Question
For the equilibrium system $$2HX\left( g \right) \rightleftharpoons {H_2}\left( g \right) + {X_2}\left( g \right)$$ the equilibrium constant is $$1.0 \times {10^{ - 5}}.$$ What is the concentration of $$HX$$ if the equilibrium concentration of $${H_2}$$ and $${X_2}$$ are $$1.2 \times {10^{ - 3}}M,$$ and $$1.2 \times {10^{ - 4}}M$$ respectively.
A.
$$12 \times {10^{ - 4}}M$$
B.
$$12 \times {10^{ - 3}}M$$
C.
$$12 \times {10^{ - 2}}M$$
D.
$$12 \times {10^{ - 1}}M$$
Answer :
$$12 \times {10^{ - 2}}M$$
Solution :
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2HX\left( g \right) \rightleftharpoons \,\,\,\,{H_2}\left( g \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,{X_2}\left( g \right) \cr
& {\text{At eqm}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{1}}{\text{.2}} \times {\text{1}}{{\text{0}}^{ - 3}}M\,\,\,\,\,\,\,\,\,1.2 \times {10^{ - 4}}M \cr
& {K_{eq}} = \frac{{\left[ {{H_2}} \right]\left[ {{X_2}} \right]}}{{{{\left[ {HX} \right]}^2}}} \cr
& {10^{ - 5}} = \frac{{1.2 \times {{10}^{ - 3}} \times 1.2 \times {{10}^{ - 4}}}}{{{{\left[ {HX} \right]}^2}}} \cr
& \left[ {HX} \right] = \sqrt {\frac{{1.2 \times 1.2 \times {{10}^{ - 7}}}}{{{{10}^{ - 5}}}}} \cr
& = 1.2 \times {10^{ - 1}} \cr
& = 12 \times {10^{ - 2}}M \cr} $$