At $20^{0} C$ 0.1 N formic acid dissociates to the extent of 4.5%. Calculate the dissociation constant of the acid.

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Solution

$C = 0.1 N$
$α = \frac{4.5}{100} = 0.045$
We know that, $K_{a} = C α^{2}$
$K_{a} = (0.1) (0.045)^{2} = 0.0002025 = 2.025 \times 10^{- 4}$

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At 200C 0.1 N formic acid dissociates to the extent of 4.5%. Calculate the dissociation constant of the acid.

C=0.1 N α=4.5100=0.045We know that, Ka=Cα2Ka=(0.1)(0.045)2=0.0002025=2.025×10−4

At $20^{0} C$ 0.1 N formic acid dissociates to the extent of 4.5%. Calculate the dissociation constant of the acid.

$C = 0.1 N$
$α = \frac{4.5}{100} = 0.045$
We know that, $K_{a} = C α^{2}$
$K_{a} = (0.1) (0.045)^{2} = 0.0002025 = 2.025 \times 10^{- 4}$