For all questions you are going to meet in this area and which do cause headscatchings to students just remember
all proceed from
just 3 principles, at a stretch 4, always the same
1.
Conservation of mass, i.e. of the total concentration of the amount of anyone species summed over all its forms; that hardly enters in this problem;
2.
Electroneutrality - equality between the sum of all the + charge (per litre let's say) OT1H and all the - charge OTOH;
3. The
equilibrium laws that apply.
All 3 embodied in equations that correspond to the case.
4. In some problems you may need also to understand what
approximations you can use (i.e. what concentrations in your equations are going to be negligible compared to others and that you can ignore to shortcut or simplify the maths.)
After a bit of practice it will become second nature and hopefully seem simple. You will no longer ask yourself the question that you did. But it is not a daft question while getting the hang of it - I had to think a bit.
Suppose I add a drop of HCl to water in amounts to make it 10^-3 M HCl. Well it's called that but it's really 10^-3 M in Cl while as the H enters into an equibrium we have to see what that is.
[Cl-] = 10^3 That would be eq. 1 corr to 1. above.
In practiced fashion I would just say [H+] is going to be 10^-3 M too, the pH will be -3.
But being more explicit stimulated by your question, I would have to say for principle 1 I have still [Cl-] = 10^3 M, there are no other forms of Cl present so it is just what I have put into the solution. For electoneutrality, there is one + species but two -, so what is your electroneutrality equation? For principle 3, you have already given the relevant equation. Now work on those to calculate [H+] exactly.
You'll find that it is not exactly 10^3 M so the pH is not exactly 3 after all, though is
as close to that as makes no practical difference so to that extent the practiced answer was right. Then you might think, ah there were H+ in the water before I added the acid so that makes the difference. But in the pure water there were 10^-7 M of H+, and your calculation will have shown that although the solution
is a tiny bit more acidic than pH3 that there are vastly less than 10^-7 M in the difference between your calculated [H+] and 10^-3 M. So I guess you
could say that you have neutralised = reacted almost all the OH- that were there before, plus you have finished with a slight excess of H+ over Cl-[/I] in order to neutralise = balance electrically the remaining OH-.
I hope this is not too confusing.
If you reduce the calcs. to a routine using the 3 or 4 principles you will find you can answer questions like yours too.
