At 250C the ionic mobility of A-, H+ are respectively 2×10-4,1.8×10-3 cm/sec. The conductivity of 0.1M HA is 19.3×10-4 Scm-1.Dissociation constant of HA is

λM=κ×1000M=193×10-5×10000.1=19.3λM0=96500(λM(H+)0+λM(A-)0)λM0=96500(1.8×10-3+0.2×10-3)=193α = λMλM0=19.3193=0.1Ka=α2c=(0.1)2(0.1)=10-3

At 250C the ionic mobility of A-, H+ are respectively 2×10-4,1.8×10-3 cm/sec. The conductivity of 0.1M HA is 19.3×10-4 Scm-1.Dissociation constant of HA is

λM=κ×1000M=193×10-5×10000.1=19.3λM0=96500(λM(H+)0+λM(A-)0)λM0=96500(1.8×10-3+0.2×10-3)=193α = λMλM0=19.3193=0.1Ka=α2c=(0.1)2(0.1)=10-3

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At 25⁰C the ionic mobility of A^-, H⁺are respectively 2 $\times$ 10^-4,1.8 $\times$ 10^-3 cm/sec. The conductivity of 0.1M HA is
19.3 $\times$ 10^-4 Scm^-1.Dissociation constant of HA is

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10^-3

10^-2

10^-4

10^-5

answer is A.

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Detailed Solution

$λ_{M} = \frac{κ \times 1000}{M} = \frac{193 \times 10^{- 5} \times 1000}{0.1} = 19.3$

$λ_{M}^{0} = 96500 (λ_{M (H^{+})}^{0} + λ_{M (A^{-})}^{0})$

$λ_{M}^{0} = 96500 (1.8 \times 10^{- 3} + 0.2 \times 10^{- 3}) = 193$

$α = \frac{λ_{M}}{λ_{M}^{0}} = \frac{19.3}{193} = 0.1$

$K_{a} = α^{2} c = {(0.1)}^{2} (0.1) = 10^{- 3}$

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At 250C the ionic mobility of A-, H+ are respectively 2×10-4,1.8×10-3 cm/sec. The conductivity of 0.1M HA is 19.3×10-4 Scm-1.Dissociation constant of HA is

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At 25⁰C the ionic mobility of A^-, H⁺are respectively 2 $\times$ 10^-4,1.8 $\times$ 10^-3 cm/sec. The conductivity of 0.1M HA is
19.3 $\times$ 10^-4 Scm^-1.Dissociation constant of HA is