A 0.170-mole quantity of NiCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Ni2 ions at equilibrium? Assume the formation constant* of Ni(NH3)62 is 5.5 × 108.
To determine the concentration of Ni2+ ions at equilibrium, we need to calculate the concentration of Ni(NH3)62+ ions formed in the solution at equilibrium.
First, let's write the balanced chemical equation for the reaction between NiCl2 and NH3:
NiCl2 + 6NH3 ⇌ Ni(NH3)62+ + 2Cl-
From the balanced equation, we can see that 1 mole of NiCl2 reacts with 6 moles of NH3 to form 1 mole of Ni(NH3)62+.
Given that we have a 0.170-mole quantity of NiCl2, this means that 0.170 moles of Ni(NH3)62+ will be formed.
Next, let's consider the volume of the solution. We have a 1.20 M (moles per liter) NH3 solution with a total volume of 1 liter. This means we have 1.20 moles of NH3 in the solution.
Now, we can calculate the concentration of Ni(NH3)62+ ions using the given formation constant of 5.5 × 10^8.
The formation constant (Kf) expression is given by:
Kf = [Ni(NH3)62+] / ([Ni2+] × [NH3]^6)
Rearranging the equation, we get:
[Ni(NH3)62+] = Kf × [Ni2+] × [NH3]^6
Substituting the given values:
[Ni(NH3)62+] = (5.5 × 10^8) × (0.170) × (1.20^6)
Calculating this product, we get:
[Ni(NH3)62+] ≈ 5.182 mol/L
Therefore, the concentration of Ni2+ ions at equilibrium is approximately 5.182 mol/L.
To find the concentration of Ni2+ ions at equilibrium, we need to consider the equilibrium reaction between NiCl2 and NH3:
NiCl2 + 6NH3 ⇌ Ni(NH3)62+ + 2Cl-
The formation constant (Kf) of Ni(NH3)62 is given as 5.5 × 10^8, which represents the equilibrium constant for the formation of the complex ion.
First, we need to determine the initial moles of Ni(NH3)62+ formed:
Initial moles of Ni(NH3)62+ = Kf * [NiCl2] * [NH3]^6
Since we have 0.170 moles of NiCl2 and a 1.20 M NH3 solution, the initial moles of Ni(NH3)62+ are:
Initial moles of Ni(NH3)62+ = 5.5 × 10^8 * 0.170 * (1.20)^6
Next, we need to calculate the final volume of the solution after adding the NiCl2:
Final volume = 1L + 1L = 2L
To determine the concentration of Ni2+ ions at equilibrium, we divide the moles of Ni(NH3)62+ by the final volume:
Concentration of Ni2+ ions at equilibrium = Initial moles of Ni(NH3)62+ / Final volume
Now you can plug in the values to calculate the concentration of Ni2+ ions at equilibrium.
You do these problems by working it twice, once forward and once in reverse like this. Since the formation constant is such a large number you assume this reaction goes to completion.
.......Ni^2+ + 6NH3 ==> Ni(NH3)6^2+
I....0.00131....1.20.....0
C...-0.00131...-0.00787...0.00131
E........0......1.20.... 0.00131
The 0.00131 is mols NiCl2/1L = 0.00131M.
Then use the same equation but work right to left. The E line from above becomes the I line for this.
.......Ni^2+ + 6NH3 ==> Ni(NH3)6^2+
I......0........1.20....0.00131
C......x........6x.......-x
E......x.......1.20+6x..0.00131-x
Then substitute the E line into the Kf expression and solve for x = (Ni^2+).