A 0.170-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 (Kf) of Co(NH3)62 is 5.0× 1031 M–6.
I don't really know how to start. At first I thought about setting up an ICE chart and getting it to equal to the Keq =Kf*Ksp, but then I can't find the Ksp for CoCl2.
To find the concentration of Co2 ions at equilibrium, we need to consider the chemical equilibrium between CoCl2 and NH3, which forms the complex ion Co(NH3)62+.
First, let's write the balanced equation for the reaction:
CoCl2 + 6NH3 ⇌ Co(NH3)62+ + 2Cl-
The formation constant (Kf) of Co(NH3)62+ is given as 5.0× 10^31 M–6.
Now, let's set up an ICE table to keep track of the changes in concentration:
Initial:
CoCl2: 0.170 mol
NH3: 1.20 M (1.20 mol/L)
Co(NH3)62+: 0 M
Cl-: 0 M
Change:
CoCl2: -x
NH3: -6x
Co(NH3)62+: +x
Cl-: +2x
Equilibrium:
CoCl2: 0.170 - x mol
NH3: 1.20 - 6x M
Co(NH3)62+: x M
Cl-: 2x M
Since the concentration of NH3 is given as 1.20 M, we assume that it remains essentially unchanged after the reaction. Therefore, x will be much smaller compared to 1.20 M, so we can neglect its effect on the concentration of NH3.
Now, we need to set up the equilibrium expression using the formation constant (Kf) of Co(NH3)62+:
Kf = [Co(NH3)62+] / ([NH3]^6 * [CoCl2])
5.0× 10^31 M–6 = x / ((1.20)^6 * (0.170 - x))
Since x is small, we can approximate (1.20 - x) as 1.20:
5.0× 10^31 M–6 = x / (1.20^6 * 0.170)
5.0× 10^31 M–6 = x / 0.02592
Multiply both sides by 0.02592:
0.02592 * 5.0× 10^31 M–6 = x
x = 1.296 × 10^30 M–5
Therefore, the concentration of Co2 ions at equilibrium is approximately 1.296 × 10^30 M–5.
To find the concentration of Co2+ ions at equilibrium, you will need to use the equilibrium expression for the formation of the complex ion Co(NH3)62+ and the given formation constant (Kf).
The reaction can be represented as follows:
CoCl2 (aq) + 6 NH3 (aq) ⇌ Co(NH3)62+ (aq) + 2 Cl- (aq)
The equilibrium constant expression for the above reaction can be written as:
Kf = [Co(NH3)62+] / ([CoCl2] * [NH3]^6)
Given:
- The formation constant (Kf) of Co(NH3)62 is 5.0× 10^31 M^–6.
- The initial concentration of CoCl2 is 0.170 moles.
- The concentration of NH3 in a 1.20 M solution is 1.20 M.
To find the concentration of [Co(NH3)62+] at equilibrium, we need to determine the concentration of CoCl2 and NH3 left at equilibrium.
1. Calculate the molar concentration of CoCl2:
Moles of CoCl2 = 0.170 moles
Volume (V) = 1 liter
Concentration of CoCl2 = Moles / Volume = 0.170 M
2. Calculate the molar concentration of NH3:
Concentration of NH3 = 1.20 M
3. Since the stoichiometric ratio of CoCl2 to Co(NH3)62+ is 1:1, the equilibrium concentration of [Co(NH3)62+] will be equal to the concentration of CoCl2 at equilibrium:
[Co(NH3)62+] = 0.170 M
Now, substitute the values into the equilibrium constant expression:
5.0× 10^31 M^–6 = [Co(NH3)62+] / (0.170 M * (1.20 M)^6)
Next, rearrange the equation to solve for [Co(NH3)62+]:
[Co(NH3)62+] = 5.0× 10^31 M^–6 * (0.170 M * (1.20 M)^6)
Calculate the concentration of Co(NH3)62+ to find the equilibrium concentration of the Co2+ ions.
CoCl2 is soluble; therefore, it has no Ksp.
The easiest way to do these is to note that with a Kf so extremely high you know the solution at equilibrium will be far to the right; therefore, we set up an ICE chart and make the reaction go 100% to the right. That is an ok assumption because of the huge Kf for the complex ion.
.......Co^2+ + 6NH3 ==> [Co(NH3)6]^2+
I.....0.170....1.20........0
C....-0.170...-1.02.......0.170
E........0.....0.18.......0.170
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Then you turn the thing around and set up and ICE chart in reverse. Completed it looks like this and you start with 0.170 of the complex and zero of Co^2+and 0.18 for NH3 (just the E line of the first equilibrium).
........Co^2+ + 6NH3 --> [Co(NH3)6]^2+
I........0.......0.18......0.170
C........x.......+6x........-x
E........x.....0.18+6x.....0.170-x
Now set up the Kf =
[{Co(NH3)6}^2+]/(Co^2+)(NH3)^6
Substitute the E line into this Kf expression and solve for x = (Co^2+).
Making the assumption that 0.170-x = 0.170 and (0.18+6x)^6 = (0.18)^6 makes it easier to work and doesn't affect the answer because x is very small.