Question

molar conductance of a 0.2 m solution of a weak acid, ha is 2.8×10 −2 sm 2 mol −1 . if the limiting molar conductance of ha is 560 s m 2 mol −1 , then the acid is

Answer 1

Answer:

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Explanation:

Degree of dissociation= α=

Limiting molar conductvity

Molar conductance of 0.2Msol

n

α=

560

2.8×10

−2

=5×10

−5

⇒HA⇌H

+

+A

−

C(1−α) Cα Cα

K

a

=

C(1−α)

C

2

α

2

⇒K

a

1−α

Cα

2

∵α<<1⇒K

a

=Cα

2

∴K

a

=0.2×(5×10

−5

)

2

=0.2×25×10

−10

=5×10

−10

Answer 2

Answer:

Degree of dissociation= α=

Limiting molar conductvity

Molar conductance of 0.2Msol

n

α=

560

2.8×10

−2

=5×10

−5

⇒HA⇌H

+

+A

−

C(1−α) Cα Cα

K

a

=

C(1−α)

C

2

α

2

⇒K

a

1−α

Cα

2

∵α<<1⇒K

a

=Cα

2

∴K

a

=0.2×(5×10

−5

)

2

=0.2×25×10

−10

=5×10

−10