You mix a 125.0-mL sample of a solution that is 0.0117 M in NiCl₂ with a 175.0-mL sample of a solution that is 0.250 M in NH₃. After the solution reaches equilibrium, what concentration of Ni²⁺(aq) remains?

• A. 0.0058 M
• B. 0.0000 M
• C. 0.0029 M
• D. 0.0117 M

Answer: (C)

Verified step by step guidance

1. First, calculate the initial moles of NiCl₂ in the 125.0 mL solution. Use the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Convert 125.0 mL to liters by dividing by 1000.
2. Next, calculate the initial moles of NH₃ in the 175.0 mL solution using the same formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Convert 175.0 mL to liters by dividing by 1000.
3. Write the balanced chemical equation for the reaction between Ni²⁺ and NH₃ to form the complex ion \([\text{Ni(NH}_3)_6]^{2+}\). This will help you understand the stoichiometry of the reaction.
4. Determine the limiting reactant by comparing the mole ratio of Ni²⁺ to NH₃ based on the balanced equation. The limiting reactant will be completely consumed, and the excess reactant will determine the equilibrium concentration.
5. Calculate the concentration of Ni²⁺ remaining at equilibrium. Subtract the moles of Ni²⁺ that reacted from the initial moles, then divide by the total volume of the mixed solution (125.0 mL + 175.0 mL converted to liters) to find the concentration.