A 0.130-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)62 is 5.0× 10^31.
To determine the concentration of Co2+ ions at equilibrium, we need to use the concept of equilibrium constant. We can start by writing the balanced equation for the formation of the complex ion Co(NH3)62+:
CoCl2 + 6NH3 -> Co(NH3)62+ + 2Cl-
Given the formation constant (Kf) of Co(NH3)62+ as 5.0× 10^31, we can write the expression for the equilibrium constant (Kc) as:
Kc = [Co(NH3)62+] / ([CoCl2] * [NH3]^6)
We are given that the initial concentration of CoCl2 is 0.130 moles, and the initial concentration of NH3 is 1.20 M.
Next, we need to determine the change in concentration for each species. Since 1 mole of CoCl2 reacts to form 1 mole of Co(NH3)62+, the concentration of CoCl2 will decrease by the same amount. Therefore, the change in [CoCl2] is -0.130 M.
Similarly, since 1 mole of Co(NH3)62+ is formed from 6 moles of NH3, the concentration of NH3 will decrease by a factor of 6. Therefore, the change in [NH3] is -6 * 1.20 M = -7.20 M.
Using the equilibrium expression, we can substitute the known values:
5.0× 10^31 = [Co(NH3)62+] / (0.130 M * (-7.20 M)^6)
Now we can solve for [Co(NH3)62+]:
[Co(NH3)62+] = 5.0× 10^31 * (0.130 M * (-7.20 M)^6)
Finally, [Co2+] is equal to the concentration of Co(NH3)62+ since Co(NH3)62+ dissociates completely:
[Co2+] = [Co(NH3)62+]
Calculate the value for [Co2+] using the given values and the expression above to find the concentration of Co2+ ions at equilibrium.