The dissociation constants of two acids HA 1 and HA 2 with equal concentrations are 3 - 10-4 and 1-8 - 10-5 respectively- The relative strength of the acids will be -A- 1- 16B- 16- 1C- 1- 4D- 4- 1

The correct option is D 4:1Given, nbsp; K_a_1(HA_1)=3×10^ 4, K_a_2(HA_2)=1.8×10^ 13 nbsp; The formula to calculate the relative strength of a weak acid is ...

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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 $H A_{2}$ with equal concentrations are $3 \times 10^{- 4}$ and $1.8 \times 10^{- 5}$ respectively. The relative strength 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 $4 : 1$
Given, $K_{a_{1}} (H A_{1}) = 3 \times 10^{- 4}, K_{a_{2}} (H A_{2}) = 1.8 \times 10^{- 13}$
The formula to calculate the relative strength of a weak acid is :
$[H A] = \sqrt{K_{a} \times C}$
The ratio of stregnth of two given acids with equal concentrations is $\frac{[H A_{1}]}{[H A_{2}]} = \sqrt{\frac{K_{a_{1}} \times C}{K_{a_{2}} \times C}} \frac{[H A_{1}]}{[H A_{2}]} = \sqrt{\frac{K_{a_{1}}}{K_{a_{2}}}} = \sqrt{\frac{3 \times 10^{- 4}}{1.8 \times 10^{- 5}}} \frac{[H A_{1}]}{[H A_{2}]} = \sqrt{\frac{16.7}{1}} = \frac{4}{1}$
So , the ratio of strength of two acids are $[H A_{1}] : [H A_{2}] = 4 : 1$

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Introduction and Degree of Hydrolysis

Standard XII Chemistry

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The dissociation constants of two acids HA1 and HA2 with equal concentrations are 3×10−4 and 1.8×10−5 respectively. The relative strength of the acids will be :

The dissociation constants of two acids $H A_{1}$ and $H A_{2}$ with equal concentrations are $3 \times 10^{- 4}$ and $1.8 \times 10^{- 5}$ respectively. The relative strength of the acids will be :