{{{K_{a2}}}}} $
Substituting the values we get,
$ \Rightarrow \dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \sqrt {\dfrac{{45}}{{18}}} $
$\dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \dfrac{5}{1}$
The relative strengths of the acid is $5:1$. So, the correct answer is “Option C”.

Note:
When the value of dissociation constant is greater, the acid will be stronger. Sulfuric acid is dissociated into ${H^ + }$ and $HS{O_4}^ - $. The single electron on the oxygen is delocalized between three atoms of oxygen, while in $S{O_4}^{2 - }$ two electrons are delocalized between four atoms of oxygen. The electron delocalization is less in $S{O_2}^{2 - }$ so the hybrid is in higher energetic state than ${\text{HS}}{{\text{O}}_{\text{4}}}^{\text{ - }}$.