The dissociation constants of two acids H A 1 and H A 2 are 3-0- 10 -4 and 1-8- 10 -5 respectively- The relative strengths of the acids will be-

The correct option is B 4:1 The dissociation constants of HA_1 and HA_2 are 3×10^ 4 #160;and 1.8×10^ 5. The strength of an acid i ...

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Standard XII

Chemistry

Ostwald Dilution Law

The dissociat...

Question

The dissociation constants of two acids $H A_{1}$ and $H A_{2}$ are $3.0 \times 10^{- 4}$ and $1.8 \times 10^{- 5}$ respectively. The relative strengths of the acids will be:

1 : 4

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4 : 1

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1 : 16

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16 : 1

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Solution

The correct option is B $4 : 1$
The dissociation constants of $H A_{1}$ and $H A_{2}$ 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:
$\frac{(A c i d i c S t r e n g t h)_{H A_{1}}}{(A c i d i c S t r e n g t h)_{H A_{2}}} = \frac{\sqrt{(D i s s o c i a t i o n C o n s t a n t)_{H A_{1}}}}{\sqrt{(D i s s o c i a t i o n C o n s t a n t)_{H A_{2}}}}$

Relative Acidic Strength= $\frac{(A c i d i c S t r e n g t h)_{H A_{1}}}{(A c i d i c S t r e n g t h)_{H A_{2}}} = \frac{\sqrt{3.0 \times 10^{- 4}}}{\sqrt{1.8 \times 10^{- 5}}} = 4.08 = 4 (a p p r o x)$

relative strengths of acids will be $4 : 1.$

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The dissociation constants of two acids HA1 and HA2 are 3.0×10−4 and 1.8×10−5 respectively. The relative strengths of the acids will be:

The dissociation constants of two acids $H A_{1}$ and $H A_{2}$ are $3.0 \times 10^{- 4}$ and $1.8 \times 10^{- 5}$ respectively. The relative strengths of the acids will be: