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 their degrees of dissociation,which is equal to the square root of the ratio of their

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(B) The relative strength of two acids is given by the ratio of their degrees of dissociation,which is equal to the square root of the ratio of their dissociation constants: $\frac{	ext{Strength}_1}{	ext{Strength}_2} = \sqrt{\frac{K_{a_1}}{K_{a_2}}}$ Given $K_{a_1} = 3.14 	imes 10^{-4}$ and $K_{a_2} = 1.96 	imes 10^{-5}$. $\frac{	ext{Strength}_1}{	ext{Strength}_2} = \sqrt{\frac{3.14 	imes 10^{-4}}{1.96 	imes 10^{-5}}} = \sqrt{\frac{31.4 	imes 10^{-5}}{1.96 	imes 10^{-5}}} \approx \sqrt{16} = 4$. Thus,the relative strength is $4:1$.

The dissociation constants of two acids $HA_1$ and $HA_2$ are $3.14 \times 10^{-4}$ and $1.96 \times 10^{-5}$ respectively. The relative strength of the acids will be approximately:

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