Molar conductivity of a monobasic acid HA having molar concentration equal to C is - and the limiting molar conductivity is - The dissociation constant of the acid is

∴ K_a=c^2 α^2c(1 α)=cα^2(1 α) ∴ K_a=C(λ/λ_∞)^21 λ/λ_∞=Cλ^2λ_∞(λ_∞ λ) ...

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Byju's Answer

Standard XII

Chemistry

Resistance and Resistivity

Molar conduct...

Question

Molar conductivity of a monobasic acid HA having molar concentration equal to $C$ is $λ$ and the limiting molar conductivity is $λ_{\infty}$ . The dissociation constant of the acid is

\frac{C λ}{λ_{\infty} - λ}

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\frac{C λ^{2}}{(λ_{\infty} + λ)}

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\frac{C λ^{2}}{λ_{\infty}^{2} - λ^{2}}

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\frac{C λ^{2}}{λ_{\infty} (λ_{\infty} - λ)}

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Solution

The correct option is D $\frac{C λ^{2}}{λ_{\infty} (λ_{\infty} - λ)}$

$∴ K_{a} = \frac{c^{2} α^{2}}{c (1 - α)} = \frac{c α^{2}}{(1 - α)}$
$∴ K_{a} = \frac{C (λ / λ_{\infty})^{2}}{1 - λ / λ_{\infty}} = \frac{C λ^{2}}{λ_{\infty} (λ_{\infty} - λ)}$

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Molar conductivity of a monobasic acid HA having molar concentration equal to C is λ and the limiting molar conductivity is λ∞. The dissociation constant of the acid is

The correct option is D Cλ2λ∞(λ∞−λ) ∴Ka=c2α2c(1−α)=cα2(1−α) ∴Ka=C(λ/λ∞)21−λ/λ∞=Cλ2λ∞(λ∞−λ)

Molar conductivity of a monobasic acid HA having molar concentration equal to $C$ is $λ$ and the limiting molar conductivity is $λ_{\infty}$ . The dissociation constant of the acid is

The correct option is D $\frac{C λ^{2}}{λ_{\infty} (λ_{\infty} - λ)}$

$∴ K_{a} = \frac{c^{2} α^{2}}{c (1 - α)} = \frac{c α^{2}}{(1 - α)}$
$∴ K_{a} = \frac{C (λ / λ_{\infty})^{2}}{1 - λ / λ_{\infty}} = \frac{C λ^{2}}{λ_{\infty} (λ_{\infty} - λ)}$