Answer:
[tex]\large \boxed{\text{0.0035 mol/L}}[/tex]
Explanation:
We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺
V/mL: 3.00 7.00
c/mol·L⁻¹: 0.050 0.20
1. Identify the limiting reactant
(a) Calculate the moles of each reactant
[tex]\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}[/tex]
(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant
(i) From Cu²⁺
[tex]\text{Moles of Cu(NH$_{3}$)$_{4}$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$^{2+}$}[/tex]
(ii) From NH₃
[tex]\text{Moles of Cu(NH$_{3}$)$_{4}$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$^{2+}$}[/tex]
NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.
(c) Concentration of the complex ion
[tex]\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}[/tex]
