InterviewSolution
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InterviewSolution
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Starting with `250 ml` of `Au^(3+)` solution `250 ml` of `Fe^(2+)` solution, the following equilibrium is established `Au^(3+) (aq) +3Fe^(2+)(aq) hArr 3Fe^(3+) (aq) +Au(s)` At equilibrium the equivalents of `Au^(3+), Fe^(2+), Fe^(3+)` and `Au` are x,y,z and w respectively. Then:A. `K_(c) = (Z^(3))/(6xy^(3))`B. `K_(c) = (3Z^(3))/(2xy^(3))`C. `K_(c) = (4Z^(3))/(9xy^(3))`D. `K_(c) = (8Z^(3))/(9 xy^(3))` |
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Answer» Correct Answer - B n factor of `Au^(+3), Fe^(+3), Fe^(+3) & Au` are 3,1,1,& 3 respectively. Total volume `500 ml = 0.5 lit` `K_(c) =((Z)^(3)(0.5))/(((x)/(3))((y)/(1))^(3))` `K_(c) = (3Z^(3))/(2xy^(3))` |
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