Question

On passing 1 mole ${\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}$ gas through ${\mathbf{1L}}$ of ${\mathbf{0}}.{\mathbf{2M}}{\text{ }}{\mathbf{CuS}}{{\mathbf{O}}_{\mathbf{4}}}$ solution, determine $\left[ {{\mathbf{C}}{{\mathbf{u}}^{{\mathbf{2}} + }}} \right]$ in resulting solution. Given ${{\mathbf{K}}_{\mathbf{f}}}$ of ${\left[ {{\mathbf{Cu}}{\text{ }}{{\left( {{\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}} \right)}_{\mathbf{4}}}} \right]^{ + {\mathbf{2}}}} = {\mathbf{2}}*{\mathbf{1}}{{\mathbf{0}}^{{\mathbf{23}}}}$. Assume no hydrolysis of ${\mathbf{C}}{{\mathbf{u}}^{{\mathbf{2}} + }}$ ion.

Answer 1

In the given question we have to determine the copper ions in the solution when 1 mole of ammonia gas is passed. We can determine copper ions in the solution by the equation of freezing point depression equation.

Complete step-by-step answer:
Copper (II) ions react with stoichiometric quantities of aqueous ammonia and form precipitate and change in color occur to light blue. Some basic salts may also form. The precipitate dissolves in excess ammonia to form a dark blue complex ion, the reaction is as follows. The precipitate does not dissolve in excess unless the $NaOH$ solution is very concentrated. However, the precipitate will dissolve upon addition of concentrated ammonia solution.

As we can say that Freezing-point depression is the reduction of the freezing point of a particular solvent on the addition of a non-volatile solute.
${K_f}$ is the molal freezing point depression constant of the solvent where m = molality = moles of solute per kilogram of solvent.
$C{u^{ + 2}} + {\text{ }}4\left( {N{H_3}} \right) \to \left[ {Cu{\text{ }}{{\left( {N{H_3}} \right)}_4}} \right]$
${K_f} = \;\dfrac{x}{{\left[ {0.2 - x} \right]{{\left[ {0.1 - x} \right]}^4}}}\;$
$2 \times {10^{23}} = \;\dfrac{x}{{\left[ {0.2 - x} \right]{{\left[ {0.1 - x} \right]}^4}}}\;$
Let $x < < 1$
$0.2 - x \equiv 0.2$
$2 \times {10^{23}} = \dfrac{x}{{\left[ {0.2} \right]{{\left[ {0.1} \right]}^4}}}{\text{ }}$
$x = 2 \times {10^{23}} \times \dfrac{{0.2{\text{ }}}}{{10}} \times {10^{ - 4}}$
$x = 4 \times {10^{18}}$

Additional information: Freezing point depression equation. Divide the freezing point depression by the molal concentration so you have: ${K_f}{\text{ }} = {\text{ }}\Delta {T_f}{\text{ per }}cm$.

Note: The $+ 2$-oxidation state is more common than the$+ 1$. Copper $\left( {II} \right)$ is commonly found as the blue hydrated ion. The freezing point depression $\Delta T{\text{ }} = {\text{ }}{K_f} \times m$ where ${K_f}$ is the molal freezing point depression constant and m is the molality of the solute. This gives the moles of the solute. By Dividing the freezing point depression by the molal concentration so you have ${K_f}{\text{ }} = {\text{ }}\Delta {T_f}{\text{ per }}cm$. Insert the values for $\Delta {T_f}$ $cm$.