  Question

A test for the complete removal of Cu2+ ions from a solution of Cu2+ (aq.) is to add NH3 (aq.). A blue colour signifies the formation of complex [Cu(NH3)4]2+ that has Kf=1.1×1013 and thus confirms the presence of Cu2+ in the solution. 250 mL of 0.1 M CuSO4 (aq.) is electrolyzed by passing a current of 3.5 ampere for 1350 seconds. Then, sufficient quantity of NH3 (aq.) is added to the electrolyzed solution maintaining [NH3]=0.1 M. If  [Cu(NH3)4]2+ is detectable upto its concentration as low as 1×10−5. Pick the correct statement(s):

A
Amount of Cu2+deposited = 1.554 g  B
Amount of Cu2+ left in the solution = 2.07×103  C
Amount of Cu2+ left in the solution = 4×103  D
Solution will show blue colour  Solution

The correct options are A Amount of Cu2+deposited = 1.554 g B Amount of Cu2+ left in the solution = 2.07×10−3 D Solution will show blue colourCu2++4NH3⇌[Cu(NH3)4]2+                Kf=[Cu(NH3)4]2+[Cu2+][NH3]4 Blue colour will not be noticed upto [Cu(NH3)4]2+ = 1×10−5 Thus at this stage Cu2+=1×10−51.1×1013×(0.1)4=9.1×10−15 M Cu deposited (w) = E.i.t.96500=63.5×3.5×13502×96500=1.5546 g millimoles of Cu2+ present initially = 250×0.1 Weight of Cu2+ present initially  = 250×0.1×63.51000  = 1.5875 g  Weight of Cu2+ left in the solution = 1.5875 - 1.5546 = 0.0329 g  Cu2+ left = 0.0329×100063.5×250=2.07×10−3 Thus solution will show blue colour, as it will provide appreciable  Cu2+ to form complex.Co-Curriculars

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