Question

A solution containing ${\text{0}}{\text{.319}}\,{\text{g}}$${\text{CrC}}{{\text{l}}_{\text{3}}}{\text{.6}}{{\text{H}}_{\text{2}}}{\text{O}}$ was passed through a cation exchange resin and acid coming out of the cation exchange resin required $28.5\,{\text{mL}}$ of ${\text{0}}{\text{.125}}\,{\text{M}}\,\,{\text{NaOH}}$. Determine the correct formula of the complex. (mo. weight of the complex is $266.5$).
A. \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]{\text{C}}{{\text{l}}_{\text{3}}}\]
B. \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_5}\,{\text{Cl}}} \right]{{\text{H}}_{\text{2}}}{\text{O}}.\,{\text{C}}{{\text{l}}_2}\]
C. \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_4}\,{\text{C}}{{\text{l}}_2}} \right]{\text{Cl}}{\text{.2}}{{\text{H}}_{\text{2}}}{\text{O}}\,\]
D. \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_3}\,{\text{C}}{{\text{l}}_3}} \right]{\text{3}}{{\text{H}}_{\text{2}}}{\text{O}}\,\]

Answer 1

Hint: First we will determine the number of moles of each, complex and base. Then by comparing the mole ratio determine the number of chloride ions and so, the formula.

Complete step by step answer:
Use the mole formula to determine the number of moles of the complex as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Substitute ${\text{0}}{\text{.319}}\,{\text{g}}$ for mass and 266.5g/mol for molar mass.
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.319}}\,{\text{g}}}}{{266.5\,{\text{g/mol}}}}$
${\text{Mole = }}\,{\text{0}}{\text{.0012}}$
Use the molarity formula to determine the mole of the base as follows:
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{{\text{Mole of solute}}}}{{{\text{Volume of solution}}}}$
Convert the volume from mL to L.
${\text{1mL}}\, = 0.001\,{\text{L}}$
$28.5\,{\text{mL}}\,{\text{ = }}\,{\text{0}}{\text{.0285}}\,{\text{L}}$
Substitute ${\text{0}}{\text{.125}}\,{\text{M}}$ for molarity and${\text{0}}{\text{.0285}}\,{\text{L}}$ for volume of the solution.
${\text{0}}{\text{.125}}\,{\text{mol/L}}\,{\text{ = }}\,\dfrac{{{\text{Mole of solute}}}}{{{\text{0}}{\text{.0285}}\,{\text{L}}}}$
${\text{Mole of NaOH}}\, = \,{\text{0}}{\text{.125}}\,{\text{mol/L}}\,\, \times {\text{0}}{\text{.0285}}\,{\text{L}}$
${\text{Mole of NaOH}}\, = \,{\text{3}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}}$
The acid coming out from the resin will contain chloride ions that is reacting with sodium hydroxide as follows:
${\text{C}}{{\text{l}}^ - }\, + \,{\text{NaOH}}\, \to \,{\text{NaCl}}\,{\text{ + }}\,{\text{O}}{{\text{H}}^ - }\,$
For the reaction of one mole of acid one mole base is required.
Compare the moles of complex and base as follows:
${\text{0}}{\text{.0012}}$ mole of complex is reacting with ${\text{3}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}}$ of base.
So, one mole of complex will react,
$ = \,\dfrac{{{\text{3}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}}\,\,{\text{base}}}}{{0.0012\,{\text{mol}}\,\,{\text{complex}}}}$
$ = 3\,{\text{mol}}\,\,{\text{base}}$
Every mole of the complex is reacting with three moles of the base means every mole of the complex is producing three-mole acid and every mole of acid has one-mole chloride ion. It means a one-mole complex has three-mole chloride ions.
So, the formula of the complex is, \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]{\text{C}}{{\text{l}}_{\text{3}}}\] as the \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]{\text{C}}{{\text{l}}_{\text{3}}}\] dissociates as,
\[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]{\text{C}}{{\text{l}}_{\text{3}}} \to {\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]^{3 + }}\, + \,3\,{\text{C}}{{\text{l}}^ - }\]

Therefore, option (A) \[\left[ {{\text{Cr}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{6}}}} \right]\] is correct.

Note: How many moles of the base are reacting with acid depends upon the balanced reaction of acid and base. One mole of the base is reacting with one mole of acid and one mole of the complex is producing three moles of acid.so, we can say three moles of the base is reacting with one mole of the complex.