# Kw of Water

So Kw = [H+][OH-] = 10^-14 is true all the time of water and aqueous solutions? When we add an acid, how does the OH- go down to maintain Kw? Is the OH- from the self-ionization of water helping to neutralize the addition of the acid?

Thanks.

Borek
Mentor
1. It is not 10-14 all the time - it depends on the temperature. See water ion product page for details.

2. It is true that Kw = [H+][OH-] all the time (although it is a simplification, see the page listed for details). If you add acid to water it will react with the OH- to maintain equilibrium. Generally speaking its the same neutralization reaction that takes place when you mix acid and base.

epenguin
Homework Helper
Gold Member
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 any one 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.

Hi epenguin,

Your post did help quite a bit, it confirmed many of the predictions I had. However, I'm still a little unclear as to how you would actually calculate the exact H+ for example you gave with the equilibrium laws. For strong acids like HCl, we know that it will virtually dissociate completely so with a concentration of 10^-3, we could expect 10^-3 H+ in ions. Though like you said, there is already 10^-7 H+ ions already existing in solution and the addition of more H+ should drive the reaction H3O+ + OH- -> H2O + H2O forwards. Would our equilibrium equation become then 10^-14 = (10^-3 + 10^-7 - x)*(10^-7 - x)?

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epenguin
Homework Helper
Gold Member
First just write the equations corresponding to principles 2 and 3.
Then we'll see.
How to go from there if you can see it. Hmm OK so I guess [H+] = [HCl] by principle 1 and Kw = [H+][OH-] = 10^-7 by equation 3...what's next?

Borek
Mentor
Hmm OK so I guess [H+] = [HCl] by principle 1

No - H+ is both from HCl and from H2O. But Cl- has only one source.

and Kw = [H+][OH-] = 10^-7 by equation 3...what's next?

think about equation 2 - electroneutrality.

Obviously you have not even tried to read about the subject, as all these equations are listed (and solved) in one of the pages linked to in this thread.

The thing is I've never learned acid/base chemistry using these three principles; this is the first time I've seen it. Currently, I'm just trying to get a basic understanding for MCAT purposes. When I asked my original question, all I wanted to know was why [OH-] goes down when we add any type of acid in order to satisfy the Kw equilibrium. I then wondered how it goes down (i.e by reacting with what) and if it had any effect on pH. We would never be asked any question using the types of calculations presented on that website. Deriving the pH from 4 different questions is just unreasonable for the MCAT.

Borek
Mentor
Well, if you don't need it, don't bother with trying to follow principles listed by epenguin, as they are exactly the same as listed in ChemBuddy lectures. You tried to and I though you want to know how it works. In fact these principles are just a general approach to ANY equilibrium calculation.

epenguin
Homework Helper
Gold Member
3 or 4 principles - it doesn't get much better!
You have half been given the answer to yr question, but if yu don't want to be floundering around:
Yeah, go back to earlier post. First just list what are the + charged and - charged species present?

I will look in tomorrow.