pK_a values of two acids A and B are 4 and 5. | vedclass

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Question

(A) The $pK_a$ value is defined as $pK_a = -\log(K_a)$.For acid $A$,$pK_a = 4$,so $K_a(A) = 10^{-4}$.For acid $B$,$pK_a = 5$,so $K_a(B) = 10^{-5}$.The

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Acid $A$ is $10$ times stronger than acid $B$

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(A) The $pK_a$ value is defined as $pK_a = -\log(K_a)$.For acid $A$,$pK_a = 4$,so $K_a(A) = 10^{-4}$.For acid $B$,$pK_a = 5$,so $K_a(B) = 10^{-5}$.The ratio of the strengths of the acids is given by the ratio of their dissociation constants: $\frac{K_a(A)}{K_a(B)} = \frac{10^{-4}}{10^{-5}} = 10$.Therefore,acid $A$ is $10$ times stronger than acid $B$.

$pK_a$ values of two acids $A$ and $B$ are $4$ and $5$. The strengths of these two acids are related as:

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