Question

A 0.15 m solution of a weak acid is 3.0 dissociated. calculate ka

Answer 1

Answer: The value of equilibrium constant is $1.39\times 10^{-4}$

Explanation:

We are given:

Percent degree of dissociation = 3.0 %

Degree of dissociation, $\alpha$ = 0.03

Concentration of weak acid ([HA]), c = 0.15 M

The chemical equation for the dissociation of weak acid follows:

                        $HA\rightleftharpoons H^++A^-$

Initial:                c          -

At Eqllm:       $c-c\alpha$      $c\alpha$   $c\alpha$

So, equilibrium concentration of HA = $c-c\alpha=[0.15-(0.15\times 0.03)]=0.1455M$

Equilibrium concentration of $[H^+]=c\alpha =[0.15\times 0.03]=0.0045M$

Equilibrium concentration of $[A^-]=c\alpha =[0.15\times 0.03]=0.0045M$

The expression of $K_{a}$ for above equation follows:

$K_{a}=\frac{[H^+][A^-]}{[HA]}$

Putting values in above equation, we get:

$K_{a}=\frac{0.0045\times 0.0045}{0.1455}\\\\K_{a}=1.39\times 10^{-4}$

Hence, the value of equilibrium constant is $1.39\times 10^{-4}$

Answer 2

If x is H+ then x/0.15 = 0.03 and x = 0.0045 M  and

Ka = x^2/(0.15 - x) = 0.0045^2/(0.15 - 0.0045) = 1.39 x 10^-4