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The dissociation constant of two acids $H{{A}_{1}}$ and $H{{A}_{2}}$are $3.14\times {{10}^{-4}}$and $1.96\times {{10}^{-5}}$respectively. The relative strength of the acid will be approximately:

Last updated date: 04th Aug 2024
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Hint: We know that ${{K}_{a}}$ is the acid dissociation constant and it is the equilibrium constant for the dissociation of acid. Whenever the compound has the tendency to readily lose its proton by dissociation, then it indicates that the compound is highly acidic.

It is known that dissociation of acid represents the release of a proton from the given acid. Therefore, the strength of the acid is easily determined by the dissociation constant. The strength of an acid is directly proportional to the square root of their dissociation constants. The dissociation constants of two acids $H{{A}_{1}}$ and $H{{A}_{2}}$are $3.14\times {{10}^{-4}}$and $1.96\times {{10}^{-5}}$ respectively.
$\dfrac{Acid\text{ }Strength[H{{A}_{1}}]}{Acid\text{ }Strength[H{{A}_{2}}]}=\dfrac{\sqrt{K{{a}_{\left[ H{{A}_{1}} \right]}}}}{\sqrt{K{{a}_{\left[ H{{A}_{2}} \right]}}}}$
Substituting the values in above equation we get;
$\Rightarrow \dfrac{\sqrt{3.14\times {{10}^{-4}}}}{\sqrt{1.96\times {{10}^{-5}}}}$
On further solving we get;
$\Rightarrow \dfrac{4}{1}$
Thus, the relative strength of two acids is in the ratio of $4:1$.

Remember that if one value of the dissociation constant is known, then other dissociation constants can be calculated. Ionic product of water is defined as the product of the molar concentration of hydroxyl ion and hydrogen ion concentration at constant temperature and its $p{{K}_{w}}$ is the sum of pH and pOH.