Question

Calculate the degree of ionization of 0.1mol l of solution of acetic acid given ka for ch3cooh is 1.8*10-5

Answer 1

Ka=x2/(concentration of acetic acid-x)

1.8 X 10-5 mol/ l.=x^2/0.1-x

X=0.00134 = the degree of ionization

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Answer 2

The degree of ionization of acetic acid is 1.33 %

Explanation:

We are given:

Concentration of acetic acid, c = 0.1 M

The chemical equation for the ionization of acetic acid follows:

                       $CH_3COOH\rightarrow CH_3COO^-+H^+$

Initial:                   c      

At eqllm:          $c-c\alpha$                   $c\alpha$              $c\alpha$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

We are given:

$K_a=1.8\times 10^{-5}$

Putting values in above expression, we get:

$1.8\times 10^-5}=\frac{(0.1\alpha)\times (0.1\alpha)}{0.1(1-\alpha)}\\\\\alpha =0.0133$

Degree of ionization = $\alpha \times 100=(0.0133\times 100)=1.33\%$

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