Question

A solution containing 0.319 g of $CrC{l}_3\cdot 6H_{2}O$ was passed through a certain exchange resin and acid coming of cation exchange resin required 28.5 mL of 0.125 M NaOH. The correct formula of complex is (mol. wt. of complex = 266.7)

• A. $[Cr(H_{2}O{)}_6]C{l}_3$
• B. $[Cr(H_{2}O{)}_{6}Cl]C{l}_2$
• C. $[Cr(H_{2}O{)}_{6}C{l}_2]Cl$
• D. $[Cr(H_{2}O{)}_{6}C{l}_3]$

Answer 1

Answer: (A)

Let the number of $C{l}^{-}$ ions outside the coordination sphere or number of chloride ions which can be ionised be n. When the solution of the complex is passed through cation exchanger, $nC{l}^{-}$ ions will combine with $H^{+}$ (of the cation exchanger) to form HCl.
$nC{l}^{-}+n{H}^{+}\to nHCl$
Thus, 1 mole of the complex will form n mole of HCl.
1 mol of complex $\equiv$ n mol HCl $\equiv$ n mol NaOH
mol of the complex $=\frac{0.319}{266.7}=0.0012$
mol of NaOH used $=\frac{28.5\times 0.125}{1000}=0.0036$ mol
$0.0012$ mol of complex $\equiv 0.0036$ mol NaOH
$\equiv 0.0036$ mol HCl
1 mol of complex $\equiv \frac{0.0036}{0.0012}=3$ mol HCl
$\therefore n=3$
Thus, all the $C{l}^{-}$ ions are outside the coordination sphere. Hence, complex is $\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3.$