**, solution of is mixed with , solution of . The resulting solution is divided into two equal parts (X) and treated with excess solution and solution respectively as shown below:
Solution (X) + solution (excess)
Solution (X) + solution (excess)
The number of moles of Y and Z respectively are:**
,
,
,
,
Correct answer:C
Standard Method
Given: of solution of is mixed with of solution of . The final solution is divided into two equal parts.
Find: The moles of precipitates Y and Z formed with excess and excess respectively.
Each original solution contains
of the coordination compound.
After mixing, the total volume becomes . Since the mixture is divided into two equal parts, each part contains half of each solute. Therefore, one part contains mol of and mol of .
For the first half:
So, Y = mol.
For the second half:
So, Z = mol.
Thus, the solution concludes that the correct answer is moles for both Y and Z. This corresponds to option C as stated in the solution, although the listed options show under option B.
Assuming the full mixed solution is used in each test. This is wrong because the resulting solution is divided into two equal parts. Use only half of the total moles in each reaction.
Treating coordinated ions as free precipitating ions. This is wrong because only the counter ion outside the coordination sphere immediately gives precipitate with the reagent.
Ignoring which ion is outside the coordination sphere in each complex. In , is outside, while in , is outside. Identify the counter ion before predicting precipitate formation.
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