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Question: The dissociation constants for two acids \(H{A_1}\) and \(H{A_2}\) are \(4.5 \times {10^{ - 4}}\) an...

The dissociation constants for two acids HA1H{A_1} and HA2H{A_2} are 4.5×10−44.5 \times {10^{ - 4}} and 1.8×10−51.8 \times {10^{ - 5}} respectively if both are having equal concentrations, then the relative strength of the acid is:
A. 5:25:2
B. 2:52:5
C. 5:15:1
D. 1:51:5

Explanation

Solution

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 HA1H{A_1} is 4.5×10−44.5 \times {10^{ - 4}}.
Dissociation constant of HA2H{A_2} is 1.8×10−51.8 \times {10^{ - 5}}.
We can give the dissociation of the acid by the equation,
HA→H++A−HA \to {H^ + } + {A^ - }
We can see that the dissociation constants for the given acids are low. The concentration of HAHA 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+{H^ + } is given as,
[H+]=Kac\left[ {{H^ + }} \right] = \sqrt {\dfrac{{{K_a}}}{c}}
Here Ka{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= = Strength of acid - 1Strength of acid - 2{\text{ = }}\dfrac{{{\text{Strength}}\,{\text{of}}\,{\text{acid - 1}}}}{{{\text{Strength}}\,{\text{of}}\,{\text{acid - 2}}}} =H+HA1H+HA2 = \dfrac{{{H^ + }_{H{A_1}}}}{{{H^ + }_{H{A_2}}}}
We have to know that concentration of hydronium ion is written as,
[H+HA1][H+HA2]=Ka1Ka2\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.
[H+HA1][H+HA2]=Ka1Ka2\dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \sqrt {\dfrac{{{K_{a1}}}}{{{K_{a2}}}}}
Substituting the values we get,
⇒[H+HA1][H+HA2]=4518\Rightarrow \dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \sqrt {\dfrac{{45}}{{18}}}
[H+HA1][H+HA2]=51\dfrac{{\left[ {{H^ + }_{H{A_1}}} \right]}}{{\left[ {{H^ + }_{H{A_2}}} \right]}} = \dfrac{5}{1}
The relative strengths of the acid is 5:15:1. 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+{H^ + } and HSO4−HS{O_4}^ - . The single electron on the oxygen is delocalized between three atoms of oxygen, while in SO42−S{O_4}^{2 - } two electrons are delocalized between four atoms of oxygen. The electron delocalization is less in SO22−S{O_2}^{2 - } so the hybrid is in higher energetic state than HSO4 - {\text{HS}}{{\text{O}}_{\text{4}}}^{\text{ - }}.