Question

A solution containing $0.319g$ of complex $CrC{l}_3.6H_{2}O$ was passed through cation exchanger and the solution given out was neutralised by $28.5mL$ of $0.125MNaOH$ . What is the correct formula of complex ?

Answer 1

The $Cl$ atoms outside the co - ordination sphere will be ionised to produce acid $HCl$

Thus , Meq. of $C{l}^{-}$ ions outside = Meq. of $HCl$ formed

= Meq. of $NaOH$ formed

= $28.5\times 0.125$

= $3.56$

$\frac{0.319}{266.5}$ mole or $1.197m$ mole of complex produce $3.56$ Meq. or millimoles small of $C{l}^{-}$ .

Thus $1$ mole of complex will give $3$ mole of $C{l}^{-}$ i.e., all the three Cl atoms are outside the co -ordination sphere .

Thus ,complex is $\left[Cr\right(H_{2}O{)}_6]C{l}_3$