When 100.0 mL of a solution that is 0.00100 M in Cu(NO₃)₂ is mixed with 100 mL of a 0.20 M solution of NH₃, what concentration of When 100.0 mL of a solution that is 0.00100 M in Cu(NO₃)₂ is mixed with 100 mL of a 0.20 M solution of NH₃, what concentration of Cu²⁺ remains in solution? remains in solution?

• A. 5.0 × 10⁻⁴M
• B. 2.9 × 10⁻¹³M
• C. 3.7 × 10⁻¹⁴M
• D. 2.9 × 10⁻¹⁶M

Answer: (B)

Verified step by step guidance

1. Determine the initial moles of Cu(NO3)2 and NH3. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For Cu(NO3)2, \( \text{moles} = 0.00100 \text{ M} \times 0.100 \text{ L} \). For NH3, \( \text{moles} = 0.20 \text{ M} \times 0.100 \text{ L} \).
2. Write the balanced chemical equation for the reaction between Cu(NO3)2 and NH3. The complex ion formed is \([\text{Cu(NH}_3)_4]^{2+}\). The balanced equation is: \( \text{Cu}^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu(NH}_3)_4]^{2+} \).
3. Calculate the limiting reactant. Compare the mole ratio from the balanced equation with the initial moles calculated. Since 4 moles of NH3 are required for every mole of Cu2+, determine which reactant is in excess.
4. Determine the moles of Cu2+ that react completely with NH3. Subtract the moles of Cu2+ that react from the initial moles to find the moles of Cu2+ remaining.
5. Calculate the concentration of Cu2+ remaining in the solution. Use the formula: \( \text{concentration} = \frac{\text{moles of Cu}^{2+} \text{ remaining}}{\text{total volume of solution}} \). The total volume is the sum of the volumes of the two solutions mixed.