Question

You mix a 150.0 −mL sample of a solution that is 0.0122 M in NiCl2 with a 190.0 −mL sample of a solution that is 0.350 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains?

Answer 1

The concentration of Ni2+(aq) remaining after the equilibrium is 0.0099 M.

Step-by-step explanation:

In order to determine the concentration of Ni2+(aq) remaining after equilibrium, we can use the principles of equilibrium constant expressions for complex ion formation. The balanced chemical equation for the reaction between Ni2+(aq) and NH3(aq) is:

`\[ Ni^(2+)(aq) + 6NH_(3)(aq) \rightleftharpoons Ni(NH_(3))_(6)^(2+)(aq) \]`

The equilibrium constant expression (K) for this reaction can be expressed as:

`\[ K = ([Ni(NH_(3))_(6)^(2+)])/([Ni^(2+)][NH_(3)]^(6)) \]`

Given that the initial concentrations of Ni2+(aq) and NH3(aq) are 0.0122 M and 0.350 M, respectively, and the volume of each solution is provided, we can set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations.

After solving for equilibrium concentrations, the concentration of Ni2+(aq) is found to be 0.0099 M. This is calculated by using the equilibrium concentrations in the equilibrium constant expression and solving for [Ni2+].

Therefore, the final concentration of Ni2+(aq) after reaching equilibrium is 0.0099 M. This result indicates the concentration of Ni2+(aq) that remains in the solution, considering the reaction between Ni2+(aq) and NH3(aq) and the establishment of equilibrium.

Answer 2

The solution reaches equilibrium, the concentration of Ni²⁺(aq) that remains is approximately 0.196 M.

To find the concentration of Ni²⁺(aq) that remains after mixing the two solutions of NiCl₂ and NH₃, we need to consider the formation of the complex ion Ni(NH₃)₆²⁺, which occurs in the equilibrium reaction:

Ni²⁺(aq) + 6NH₃(aq) ⇌ Ni(NH₃)₆²⁺(aq)

Let's use the method of initial concentrations and change in concentrations to solve this problem step by step:

1. Calculate the moles of NiCl₂ initially in the 150.0 mL solution:

Moles of NiCl₂ = Molarity × Volume (in liters)

Moles of NiCl₂ = 0.0122 M × (150.0 mL / 1000 mL/L) = 0.00183 moles

2. Calculate the moles of NH₃ initially in the 190.0 mL solution:

Moles of NH₃ = Molarity × Volume (in liters)

Moles of NH₃ = 0.350 M × (190.0 mL / 1000 mL/L) = 0.0665 moles

3. Based on the balanced equation, we see that 1 mole of NiCl₂ reacts with 6 moles of NH₃ to form 1 mole of Ni(NH₃)₆²⁺.

4. Calculate the limiting reactant:

Since 1 mole of NiCl₂ reacts with 6 moles of NH₃, and you have 0.00183 moles of NiCl₂ and 0.0665 moles of NH₃, NH₃ is the limiting reactant.

5. Determine the moles of Ni(NH₃)₆²⁺ formed:

Moles of Ni(NH₃)₆²⁺ = Moles of NH₃ (since they react in a 1:1 ratio)

Moles of Ni(NH₃)₆²⁺ = 0.0665 moles

6. Calculate the volume of the final solution:

The total volume of the final solution is the sum of the volumes of the two solutions: 150.0 mL + 190.0 mL = 340.0 mL or 0.340 L.

7. Calculate the concentration of Ni²⁺(aq) that remains:

Concentration = Moles / Volume

Concentration of Ni²⁺(aq) = 0.0665 moles / 0.340 L ≈ 0.196 M

So, after the solution reaches equilibrium, the concentration of Ni²⁺(aq) that remains is approximately 0.196 M.