Question

0.001 mol of $Co\left(N{H}_3{)}_5\right(N{O}_3\left)\right(S{O}_4)$ was passed through a cation exchanger and the acid coming out of it required 20 ml of 0.1M NaOH for neutralisation. Hence, the complex is:

• A. $\left[Co\right(N{H}_3{)}_{5}S{0}_4]N{O}_3$
• B. $\left[Co\right(N{H}_3{)}_{5}N{O}_3]S{O}_4$
• C. $\left[Co\right(N{H}_3{)}_5\left]\right(S{O}_3\left)\right(N{O}_3)$
• D. none of these

Answer 1

Answer: (B)

$\left[Co\right(N{H}_3{)}_{5}N{O}_3]S{O}_4$

2 moles of $NaOH$ is required to neutralise 1 mol of $H_{2}S{O}_4$.

Moles of $NaOH$ used = $\frac{20}{1000}\times 0.1$ = $0.002$ mol

Moles of $Co\left(N{H}_3{)}_5\right(N{O}_3\left)\right(S{O}_4)$ =$0.001$ mol

Thus $S{O}_4$ should be outside the complex ring.