Question

A test for complete removal of Cu$^{2+}$ ions from a solution of Cu$^{2+}$ is to add NH$_3$(aq). A blue colour signifies the formation of complex [Cu(NH$_3$)$_4$]$^{2+}$ having K$_f$ = 1.1 × 10$^{13}$ and thus confirms the presence of Cu$^{2+}$ in solution. 250 ml of 0.1 M – CuSO$_4$ is electrolysed by passing a current of 5 A for 1351 s. After passage of this charge, sufficient quantity of NH$_3$ is added to electrolysed solution maintaining [NH$_3$] = 0.10 M. If [Cu(NH$_3$)$_4$]$^{2+}$ is detectable up to its concentration as low as 1 × 10$^{-5}$ M, would a blue colour be shown by the electrolysed solution on addition of NH$_3$. Mark '1', if the answer is 'yes' and mark '2', if the answer is 'no'.

• A. 1
• B. 2

Answer 1

Answer: (B)

2

Answer 2

The charge passed during electrolysis is sufficient to reduce all initial Cu$^{2+}$ ions to copper metal, leaving a residual [Cu$^{2+}$] of 0 M. With no free Cu$^{2+}$ ions, no complex $[Cu(NH_3)_4]^{2+}$ can form. Since the concentration of the complex is 0 M, it falls below the detection limit of $1 \times 10^{-5}$ M, and thus no blue colour is observed.