Question

The conductivity of 0.001 M acetic acid is 5×10^-5 and^is 390.5 s cm2 mol then the calculated value of dissociation constant of acetic acid

Answer 1

Thus the value of dissociation constant is K = 1.88 x 10^-5

Explanation:

Given data:

• Molarity = 0.001 M
• Conductivity =  5×10^-5 and 390.5 S.cm^2 / mol
• ∧m = 1000 x K / C = 1000 x 5 x 10^-5 / 0.001 = 50 cm^2 / mol

Degree of dissociation = ∧m / ∧mo = 50 / 390.5

Degree of dissociation = 1.28 x 10^-1

CH3COOH ⇆ CH3COO- + H+

K = C^2 . α^2 / C ( 1 - α)   => Cα^2 / 1 - α

K = 10^-3 (1.28 x 10^-1 )^2 / 1 - 0.128

K = 1.88 x 10^-5

Thus the value of dissociation constant is K = 1.88 x 10^-5

Answer 2

Given :

Concentration of acetic acid , $c=0.001 \ M$ .

Conductivity , $k=5\times 10^{-5}\ S\ cm^{-1}$ .

$\Lambda^o_m=390.5\ S\ cm^2\ mol^{-1}$ .

To find :

Value of dissociation constant of acetic acid .

Solution :

We know ,

$\Lambda_m=\dfrac{k\times1000 }{C}\\\\\Lambda_m=\dfrac{5\times 10^{-5}\times1000 }{10^{-3}}\\\\\Lambda_m=50\ S\ cm^2\ mol^{-1}$

We know , degree of dissociation is :

$\alpha=\dfrac{\Lambda_m}{\Lambda^o_m}\\\\\alpha=\dfrac{50}{390.5}\\\\\alpha=0.128$

Now , dissociation constant of acetic acid is :

$K=\dfrac{C \alpha^2}{1-\alpha}\\\\K=\dfrac{0.001 \times 0.128^2}{1-0.128}\\\\K=18.8\times 10^{-6}$

Therefore , value of dissociation constant of acetic acid is $18.8\times 10^{-6}$ .

Learn More :

Conductivity

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