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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
$1 : 4$
$4 : 1$
$1 : 16$
$16 : 1$
Solution
The dissociation constants of $HA _1$ and $HA _2$ are $3 \times 10^{-4}$ and $1.8 \times 10^{-5}$.
The strength of an acid is directly proportional to square root of dissociation constants of acids. So relative strength of the given acids are:
$\frac{(\text { Acidic Strength })_{ HA _1}}{\text { (Acidic Strength })_{ HA _2}}=\frac{\sqrt{(\text { Dissociation Constant })_{ HA _1}}}{\sqrt{(\text { Dissociation Constant })_{ HA _2}}}$
Relative Acidic Strength $=\frac{(\text { Acidic Strength })_{ HA _1}}{\left(\text { Acidic Strength } HA _2\right.}=\frac{\sqrt{3.0 \times 10^{-4}}}{\sqrt{1.8 \times 10^{-5}}}=4.08=$
$4$ (approx)
relative strengths of acids will be $4: 1$.
