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Standard XII

Chemistry

Conductivity of Ionic Solutions

The equivalen...

Question

The equivalent conductance of an infinitely dilute solution $N H_{4} C l$ is $150$ and that of $C l^{-}$ and $O H^{-}$ is 76 and 198 respectively. The equivalent conductance of a $0.01 N$ solution $N H_{4} O H$ is $9.6$ . If its degree of dissociation of $N H_{4} O H$ is $353 \times 10^{- x}$ , then what is the value of $x$ ?

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Solution

$Δ_{N H_{4} C l}^{\circ} = λ_{N H_{4}^{+}}^{\circ} + λ_{C l^{-}}^{\circ}$
$∴ λ_{N H_{4}^{+}}^{\circ} = Δ_{N H_{4} C l}^{\circ} - λ_{C l^{-}}^{\circ} = 150 - 76 = 74$
$∴ Δ_{N H_{4} O H}^{\circ} = λ_{N H_{4}^{+}}^{\circ} + λ_{O H^{-}}^{\circ} = 74 + 198 = 272$
Further,
Degree of dissociation, $α = \frac{Δ_{N H_{4} O H}}{Δ_{N H_{4} O H}^{\infty}} = \frac{9.6}{272} = 0.0353 = 353 \times 10^{- 4}$

$∴ x = 4$

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The equivalent conductance of an infinitely dilute solution NH4Cl is 150 and that of Cl− and OH− is 76 and 198 respectively. The equivalent conductance of a 0.01N solution NH4OH is 9.6. If its degree of dissociation of NH4OH is 353×10−x, then what is the value of x?

Δ∘NH4Cl=λ∘NH+4+λ∘Cl−∴λ∘NH+4=Δ∘NH4Cl−λ∘Cl−=150−76=74∴Δ∘NH4OH=λ∘NH+4+λ∘OH−=74+198=272Further,Degree of dissociation, α=ΔNH4OHΔ∞NH4OH=9.6272=0.0353=353×10−4∴x=4