InterviewSolution
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InterviewSolution
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If the concentration of `OH^(-)` ions in the reaction `Fe(OH)_(3)(s)hArrFe^(3+)(aq.)+3OH^(-)(aq.)` is decreased by `1//4` times, then the equilibrium concentration of `Fe^(3+)` will increase byA. 8 timesB. 16 timesC. 64 timesD. 4 times |
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Answer» Correct Answer - C `Fe(OH)_3(s)hArrFe^(3+)(aq)+3OH^(-)(aq)` `K=([Fe^(3+)][OH^(-)]^(3)]/([Fe(OH)_(3)])` To maintain equilibrium constant, let the concentration of `Fe^(3+)` is increased x times, on decreasing the concentration of `OH^(-)` by `(1)/(4)` times `K=([xFe^(3+)][OH^(-)]^(3))/([Fe(OH)_(3)])`.....(ii) By dividing eq. (ii) by (i) we get `(1)/(64)xxX=1rArrX=64` times. |
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