Question

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

• A. $\left[Co\right(N{H}_3{)}_5\left(N{O}_3\right)]S{O}_4$
• B. $\left[Co\right(N{H}_3{)}_5\left(S{O}_4\right)]N{O}_3$
• C. $\left[Co\right(N{H}_3{)}_5]N{O}_3\cdot S{O}_4$
• D. none of the above

Answer 1

Answer: (A)

$\left[Co\right(N{H}_3{)}_5\left(N{O}_3\right)]S{O}_4$

Number of moles of $NaOH$ $=0.1mol\times \frac{20mL}{1,000mL/L}=0.002$ moles.
0.002 moles of $NaOH$ combines with 0.001 moles of the complex.

The basicity of complex is $\frac{0.002}{0.001}=2$.

It is a dibasic complex.

Hence, its structure is $\left[Co\right(N{H}_3{)}_5\left(N{O}_3\right)]S{O}_4$.