Question

A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Assuming all the Cu2 is complexed with NH3to form the [Cu(NH3)4]2 ion, determine the concentration of the complex in the solution.

Answer 1

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

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  

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH _{3} ) _{4} ^{2+} } = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH _{3} ) _{4} ^{2+} }}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH _{3} ) _{4} ^{2+} }$

(ii) From NH₃

$\text{Moles of Cu(NH _{3} ) _{4} ^{2+} } = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH _{3} ) _{4} ^{2+} }}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH _{3} ) _{4} ^{2+} }$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH _{3} ) _{4} ^{2+} in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is \large \boxed{\textbf{0.0035 mol/L}} }$