I have learnt the theory - as a fact - that when you want to combine equilibria you "add" them or subtract them, or manipulate as you will, and then the overall equilibrium constant for the expression you end up with is equivalent to the constants for each reaction you had before, to the power of their coefficient in my manipulation of them. For example:
MoS
42– + H2O(l)
MoOS
32– + H
2S(aq) K1 = 1.3×10
–5MoOS
32– + H2O(l)
MoO
2S
22– + H
2S(aq) K2 = 1.0×10
–5MoO
2S
22– + H2O(l)
MoO
3S
2– + H2S(aq) K3 = 1.6×10
–5MoO
3S
2– + H2O(l)
MoO
42– + H
2S(aq) K4 = 6.5×10
–6If at equilibrium a solution contains 1×10
–7 M MoO
42– and 1×10
–6 M H
2S(aq), what would be the concentration of MoS
42–?
My solution:
You must simply add the reactions together:
R(1)+R(2)+R(3)+R(4) produces R(overall): MoS
42– + 4 H2O (l)
MoO
42- + 4 H
2S
K(overall)= ([MoO
42-]*[H
2S]
4)/([MoS
42–]*[H2O (l)]
4)
H2O is excluded because it is so large that the change is negligible:
K(overall)= ([MoO
42-]*[H
2S]
4)/[MoS
42–]
[MoS
42–]=([MoO
42-]*[H
2S]
4)/K(overall)
So now comes in my point. We have added the reactions plainly, as R(1)+R(2)+R(3)+R(4), so K(overall)=K
11*K
21*K
31*K
41. The powers are because the coefficients on the reactions as we add/manipulate them are 1. So now:
[MoS
42–]=([MoO
42-]*[H
2S]
4)/(K
11*K
21*K
31*K
41)
Plug in the numbers and get an answer of: 7.396*10
-12 moldm-3, roughly 7.4*10
-12.
What's my question?
Why? Why is it then when we take R(1)+R(2)+R(3)+R(4) we get K
11*K
21*K
31*K
41; and if we take R(1)+2*R(2)-1/3*R(3)+5*R(4) we get K
11*K
22*K
3-(1/3)*K
45? The maths behind this bit is unclear to me.