Question

Percent dissociation of 0.024m solution of weak acid ha (ka = 2 × 10-3) is

Answer 1

Percent dissociation of 0.024 m solution of weak acid HA is 25%

Explanation:

The dissociation of the weak acid is represented by:

$HA\rightarrow H^+A^-$

 c               0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given:

c= 0.024 M and $\alpha$ = ?

$K_a=2\times 10^{-3}$

Putting in the values we get:

$2\times 10^{-3}=\frac{(0.024\times \alpha)^2}{(0.024-0.024\times \alpha)}$

$(\alpha)=0.25$

$\%(\alpha)={0.25}\times {100}=25\%$

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