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For the equilibrium system,
$2 HX (g) ⇌ H_{2} (g) + X_{2} (g)$
the equilibrium constant is $1.0 \times 10^{- 5}$ .What will be the concentration of HX if the equilibrium concentrations of $H_{2} and X_{2}$ are $1.2 \times 10^{- 3} M$ and $1.2 \times 10^{- 4} M$ respectively?

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$1.2 \times 10^{- 4} M$

$1.2 \times 10^{- 3} M$

$1.2 \times 10^{- 2} M$

$1.2 \times 10^{- 1} M$

answer is D.

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Detailed Solution

$K = \frac{[H_{2}] [X_{2}]}{[HX]^{2}}$

$10^{- 5} = \frac{(1.2 \times 10^{- 3}) (1.2 \times 10^{- 4})}{[HX]^{2}}$

or, $[HX]^{2} = (1.2)^{2} \times 10^{- 2}$

or, $[HX] = 1.2 \times 10^{- 1} M$

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