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 approximately

The correct option is B 4 : 1Relative strength = √( K _ a _ 1 / K _ a _ 2 #160;) = √( #160; 3× 10 ^ 4 / 1.8× 10 ^ 5 #160;) ⇒ 4 : 1 Hence, t ...

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Standard XII

Chemistry

Introduction and Degree of Hydrolysis

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 approximately

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
Relative strength $= \sqrt{\frac{K_{a_{1}}}{K_{a_{2}}}} = \sqrt{\frac{3 \times 10^{- 4}}{1.8 \times 10^{- 5}}}$
$\Rightarrow 4 : 1$

Hence, the correct option is $B$

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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 approximately

The correct option is B 4 : 1Relative strength =√Ka1Ka2=√3×10−41.8×10−5⇒4:1Hence, the correct option is B

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 approximately

The correct option is B 4 : 1
Relative strength $= \sqrt{\frac{K_{a_{1}}}{K_{a_{2}}}} = \sqrt{\frac{3 \times 10^{- 4}}{1.8 \times 10^{- 5}}}$
$\Rightarrow 4 : 1$

Hence, the correct option is $B$