Question

A solution containing 0.319 g of $CrCl_3·6H_2O$ 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. $[Cr(H_2O)_6]Cl_3$
• B. $[Cr(H_2O)_6Cl]Cl_2$
• C. $[Cr(H_2O)_6Cl_2]Cl$
• D. $[Cr(H_2O)_6Cl_3]$

Answer 1

Answer: (A)

Let the number of $Cl^-$ 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, $nCl^-$ ions will combine with $H^+$ (of the cation exchanger) to form HCl.
$nCl^- + nH^+ \to nHCl$
Thus, 1 mole of the complex will form n mole of HCl.
1 mol of complex $≡$ n mol HCl $≡$ n mol NaOH
mol of the complex $= \frac{0.319}{266.7} = 0.0012$
mol of NaOH used $= \frac{28.5\times0.125}{1000} = 0.0036$ mol
$0.0012$ mol of complex $≡ 0.0036$ mol NaOH
$≡ 0.0036$ mol HCl
1 mol of complex $≡ \frac{0.0036}{0.0012} = 3$ mol HCl
$\therefore \quad n = 3$
Thus, all the $Cl^{-}$ ions are outside the coordination sphere. Hence, complex is $\left[Cr\left(H_{2}O\right)_{6}\right]Cl_{3}.$