Question

1. Using the${\mathit{K}}_{\mathbf{s}\mathbf{p}}$for $\mathit{C}\mathit{u}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{)}$and the overall formation constant for $\mathit{C}\mathit{u}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{)}$, calculate a value for the equilibrium constant for the reaction

$\mathit{C}\mathit{u}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{+}\mathbf{4}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{C}\mathit{u}{{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

1. Use the value of the equilibrium constant you calculated in part a to calculate the solubility of$\mathit{C}\mathit{u}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$in $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{M}\mathbf{}\mathit{N}{\mathit{H}}_{\mathbf{3}}$. In 5.0 M $\mathit{N}{\mathit{H}}_{\mathbf{3}}$, the concentration of $\mathit{O}{\mathit{H}}^{\mathbf{-}}$is 0.0095 M.

Short answer

Answer

1. The value for the equilibrium constant for the reaction is $\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}$.

2. The value of solubility is 0.056 mol/L.

Step-by-step solution

Definition of the equilibrium constant.

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature is defined as the equilibrium constant.

Subpart (a) The equilibrium constant.

$$\begin{gathered} \mathit{K}\mathbf{=}\frac{\left[Cu\left(N{H}_4^{2+}\right)\right]{\left[O{H}^{-}\right]}^{\mathbf{2}}}{\left[N{H}_3\right]} \\ \mathit{K}\mathbf{=}{\mathit{K}}_{\mathbf{s}\mathbf{p}}\mathbf{\times }{\mathit{K}}_{\mathbf{f}} \\ \mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}} \\ \mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}} \end{gathered}$$

Subpart (a) The equilibrium

The reaction is

$\mathit{C}\mathit{u}{\left(OH\right)}_{\mathbf{2}}\left(s\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathit{N}{\mathit{H}}_{\mathbf{3}}\left(aq\right)\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{C}\mathit{u}{\left(N{H}_3\right)}_{\mathbf{4}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}\left(aq\right)$

The ICE table for the ions in the reaction is,

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{4}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{C}\mathit{u}{\left(N{H}_3\right)}_{\mathbf{4}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}\left(aq\right) \\ \mathit{i}\mathit{n}\mathit{t}\mathit{i}\mathit{a}\mathit{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0095} \\ \mathit{c}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{4}\mathit{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathit{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{2}\mathit{s} \\ \mathit{e}\mathit{q}\mathit{u}\mathit{i}{\mathit{l}}^{\mathbf{m}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{4}\mathit{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(0.0095+2s\right) \end{gathered}$$

Substituting the values,

\$\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{=}\frac{\mathbf{s}{\left(0.0095+2s\right)}^{\mathbf{2}}}{{\left(5.0-4s\right)}^{\mathbf{4}}}$

The value of s is 0.056 mol/L.