The equilibrium constant values for,HF⇌H++F−and HF2−⇌HF+F−Are respectively 7×10-4 mol lit-1 and 0.2 mol lit-1. The equilibrium constant values for2HF⇌H++HF2−and HF2−⇌H++2F−respectively are :

Here, we have; (p) ⇒RxnI−RxnIIK=7×10−42×10−1=3.5×10−3 (q) ⇒RxnI+RxnIIK=7×10−4×0.2=1.4×10−4

The equilibrium constant values for,HF⇌H++F−and HF2−⇌HF+F−Are respectively 7×10-4 mol lit-1 and 0.2 mol lit-1. The equilibrium constant values for2HF⇌H++HF2−and HF2−⇌H++2F−respectively are :

Here, we have; (p) ⇒RxnI−RxnIIK=7×10−42×10−1=3.5×10−3 (q) ⇒RxnI+RxnIIK=7×10−4×0.2=1.4×10−4

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The equilibrium constant values for,
$HF ⇌ H^{+} + F^{-} and {HF}_{2}^{-} ⇌ HF + F^{-}$
Are respectively $7 \times 10^{- 4} m o l l i t^{- 1}$ and $0.2 m o l l i t^{- 1}$ . The equilibrium constant values for
$2 HF ⇌ H^{+} + {HF}_{2}^{-} and$
${HF}_{2}^{-} ⇌ H^{+} + 2 F^{-} respectively are :$

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$3.5 \times 10^{- 3} and 1.4 \times 10^{- 4}$

$3.5 \times 10^{- 4} and 1.4 \times 10^{- 4}$

$3.5 \times 10^{- 4} and 1.4 \times 10^{- 3}$

$3.5 \times 10^{- 3} and 1.4 \times 10^{- 3}$

answer is D.

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Detailed Solution

Here, we have;

$(p) \Rightarrow R x^{n} I - R x^{n} II$

$K = \frac{7 \times 10^{- 4}}{2 \times 10^{- 1}} = 3.5 \times 10^{- 3}$

$(q) \Rightarrow R x^{n} I + R x^{n} II$

$K = 7 \times 10^{- 4} \times 0.2 = 1.4 \times 10^{- 4}$

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The equilibrium constant values for,HF⇌H++F−and HF2−⇌HF+F−Are respectively 7×10-4 mol lit-1 and 0.2 mol lit-1. The equilibrium constant values for2HF⇌H++HF2−and HF2−⇌H++2F−respectively are :

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