A solution containing 0.319 g of CrCl_(3).6 H_(2)O was passed through a cation exchange resin and acid coming out of the cation exchange resin required 28.5 ml of 0.125 M NaOH. Determine correct formula of the complex [Mol. Wt. of the complex = 266.5]

A solution containing 0.319 g of CrCl_(3).6 H_(2)O was passed through

1.	A solution containing 0.319 g of CrCl_(3).6 H_(2)O was passed through a cation exchange resin and acid coming out of the cation exchange resin required 28.5 ml of 0.125 M NaOH. Determine correct formula of the complex [Mol. Wt. of the complex = 266.5]
Answer» `[Cr(H_(2)O)_(6)]Cl_(3)` `[Cr(H_(2)O)_(5)CL]H_(2)O.Cl_(2)` `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O` `[Cr(H_(2)O)_(3)Cl_(3)].3 H_(2)O` Solution :Suppose the number of ionizable `Cl^(-)` ions (present outside the coordination sphere = n). When the solution of the COMPLEX passes through the cation exchanger, these `Cl^(-)` ions will combine with `H^(+)` ions of the cation exchanger to form HCl `nCl^(-)+nH^(+)RARR nHCl` Thus, 1 mole of the complex will form n moles of HCl. 1 mole of the complex`-=` n moles of HCl `-=` n moles of NaOH Moles of the comples `= (0.319)/(266.5)` =0.0012 mole Moles of NaOH used `= (28.5xx0.125)/(1000)` =0.0036 mole 0.0012 mole of the complex = 0.0036 mole NaOH `-=` 0.0036 mole HCl `therefore` 1 mole of the complex `= (0.0036)/(0.0012)=3` mole of HCl `-=` 3 moles of `Cl^(-)` ions Thus, all the `Cl^(-)` ions are outside the coordination sphere. Hence, the complex is `[Cr(H_(2)O)_(6)]Cl_(3)`.