The dissociation constant of two acids HA 1 and HA 2 are 3-14 - 10-4 and 1-96 - 10-5 respectively- The relative strength of the acids will be approximatelyA- 1- 4B- 4- 1C- 1- 16D- 16- 1

The correct option is B 4:1 α_1/α2 = √(K_a1/K_a2) = √(3.14 × 10^ 4/1.6 × 10^ 5) = 4 : 1 ...

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Standard XII

Chemistry

Ostwald Dilution Law Mathematical Interpretation

The dissociat...

Question

The dissociation constant of two acids $H A_{1}$ and $H A_{2}$ are $3.14 \times 10^{- 4}$ and $1.96 \times 10^{- 5}$ respectively. The relative strength of the acids will be approximately

1:4

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1:16

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Solution

The correct option is B
4:1

$\frac{α_{1}}{α 2} = \sqrt{\frac{K_{a 1}}{K_{a 2}}} = \sqrt{\frac{3.14 \times 10^{- 4}}{1.6 \times 10^{- 5}}}$ = $4 : 1$

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The dissociation constant of two acids HA1 and HA2 are 3.14 × 10−4 and 1.96 × 10−5 respectively. The relative strength of the acids will be approximately

The correct option is B 4:1 α1α2 = √Ka1Ka2 = √3.14 × 10−41.6 × 10−5 = 4 : 1

The dissociation constant of two acids $H A_{1}$ and $H A_{2}$ are $3.14 \times 10^{- 4}$ and $1.96 \times 10^{- 5}$ respectively. The relative strength of the acids will be approximately

The correct option is B
4:1

$\frac{α_{1}}{α 2} = \sqrt{\frac{K_{a 1}}{K_{a 2}}} = \sqrt{\frac{3.14 \times 10^{- 4}}{1.6 \times 10^{- 5}}}$ = $4 : 1$