Chemistry, asked by vedasmiritha, 1 year ago

The conductivity of a 0.01M solution of acetic acid at 298K is 1.65×10-4S/cm.calculate the molar conductivity of the solution , degree of dissociation of acetic acid , dissociation constant of acetic acid.

Answers

Answered by Unknown135
116
hey your question is not complete you should give the molar conductance at infinite dilution for values of disassociation constant and degree of disassociation but I knew the question and sent you the answer with the unknown value
so hope it helps you
Attachments:
Answered by AnirudhaM5
4

Explanation:

The specific conductance of a 0.01M solution of acetic acid at 298K is 1.65×10

−4

ohm

−1

cm

−1

. The molar conductance at infinite dilution for H

+

ion and CH

3

COO

ion are 349.1ohm

−1

cm

2

mol

ANSWER

(i) Molar conductance Λ

m

=

C

K×1000

=

0.01

1.65×10

−4

×1000

=16.5 ohm

−1

cm

2

mol

−1

(ii) Degree of dissociation α=

Λ

m

Λ

m

Λ

m

=16.5ohm

−1

cm

2

mol

−1

Λ

m(CH

3

COOH)

[H

+

]+Λ

[CH

3

COOH]

=349.1+40.9=390ohm

−1

cm

2

mol

−1

α=

390

16.5

=0.0423

(iii)

0.01M0.01(1−α)

CH

3

COOH

00.01α

CH

3

COO

+

00.01α

H

+

Dissociation constant K

d

=

[CH

3

COO

]

[CH

3

COO

][H

+

]

=

0.01(1−α)

0.01α×0.01α

=

1−α

0.01α

2

=0.01α

2

[Taking 1−α=1, as α is negligible as compared to 1]

=0.01×(0.0423)

2

K

d

=1.86×10

−5

Hence, the dissociation constant (K

d

) for acetic acid is 1.86×10

−5

.

t

Try solving tr the question according to this.

HOPE IT HELPS

−1

and 40.9ohm

−1

cm

2

mol

−1

respectively.

Calculate:

(i) Molar conductance of the solution.

(ii) Degree of dissociation of CH

3

COOH.

(iii) Dissociation constant for acetic acid.

medium

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