Problem: A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of 2.0 X 10  -3 M CuNO3. Assume that Cu+ form complex ions with X - as follow:Cu+(aq) + X -(aq) ⇌ CuX(aq)             K  1 = 1.0 X 10 2CuX(aq) + X -(aq) ⇌ CuX2-(aq)          K 2 = 1.0 X 104CuX2-(aq) + X -(aq) ⇌ CuX32-(aq)     K 3 = 1.0 X 103with an overall reactionCu+(aq) + 3X-(aq) ⇌ CuX32-(aq)       K = 1.0 X 10 9Calculate the following concentration at equilibrium.c. Cu+

Cu⁺(aq) + 3 X^-(aq) ⇌ CuX₃^2-(aq)       K = 1.0 x 10⁹

Calculate CuX₃^2- formed from the reactants.

$\boxed{\mathbf{M}\mathbf{=}\frac{\mathbf{mol}}{\mathbf{L}}}$

50.0 mL of 10.0 M NaX

NaX(aq) → Na⁺(aq) + X^-(aq)

$\mathbf{moles}\mathbf{ }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\mathbf{\times }\frac{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{X}}^{\mathbf{-}}}}{\cancel{\mathbf{L}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}}{\mathbf{3}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{X}}^{\mathbf{-}}}}$

moles CuX₃^2- = 0.1667 mol

50.0 mL of 2.0 X 10^-3 M CuNO₃

CuNO₃(aq) → Cu⁺(aq) + NO₃^-(aq)

$\mathbf{moles}\mathbf{ }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{Cu}}^{\mathbf{+}}}}{\cancel{\mathbf{L}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{Cu}}^{\mathbf{+}}}}$

moles CuX₃^2- = 1.0x10^-4 mol

CuNO₃ → produces a smaller amount of the product → limiting reactant

Calculate X^-(aq) left in the reaction:

$\mathbf{moles}\mathbf{ }{\mathbf{X}}^{\mathbf{-}}\mathbf{ }\mathbf{initial}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\mathbf{\times }\frac{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{X}}^{\mathbf{-}}}{\cancel{\mathbf{L}}}$

moles X^- initial = 0.500 mol

$\mathbf{moles}\mathbf{ }{\mathbf{X}}^{\mathbf{-}}\mathbf{ }\mathbf{reacted}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}}\mathbf{\times }\frac{\mathbf{3}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{X}}^{\mathbf{-}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{{\mathbf{CuX}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}}}$

moles X^- reacted = 3.0x10^-4 mol

moles X^- left = 0.500 mol - 3.0x10^-4 mol

moles X^- left = 4.997 mol

Calculate the concentrations of CuX₃^2-(aq) and X^-(aq):

$$\begin{gathered} \mathbf{final}\mathbf{ }\mathbf{volume}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mL}\mathbf{+}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mL} \\ \mathbf{final}\mathbf{ }\mathbf{volume}\mathbf{=}\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{L}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}} \end{gathered}$$

final volume = 0.100 L

${{\mathbf{\text{[CuX}}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{\text{]}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{L}}$

[CuX₃^2-] = 1.00x10^-3 M

${\mathbf{\text{[X}}}^{\mathbf{-}}\mathbf{\text{]}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{4497}\mathbf{ }\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{L}}$

[X^-] = 4.997 M

Calculate the concentrations of Cu⁺(aq):