I started to try to solve it like this:
I'm simplifying the situation: my goal is to get a mixture of H2SO4(aq) and NH3(aq) that will give me a pH of 7. Specifically, I want a formula that will calculate [NH3]initial (where [NH3]initial is the concentration of aqua ammonia I need to add), for any given initial concentration of sulfuric acid [H2SO4]initial, in order to get pH = 7.
So, I know [H2SO4]initial (which is the same as [HSO4−]initial or [H+]initial, since I assume complete dissociation of the first proton), and I know the final proton concentration that I want: [H+] = 10−7.
I have five unknowns: [HSO4−], [SO42−], [NH3], [NH4+], and [NH3]initial (which is the thing I'm trying to solve for), so I need five equations.
The first four are obvious:
(1) Ka(HSO4−) [HSO4−] = [SO42−] [H+] (definition of Ka),
and since [H+] = 10−7, we have Ka(HSO4−)[HSO4−] = 10−7 [SO42−]
(2) Ka(NH4+) [NH4+] = [NH3] [H+] (definition of Ka)
and since [H+] = 10−7, we have Ka(NH4+) [NH4+] = 10−7 [NH3]
(3) [H2SO4]initial = [HSO4−] + [SO42−] (balance S)
(4) [NH3]initial = [NH3] + [NH4+] (balance N)
I need one more equation. I was thinking this, is this valid: ??
(5?) [NH4+] − [NH3] = [H2SO4]initial + [HSO4−] − [SO42−]
which describes the claim that the net H+'s gained by NH3 (to make NH4+) came from either the initial [H+]initial (which equals [H2SO4]initial), or from the dissociation of the second proton ([HSO4−] − [SO42−]). Is this 5th equation valid? If so, I can combine them all and solve for [NH3]initial.