Hydrofluoric acid is a weak acid- At 25-C- the molar conductivity of 0-002 M HF is 176-2 ohm-1 cm2 mol-1- If its -m- - 405-2 ohm-1 cm2 mol-1- calculate its degree of dissociation and equilibrium constant at the given concentration-

α = Λ_cΛ_m^0= 176.2/405.2=0.435 K= [H^+][F^ ][HF] = Cα^2 1 α = 0.002 × 0.435^21 0.435 = 6.70 × 10^ 4M ...

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Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

Hydrofluoric ...

Question

Hydrofluoric acid is a weak acid. At $25^{\circ} C$ , the molar conductivity of $0.002 M H F$ is $176.2 o h m^{- 1} c m^{2} m o l^{- 1}$ . If its $\land_{m}^{\infty} = 405.2 o h m^{- 1} c m^{2} m o l^{- 1}$ , calculate its degree of dissociation and equilibrium constant at the given concentration.

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Solution

$α = \frac{Λ_{c}}{Λ_{m}^{0}} = 176.2 / 405.2 = 0.435$
$K = \frac{[H^{+}] [F^{-}]}{[H F]} = \frac{C α^{2}}{1 - α} = \frac{0.002 \times {0.435}^{2}}{1 - 0.435} = 6.70 \times 10^{- 4} M$

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Hydrofluoric acid is a weak acid. At 25∘C, the molar conductivity of 0.002 M HF is 176.2 ohm−1cm2mol−1. If its ∧∞m=405.2 ohm−1cm2mol−1, calculate its degree of dissociation and equilibrium constant at the given concentration.

α=ΛcΛ0m=176.2/405.2=0.435K=[H+][F−][HF]=Cα21−α=0.002×0.43521−0.435=6.70×10−4M

Hydrofluoric acid is a weak acid. At $25^{\circ} C$ , the molar conductivity of $0.002 M H F$ is $176.2 o h m^{- 1} c m^{2} m o l^{- 1}$ . If its $\land_{m}^{\infty} = 405.2 o h m^{- 1} c m^{2} m o l^{- 1}$ , calculate its degree of dissociation and equilibrium constant at the given concentration.

$α = \frac{Λ_{c}}{Λ_{m}^{0}} = 176.2 / 405.2 = 0.435$
$K = \frac{[H^{+}] [F^{-}]}{[H F]} = \frac{C α^{2}}{1 - α} = \frac{0.002 \times {0.435}^{2}}{1 - 0.435} = 6.70 \times 10^{- 4} M$