The limiting molar ionic conductivities of H^+ and A^- ions are 350 ~S~cm ^2 ~mol ⁻¹ and 50 ~S~cm ^2 ~mol ⁻¹ respectively. If the molar conductivity of a 0.01 ~M solution of the weak acid HA is 32 ~S~cm ^2 ~mol ⁻¹ , the percentage degree of

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The limiting molar ionic conductivities of H^+ and A^- ions are 350 ~S~cm ^2 ~mol ⁻¹ and 50 ~S~cm ^2 ~mol ⁻¹ respectively. If the molar conductivity of a 0.01 ~M solution of the weak acid HA is 32 ~S~cm ^2 ~mol ⁻¹ , the percentage degree of dissociation of the acid is :

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Correct answer: D. 8 %

First, calculate the limiting molar conductivity of the weak acid HA using Kohlrausch's law: ^ _ m (HA) = ^ _ m (H^+) + ^ _ m (A^-) ^ _ m (HA) = 350 + 50 = 400 ~S~cm ^2 ~mol ⁻¹ The degree of dissociation ( ) is the ratio of molar conductivity at a given concentration to the limiting molar conductivity: = _ m ^ _ m = 32 400 = 0.08 To find the percentage degree of dissociation, multiply by 100 : Percentage dissociation = 0.08 100 = 8 % Answer: 8 %

The limiting molar ionic conductivities of H^+ and A^- ions are 350 ~S~cm ^2 ~mol ⁻¹ and 50 ~S~cm ^2 ~mol ⁻¹ respectively. If the molar conductivity of a 0.01 ~M solution of the weak acid HA is 32 ~S~cm ^2 ~mol ⁻¹ , the percentage degree of dissociation of the acid is :

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