Question

A solution containing 0.319 g of $CrC{l}_3.6H_{2}O$ was passed through a cation exchange resin in the acid form, and the acid liberated was titrated with a standard solution of NaOH. This required $28.5c{m}^3$ of 0.125 M NaOH. The correct formula of the Cr ( III ) complex is:

• A. $\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3$
• B. $\left[CrCl{\left(H_{2}O\right)}_5\right]C{l}_2.H_{2}O$
• C. $\left[CrC{l}_2{\left(H_{2}O\right)}_5\right]Cl.2H_{2}O$
• D. $[CrC{l}_3.{\left(H_{2}O\right)}_3].3H_{2}O$

Answer 1

The correct option is A $\left[Cr{\left(H_{2}O\right)}_6\right]C{l}_3$

No. of moles of given complex $=(0.319/248.5=0.0012)$. And no . of millimoles of NaOH used with acid $=0.125\times 28.5=3.6$ millimole. That means every mole of complex give three moles of acid so three chloride ions are outside complex sphere. So compound is - $\left[Cr\right(H_{2}O{)}_6]C{l}_3$.