A) 8 times
B) 16 times
C) 64 times
D) 4 times
Correct Answer: B
Solution :
| For this reaction \[{{K}_{eq.}}\] is given by \[K=\frac{\left[ F{{e}^{3+}} \right]{{\left[ O{{H}^{-}} \right]}^{3}}}{\left[ Fe{{\left( OH \right)}_{3}} \right]}\] |
| \[=\left[ F{{e}^{3+}} \right]\,{{\left[ O{{H}^{-}} \right]}^{3}}\] \[\left[ \because \,\,\left[ soild \right]=1 \right]\] |
| If \[\left[ O{{H}^{-}} \right]\] is decreased by \[\frac{1}{4}\] times then for reaction equilibrium constant to remain constant, we have to increase the concentration of \[\left[ F{{e}^{3+}} \right]\] by a factor of \[{{4}^{3}}\] i.e., \[4\times 4\times 4=64.\] |
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