Then we have:
Kb = (1-x)(10^-6)/(x)
Which is essentially what I just found to be incorrect :/
No, it gives completely different concentration of NH3.
Don't know how I missed that -_-
OK, so solving for x, x = 0.0593 so [NH3] at equilibrium = .0593 at pH 8, which seems reasonable (:
Now to work through the reactions
Mn 2+ + NH3 <---> [Mn(NH3)]2+ log K = 1.00
Mn 2+ + 2NH3 <---> [Mn(NH3)2]2+ log K = 1.54
Mn 2+ + 3NH3 <---> [Mn(NH3)3]2+ log K = 1.70
Mn 2+ + 4NH3 <---> [Mn(NH3)4]2+ log K = 1.30
I crunched a few numbers, and the results I am getting are very, very close to being correct. However, the results still seem to be a bit high at the central peak of [Mn(NH3)n]2+.
Specifically:
@ pH 7.5 -> [Mn(NH3)n]2+ reaches a near maximum slightly greater than 0.2.
However, working this out:
[Mn 2+] = Ksp / [OH-]^2 = 10^(-12.7) / ( 10^(-(14-7.5)) )^2
[Mn 2+] = 1.9952
Kb = 10^(-4.80) = (1-x)(10^-(14-7.5))/(x)
x = [NH3] = .01956
Equilibrium Balances:
[Mn(NH3)]2+ = 10^1 * 1.99526 * 0.01956 = 0.3903
[Mn(NH3)2]2+ = 10^1.54 * 1.99526 * 0.01956^2 = 0.02648
[Mn(NH3)3]2+ = 10^1.70 * 1.99526 * 0.01956^3 = 7.486*10^-4
[Mn(NH3)4]2+ = 10^1.3 * 1.99526 * 0.01956^4 = 5.83*10^-6
[Mn(NH3)n]2+ = .4175
This is about twice what it should be. However, values farther away from the maximum concentration seem to be more accurate. Do all of my calculations seem fine, and is the figure I am looking at possibly flawed, or am I missing an important step? I've also noticed that the last reaction, for [Mn(NH3)4]2+, hardly seems to contribute to the total concentration at all. Is this normal, assuming that [Mn 2+] will react with a smaller number of ammonia molecules far more than with multiple molecules?
Thanks very much for your help. I've made much more progress in the last two days than in the previous week and a half.