A test for complete removal of `Cu^(2+)` ions from a solution of `Cu^(2+)` (aq) is to add `NH_3`(aq). A blue colour signifies the formation of complex `[Cu(NH_3)_4]^(2+)` having `K_f = 1.1 xx 10^13` and thus confirms the presence of `Cu^(2+)` in solution. 250 mL of 0.1 M `CuSO_4` (aq.) is electrolysed by passing a current of 3.512 ampere for 1368 second. After passage of this charge sufficient quantity of `NH_3`(aq) is added to electrolysed solution maintianing `[NH_3]` = 0.10 M. If `[Cu(NH_3)_4]^(2+)` is detectable upto its concentration as low as `1 xx 10^(–5)`, would a blue colour be shown by the electrolysed solution on addition of `NH_3` ?

Answer

To solve the problem step by step, we need to determine if the concentration of `Cu^(2+)` ions remaining in the solution after electrolysis is sufficient to form the complex `[Cu(NH_3)_4]^(2+)` and thus produce a blue color when `NH_3` is added.

### Step 1: Calculate the total charge passed during electrolysis

The total charge \( Q \) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I = 3.512 \, \text{A} \) (current)
- \( t = 1368 \, \text{s} \) (time)

Calculating \( Q \):
\[ Q = 3.512 \, \text{A} \times 1368 \, \text{s} = 4,804.416 \, \text{C} \]

### Step 2: Calculate the amount of copper deposited

The equivalent weight of copper (Cu) is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass of Cu}}{n} = \frac{63.5 \, \text{g/mol}}{2} = 31.75 \, \text{g/equiv} \]
where \( n = 2 \) because each \( Cu^{2+} \) ion requires 2 electrons for reduction.

Now, using Faraday's law:
\[ \text{Mass of Cu deposited} = \frac{Q}{F} \times \text{Equivalent weight} \]
where \( F = 96500 \, \text{C/mol} \) (Faraday's constant).

Calculating the mass of copper deposited:
\[ \text{Mass of Cu} = \frac{4804.416 \, \text{C}}{96500 \, \text{C/mol}} \times 31.75 \, \text{g/equiv} \approx 1.5807 \, \text{g} \]

### Step 3: Convert the mass of copper deposited to moles

To find the number of moles of copper deposited:
\[ \text{Moles of Cu} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.5807 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.0249 \, \text{mol} \]

### Step 4: Calculate the initial moles of `Cu^(2+)` in the solution

The initial concentration of `CuSO_4` is \( 0.1 \, \text{M} \) in a volume of \( 250 \, \text{mL} \):
\[ \text{Initial moles of } Cu^{2+} = 0.1 \, \text{mol/L} \times 0.250 \, \text{L} = 0.025 \, \text{mol} \]

### Step 5: Calculate the remaining moles of `Cu^(2+)` after electrolysis

The remaining moles of `Cu^(2+)` after electrolysis:
\[ \text{Remaining moles of } Cu^{2+} = \text{Initial moles} - \text{Moles deposited} = 0.025 \, \text{mol} - 0.0249 \, \text{mol} \approx 0.0001 \, \text{mol} \]

### Step 6: Calculate the concentration of `Cu^(2+)` remaining in the solution

The concentration of `Cu^{2+}` after electrolysis in the total volume of \( 250 \, \text{mL} \):
\[ \text{Concentration of } Cu^{2+} = \frac{0.0001 \, \text{mol}}{0.250 \, \text{L}} = 0.0004 \, \text{M} = 4 \times 10^{-4} \, \text{M} \]

### Step 7: Compare with the detection limit

The detection limit for the complex formation is \( 1 \times 10^{-5} \, \text{M} \). Since \( 4 \times 10^{-4} \, \text{M} \) is greater than \( 1 \times 10^{-5} \, \text{M} \), the solution will show a blue color upon addition of `NH_3`.

### Conclusion

Yes, a blue color will be shown by the electrolyzed solution on the addition of `NH_3`.

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