Question

A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of $\mathbf{\text{2.0}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{\hspace{0.17em}\hspace{0.17em}CuNO}}}_{\mathbf{3}}$. Assume that Cu(I) forms complex ions with ${\mathbf{\text{X}}}^{\mathbf{-}}$as follows:

role="math" localid="1658120373955" $\begin{array}{l}{\mathbf{Cu}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{X}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{CuX}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\\ \mathbf{CuX}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{X}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftharpoons }{{\mathbf{CuX}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\\ {{\mathbf{CuX}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{X}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftharpoons }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{}{\mathbf{K}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\end{array}$

Calculate the following concentrations at equilibrium.
a.${\mathbf{\text{CuX}}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}$
b.${\mathbf{\text{CuX}}}_{\mathbf{2}}^{\mathbf{-}}$
c.${\mathbf{\text{Cu}}}^{\mathbf{+}}$

Short answer

Theconcentration at equilibrium is

1. $\left[{{\mathrm{CuX}}_3}^{2-}\right]=0.001\mathrm{M}$
   
2. $\left[{{\mathrm{CuX}}_2}^{-}\right]=2\times {10}^{-7}\mathrm{M}$
   
3. $\left[{\mathrm{Cu}}^{+}\right]=8\times {10}^{-15}$
   

Step-by-step solution

Subpart (a) The concentration of CuX32−

The initial concentration can be denoted as,

$\begin{array}{c}{\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2\\ {\mathrm{X}}^{-}=\frac{10\mathrm{mol}}{100\mathrm{mL}}\times 50\mathrm{mL}\\ =5\mathrm{M}\end{array}$

So,

${\mathrm{Cu}}^{2+}=\frac{0.002\mathrm{mol}}{100\mathrm{mL}}\times 50\mathrm{mL}=0.001\mathrm{M}$

The final reaction can be denoted as,

$\begin{array}{c}{\mathrm{K}}_{\mathrm{overall}}={\mathrm{K}}_1{\mathrm{K}}_2{\mathrm{K}}_3\\ =100\times 10000\times 1000\\ {\mathrm{K}}_{\mathrm{overall}}=1\times {10}^9\end{array}$

Also,

$\begin{array}{c}{\mathrm{K}}_{\mathrm{overall}}=\frac{\left[{{\mathrm{CuX}}_3}^{2-}\right]}{\left[{\mathrm{Cu}}^{+}\right]{\left[{\mathrm{X}}^{-}\right]}^3}\\ 1\times {10}^9=\frac{0.001}{\mathrm{x}(4.997+{\mathrm{x}}^2)}\end{array}$

Hence,

$\mathrm{x}=8\times {10}^{-15}$

And,

$\begin{array}{c}\left[{{\mathrm{CuX}}_3}^{2-}\right]=0.001-8\times {10}^{-15}\\ =0.001\mathrm{M}\end{array}$

Subpart (b) The concentration of CuX2−

The concentration can be denoted as,

$\begin{array}{c}{\mathrm{K}}_3=\frac{\left[{{\mathrm{CuX}}_3}^{2-}\right]}{\left[{{\mathrm{CuX}}_2}^{-}\right]\left[{\mathrm{X}}^{-}\right]}\\ 1000=\frac{0.001}{\left[{{\mathrm{CuX}}_2}^{-}\right]\times 5}\end{array}$

Hence,

$\left[{{\mathrm{CuX}}_2}^{-}\right]=2\times {10}^{-7}\mathrm{M}$

Subpart (c) The concentration of Cu+

The concentration can be denoted as,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{overall}}=\frac{\left[{{\mathrm{CuX}}_3}^{2-}\right]}{\left[{{\mathrm{CuX}}_2}^{-}\right]\left[{\mathrm{X}}^{-}\right]}\\ 1\times {10}^9=\frac{(0.001-\mathrm{x})}{\mathrm{x}{(4.997+\mathrm{x})}^3}\end{array}$

Hence,

$\mathrm{x}=\left[{\mathrm{Cu}}^{+}\right]=8\times {10}^{-15}$