A 0.140-mole quantity of CoCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Co2+ ions at equilibrium? Assume the formation constant of Co(NH3)6^2+ is 5.0e31.
Please help, I don't know how to do this problem.
Here is how you do this with a little chicanery. First, we assume the reaction goes to completion to the right. That's not a bad assumption considering the size of the formation constant.
.........Co^2+ + 4NH3 ==> Co(NH3)4^2+
Initial.0.140....1.2........0
change...-0.140.-4*1.2.....0.140
equil.....0.......0.64......0.140
Actually, this isn't at equilibrium since we drove it to completion. Therefore, the numbers above are a little TOO far for equilibrium; therefore, it must move back to the left. How much will it move? We set up another set of conditions but start with where we ended above.
..........Co^2+ + 4NH3 ==> Co(NH3)4^2+
I.........0.......0.64......0.140
C.........+x......+4x.......-x
E.........x.....0.64+4x.....-x
Substitute into K.
K = [Co(NH3)4^2+]/[Co^2+][MH3]^4
5E31 = [0.140-x][x][0.64+4x]^4
Now assume 0.140-x = 0.140 and assume 0.64+4x = 0.64 and solve for x = (Co^2+). It's a very small number.
To find the concentration of Co2+ ions at equilibrium, we need to consider the following equilibrium reaction:
CoCl2 + 6 NH3 ⇌ Co(NH3)6^2+ + 2 Cl-
First, we need to calculate the initial concentration of Co2+ ions in the solution.
Given that you have added 0.140 moles of CoCl2 to a liter of 1.20 M NH3 solution, we can assume that the volume remains constant, and the number of moles of NH3 remains the same after the addition of CoCl2.
Hence, the initial concentration of NH3 is still 1.20 M.
Now, let's focus on calculating the concentration of Co2+ at equilibrium.
We know that the formation constant (Kf) for Co(NH3)6^2+ is 5.0 x 10^31. The formation constant represents the equilibrium constant for the formation of a complex ion.
The equilibrium constant expression is given as follows:
Kf = ([Co(NH3)6^2+])/([CoCl2] x [NH3]^6)
We are given the formation constant (Kf), and we can assume that the initial concentration of Co(NH3)6^2+ and Cl- ions is negligible compared to the initial concentration of NH3.
Let's assume the equilibrium concentration of Co(NH3)6^2+ is x M, then the equilibrium concentration of Cl- ions will be 2x M.
The concentration of NH3 will remain the same (1.20 M).
Using the equilibrium constant expression:
5.0 x 10^31 = (x)/((0.140 M) x (1.20 M)^6)
Now, let's solve for x:
5.0 x 10^31 = (x)/(0.140 M x 1.20^6)
Multiplying the denominator and numerator to one side:
5.0 x 10^31 = (x)/(0.140 M x 1.20^6)
5.0 x 10^31 x 0.140 M x 1.20^6 = x
x ≈ 1.008 M
Therefore, the concentration of Co2+ ions at equilibrium is approximately 1.008 M.
To solve this problem, we need to use the concept of equilibrium and the formation constant.
First, let's write out the balanced equation for the reaction between CoCl2 and NH3:
CoCl2 + 6NH3 → [Co(NH3)6]2+ + 2Cl-
Based on the given information, we start with a 0.140 mole quantity of CoCl2 and a 1.20 M NH3 solution. This means that we have 1.20 moles of NH3 in 1 liter of solution.
Now, let's calculate the initial concentration of Co2+ ions. Since 1 mole of CoCl2 produces 1 mole of Co2+ ions, the initial concentration of Co2+ ions is also 0.140 M.
To find the equilibrium concentration of Co2+ ions, we need to consider the formation constant (Kf). The formation constant is defined as the equilibrium constant for the formation of a complex ion from its constituent ions.
In this case, the formation constant (Kf) for the formation of Co(NH3)6^2+ is given as 5.0e31. This means that the higher the value of Kf, the more stable the complex ion Co(NH3)6^2+ is, and the higher the concentration of the complex ion at equilibrium.
To calculate the equilibrium concentration of Co2+ ions, we can use the following expression:
[Co2+]eq = ([Co(NH3)6]2+)(2[Cl-])/([CoCl2])
Since we started with 0.140 moles of CoCl2, the initial concentration of CoCl2 is 0.140 M. The concentration of [Cl-] in the solution is equal to 2 times the concentration of CoCl2 since the reaction produces 2 moles of Cl- ions for every 1 mole of CoCl2. Therefore, [Cl-] is 2 times 0.140 M, which is 0.280 M.
Next, we need to calculate the concentration of [Co(NH3)6]2+ at equilibrium. To do this, we use the equation:
[Co(NH3)6]2+ = (Kf)([Co2+]) / ([NH3]^6)
Substituting the values we have, we can calculate:
[Co(NH3)6]2+ = (5.0e31)(0.140 M) / (1.20 M^6)
Now we have the concentration of [Co(NH3)6]2+, which is 5.83e23 M.
Finally, substituting the values back into the first equation, we can solve for the equilibrium concentration of [Co2+] ions:
[Co2+]eq = (5.83e23 M)(2*0.280 M) / (0.140 M)
After performing the calculation, we find that the equilibrium concentration of Co2+ ions is 2.34e24 M.
Therefore, the concentration of Co2+ ions at equilibrium is 2.34e24 M.