A solution containing `0.319` of complex `CrCl_3. 6 H_2 O` was passed through cation exchange and the solution given out was neutralised by `2.85` mL of `0.125 M` NaOH. What is the correct formula of the complex.

Options

A. `[Cr(H_2O)_6]Cl_3`
B. `[Cr(H_2O)_5]H_2O. Cl_2`
C. `[Cr(H_2O)_4Cl_2]Cl. 2H_2O`
D. `[Cr(H_2O)_3 Cl_3 ] . 3H_2O`

Answer

A

Explanation

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 HCI
`n Cl^(-) + nH^(+) to 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 complex ` = 0.319/(266.5) = 0.0012 ` mole
Moles of NaOH used `= (28.5 xx 0.125)/1000 = 0.0036` mole
`0.0012` mole of the complex =0.0036 mole NaOH =0.0036 mole HCl
`:. `1 mole of the complex `= 0.0036/0.0012 = 3` moles of HCl = 3 moles of `Cl^(-)` ions.
Thus, all the `Cl^(-)` ions are outside the coordination sphere. Hence, the complex is `[Cr(H_2O)_6]Cl_3`