Question

A test for the complete removal of $C{u}^{2+}$ ions from a solution of $C{u}^{2+}(aq.)$ is to add $N{H}_3(aq.)$. A blue colour signifies the formation of complex $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ that has $K_f=1.1\times {10}^{13}$ and thus confirms the presence of $C{u}^{2+}$ in the solution. 250 mL of 0.1 M $CuS{O}_4(aq.)$ is electrolyzed by passing a current of 3.5 ampere for 1350 seconds. Then, sufficient quantity of $N{H}_3(aq.)$ is added to the electrolyzed solution maintaining $\left[N{H}_3\right]=0.1M$. If $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ is detectable upto its concentration as low as $1\times {10}^{-5}$. Pick the correct statement(s):

• A. Amount of $C{u}^{2+}$deposited = 1.554 g
• B. Amount of $C{u}^{2+}$ left in the solution = $2.07\times {10}^{-3}$
• C. Amount of $C{u}^{2+}$ left in the solution = $4\times {10}^{-3}$
• D. Solution will show blue colour

Answer 1

Answer: (A), (B), (D)

The correct options are
A Amount of $C{u}^{2+}$deposited = 1.554 g
B Amount of $C{u}^{2+}$ left in the solution = $2.07\times {10}^{-3}$
D Solution will show blue colour

$C{u}^{2+}+4N{H}_3\rightleftharpoons \left[Cu\right(N{H}_3{)}_4{]}^{2+}$
$K_f=\frac{\left[Cu\right(N{H}_3{)}_4{]}^{2+}}{\left[C{u}^{2+}\right][N{H}_3{]}^4}$

Blue colour will not be noticed upto $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ = $1\times {10}^{-5}$

Thus at this stage
$C{u}^{2+}=\frac{1\times {10}^{-5}}{1.1\times {10}^{13}\times (0.1{)}^4}=9.1\times {10}^{-15}M$

$Cu$ deposited (w)
= $\frac{E.i.t.}{96500}=\frac{63.5\times 3.5\times 1350}{2\times 96500}=1.5546g$

millimoles of $C{u}^{2+}$ present initially
= $250\times 0.1$

Weight of $C{u}^{2+}$ present initially
= $\frac{250\times 0.1\times 63.5}{1000}$
= 1.5875 g

Weight of $C{u}^{2+}$ left in the solution
= 1.5875 - 1.5546 = 0.0329 g

$C{u}^{2+}$ left = $\frac{0.0329\times 1000}{63.5\times 250}=2.07\times {10}^{-3}$

Thus solution will show blue colour, as it will provide appreciable $C{u}^{2+}$ to form complex.