# Salts in solution

If you use Ksp, you shouldn't need any other equation. Let's say I've got a solution with Na2SO3 and NaHSO3 both put in, of which Na2SO3 (let's say) is in up to saturation (but NaHSO3 is still entirely dissolved at this point). Then my equations are:

Mass Balance for S: c0[Na2SO3] + c0[NaHSO3] = [SO2] + [HSO3-] + [SO32-]
Charge Balance: [H3O+] + [Na+] = [HSO3-] + 2[SO32-] + [OH-]
Equilibria: SO2 (aq) + H2O (l) ⇔ HSO3- (aq) + H3O+ (aq) ... K1
HSO3- (aq) + H2O (l) ⇔ SO32- (aq) + H3O+ (aq) ... K2
2H2O (l) ⇔ H3O+ (aq) + OH- (aq) ... Kw
Na2SO3 (s) ⇔ 2Na+ (aq) + SO32- (aq) ... Ksp

There are 6 equations there and 6 ([SO2], [HSO3-], [SO32-], [OH-], [H3O+], [Na+]) variables. This is solved, no need for a Na mass balance. So then, am I right to say that, if I ever have to move from saying "this salt dissolves completely in solution" to treating the case where "some of the salt precipitates in solution", the only change I need to make is throw out the old mass balance for Na (if the dissolution was complete we'd have written 2c0[Na2SO3] + c0[NaHSO3] = [Na+], but it's not so we have to throw this equation out) and include the Ksp equation instead?

DrDu
Mass Balance for S: c0[Na2SO3] + c0[NaHSO3] = [SO2] + [HSO3-] + [SO32-]
And how do you know c_0(Na2SO3) which must be the amount of dissolved Na2SO3 in this equation?

I see, good point. So what do you do then? I'll assume my other 5 equations are the correct ones for the 6 variables, to which one equation (the form of which you might suggest) would be added for a complete set.

DrDu
I'd take the conservation of Na to eliminate the dissolved c(Na2SO4) from the conservation of S.

What do you mean? Can you just write the one equation?

The problem with Na+ conservation is that Na+ is also contributed to by another salt.

DrDu
Come on, try to work out something yourself!
What do you mean? Can you just write the one equation?

The problem with Na+ conservation is that Na+ is also contributed to by another salt.
Of course, but ##c_0(NaHSO3)## is known as it is completely dissolved, at least, that's what you were assuming.

I'm sorry, I've tried and I don't know the answer.

DrDu