For the equilibrium system $2 \mathrm{HX}(\mathrm{g}) ightleftharpoons…
the equilibrium constant is $1.0 \times 10^{-5} .$ What is the concentration of HX if the equilibrium concentration of $\mathrm{H}_{2}$ and $\mathrm{X}_{2}$ are $1.2 \times 10^{-3} \mathrm{M}$, and $1.2 \times 10^{-4} \mathrm{M}$ respectively.
- $12 \times 10^{-4} \mathrm{M}$
- $12 \times 10^{-3} \mathrm{M}$
- $12 \times 10^{-2} \mathrm{M}$
- $12 \times 10^{-1} \mathrm{M}$
Solution

At eqm.
$\mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{H}_{2}ight]\left[\mathrm{X}_{2}ight]}{[\mathrm{HX}]^{2}}$
$10^{-5}=\frac{1.2 \times 10^{-3} \times 1.2 \times 10^{-4}}{[\mathrm{H} \mathrm{X}]^{2}}$
$[\mathrm{HX}]=\sqrt{\frac{1.2 \times 1.2 \times 10^{-7}}{10^{-5}}}$
$=1.2 \times 10^{-1}$
$=12 \times 10^{-2} \mathrm{M}$ ~
Asked in: JEE-TOPICTESTS-CHEMISTRY