Question

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

Answer 1

After the solution reaches equilibrium, no Ni2+(aq) remains in the solution due to the reaction between NiCl2 and NH3.

Step-by-step explanation:

To determine the concentration of Ni2+ remaining in the solution after equilibrium, we need to consider the reaction that occurs between NiCl2 and NH3.

The balanced chemical equation is:

NiCl2 + 4NH3 → Ni(NH3)4^2+ + 2Cl-

From the balanced equation, we can see that 1 mole of NiCl2 reacts with 4 moles of NH3 to form 1 mole of Ni(NH3)4^2+.

Since the solutions have different volumes, we need to calculate the moles of each substance:

Moles of NiCl2 = 0.0117 M * 0.125 L = 0.00146 mol

Moles of NH3 = 0.250 M * 0.175 L = 0.04375 mol

Since 1 mole of NiCl2 reacts with 4 moles of NH3, we can determine the limiting reactant by comparing the moles.

In this case, NiCl2 is the limiting reactant because it has fewer moles.

Since NiCl2 is the limiting reactant, all of it will react with NH3 to form Ni(NH3)4^2+.

Therefore, there will be no Ni2+ remaining in the solution after equilibrium.

Answer 2

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.