The dissociation constants of two acids HA_1 | vedclass

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Question

The dissociation constants of $HA _1$ and $HA _2$ are $3 	imes 10^{-4}$ and $1.8 	imes 10^{-5}$.

The strength of an acid is directly proportional

Vedclass · Accepted Answer

$4 : 1$

Vedclass · Answer

The dissociation constants of $HA _1$ and $HA _2$ are $3 	imes 10^{-4}$ and $1.8 	imes 10^{-5}$.

The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:

$\frac{(	ext { Acidic Strength })_{ HA _1}}{	ext { (Acidic Strength })_{ HA _2}}=\frac{\sqrt{(	ext { Dissociation Constant })_{ HA _1}}}{\sqrt{(	ext { Dissociation Constant })_{ HA _2}}}$

Relative Acidic Strength $=\frac{(	ext { Acidic Strength })_{ HA _1}}{\left(	ext { Acidic Strength } HA _2ight.}=\frac{\sqrt{3.0 	imes 10^{-4}}}{\sqrt{1.8 	imes 10^{-5}}}=4.08=$

$4$ (approx)

relative strengths of acids will be $4: 1$.

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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