The dissociation constants of two acids HA_1 | vedclass

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(B) The relative strength of two acids is given by the ratio of the square roots of their dissociation constants $(K_a)$: $	ext{Relative Strength} =

Vedclass · Accepted Answer

$4 : 1$

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(B) The relative strength of two acids is given by the ratio of the square roots of their dissociation constants $(K_a)$: $	ext{Relative Strength} = \sqrt{\frac{K_{a1}}{K_{a2}}} = \sqrt{\frac{3.0 	imes 10^{-4}}{1.8 	imes 10^{-5}}} = \sqrt{\frac{30 	imes 10^{-5}}{1.8 	imes 10^{-5}}} = \sqrt{\frac{30}{1.8}} = \sqrt{16.66} \approx 4.08 \approx 4$. Thus,the ratio of the strengths of $HA_1$ to $HA_2$ is $4 : 1$.

The dissociation constants of two acids $HA_1$ and $HA_2$ are $3.0 \times 10^{-4}$ and $1.8 \times 10^{-5}$ respectively. The relative strengths of the acids will be

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