Question

A test for complete removal of $C{u}^{2+}$(aq) is to add $N{H}_3$(aq). A blue colour signifies the formation of complex $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ having $K_f$ = 1.1$\times$ ${10}^{13}$ and thus confirms the presence of $C{u}^{2+}$ in solution. 250 mL of 0.1 M $CuS{O}_4$(aq) is electorlysed by passing a current of 3.512 ampere for 1368 second. After passage of this charge sufficient quantity of $N{H}_3$(aq) is added to electrolysed solution maintaining $\left[N{H}_3\right]$=0.10M. If $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ is detectable upto its concentration as low as 1$\times {10}^{-5}$, would a blue colour be shown by the electrolysed solution on addition of $N{H}_3$?

Answer 1

${Cu}^{+2}+4N{H}_3\rightleftharpoons {\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$
$\therefore k_f=\frac{{\left[Cu{\left(N{H}_2\right)}_4\right]}^{+2}}{\left[{Cu}^{+2}\right]{\left[N{H}_3\right]}^4}$
The blue color will be noticed upto ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}=1\times {10}^{-5}$
Thus, at this stage,
$\left[{Cu}^{+2}\right]=\frac{1.0\times {10}^{-5}}{1.1\times {10}^{13}\times {\left(0.1\right)}^4}=9.1\times {10}^{-15}M$
Also Cu deposited (w)$=\frac{E.i.t}{96500}$
$=\frac{63.5\times 3.512\times 1368}{2\times 96500}$
$=1.5807ℊ$
$\therefore \left[{Cu}^{+2}\right]$ lost $=\frac{1.581\times 1000}{63.5\times 250}=9.96\times {10}^{-2}M$
$\left[{Cu}^{+2}\right]$ present initially $=0.1$ or $10\times {10}^{-2}M$
$\therefore \left[{Cu}^{+2}\right]$ left $=10\times {10}^{-2}-9.96\times {10}^{-2}M$
$=4\times {10}^{-4}M$
Thus, solution will show blue color as it will provide appreciable ${Cu}^{+2}$ to form complex.