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Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium
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Q)

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

(a) 1 : 4

(b) 4 : 1

(c) 1 : 16

(d) 16 : 1

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A)
Answer: 4 : 1
 
$\frac{\text{Strength of acid HA_1}}{\text{Strength of acid HA_2}} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{3.0 \times 10^{-4}}{1.8 \times 10^{-5}}} = 4 : 1$

 

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