Question

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

Answer 1

Answer:

=$\frac{strength of acid HA_1}{strength of acid HA_2}$

=$\sqrt{\frac{K_{1} }{K_{2} } }$

=$\sqrt{\frac{3.0*10^{-4} }{1.8*10^{-5} } }$

=4:1

Answer 2

The relative strengths of the acids will be, 4 : 1

Explanation :

As we know that:

$\alpha=\sqrt{\frac{K_a}{C}}$

where,

$\alpha$ = degree of dissociation

$K_a$ = dissociation constant

C = concentration

As the concentration is same. So the expression will be:

$\frac{\alpha_1}{\alpha_2}=\sqrt{\frac{K_{a1}}{K_{a2}}}$

$\frac{\text{Strength of }HA_1}{\text{Strength of }HA_2}=\sqrt{\frac{K_{a1}}{K_{a2}}}$

$\frac{\text{Strength of }HA_1}{\text{Strength of }HA_2}=\sqrt{\frac{3.0\times 10^{-4}}{1.8\times 10^{-5}}}$

$\frac{\text{Strength of }HA_1}{\text{Strength of }HA_2}=\frac{4}{1}$

Thus, the relative strengths of the acids will be, 4 : 1

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