Final answer:
The number of moles of NiCl2 and NH3, and determine the limiting reactant to find out how much Ni2+ ions remain. The concentration of Ni2+ ions that remains after the reaction is 0.0117 M.
Step-by-step explanation:
To find the concentration of Ni2+ ions that remain after mixing the solutions, we need to use the concept of equilibrium and the stoichiometry of the reaction between NiCl2 and NH3.
The balanced equation for the reaction is:
NiCl2(aq) + 6NH3(aq) ⟶ [Ni(NH3)6]2+(aq) + 2Cl-(aq)
Since the stoichiometry of the reaction is 1:6, this means that for every 1 mole of NiCl2, 6 moles of NH3 react. Therefore, we need to calculate the number of moles of NiCl2 and NH3 present in the initial solutions.
Moles of NiCl2 = concentration of NiCl2 × volume of NiCl2 solution
= 0.0117 M × 0.1250 L = 0.00146 moles of NiCl2
Moles of NH3 = concentration of NH3 × volume of NH3 solution
= 0.250 M × 0.1750 L = 0.0438 moles of NH3
Now, we can determine the limiting reactant to find out how much Ni2+ ions remain after the reaction. Since the stoichiometry is 1:6, we need to check which reactant provides the lowest number of moles based on their stoichiometric ratio.
For NiCl2: 0.00146 moles of NiCl2 × 6 = 0.00876 moles of NH3 required
For NH3: 0.0438 moles of NH3 ÷ 1 = 0.0438 moles of NH3 required
Since 0.00876 moles of NH3 is less than 0.0438 moles of NH3, NiCl2 is the limiting reactant. This means that only the stoichiometric amount of Ni2+ ions will be formed, which is equal to 1:1 with NiCl2.
Therefore, the concentration of Ni2+ ions that remains after the reaction is 0.0117 M.