1. Hi,

I know that, if we add a strong acid to water, water contribution to give H+ is very minor, so that we ignore it and we just take acid's when we want to calculate the pH of the solution.

But what if I mix 2 strong acids? Will their [H] contributions add, making pH lower?

Also, if I dissolve, say 2 tablets of aspirin (weak acid) in water. Should I consider the volume of water in my calculations?
I dunno, I'm confused. I think it will affect the molarity of the acid and thus the pH. But in the same time, we keep saying that water contribution do not affect pH. :?

Thanks  2.

3. i think you do count the water, as far as i know.

best bet. do both

which one comes out closest?  4. Everything about acids is bit far for me.
But if I remember correct the pH formulas are all meant to be used in strong dilution in water.

pH 1 being 0,1M (lowest) so if you ad 1 mol of HCl in 1l you can forget all the formulas and everything become a different ball game using activities and so on.

So if you have two strong acids (totally loose H+ in pH=1) you can add their contribution as long as you don’t get under this pH=1

For the asperine part, I can only say water must be the molecule dissolving the other one, It must be used in large over mate (English?) otherwise the formulas to not hold either.  5. When calculating molarity you always include the volume of the solvent and the solute. In other words you take the volume of the total solution.
M=n/L where n= moles of what your measuring, in this case aspirin, and L is the volume os solution in leters. As far as pH goes you normally only cosider whats being disolve because when H2O disociates it forms OH- and H+ which still cancel each other out.  6. Uh, I don't actually know the correct final result if that's what you mean by "which one comes out closest?"

About the aspirin, you mean that there must be just enough water to dissolve it? so that, if I put the apirin in, say 250 ml, the pH will be the same as 500 ml for instance?

But in this case, why do we take molarities? why not simply the no. of moles of the acid?  7. You usually neglect the H+ contribution from water and just use the Ka and molarity of the weak acid. Unless the Ka of your acid is down around 10^-14, the effects of the weak acid on the pH will be so much greater that you can just ignore the H+ from the water.  8. Yes, I understand this. My only problem now is, when I dissolve a solid or gas in the solution (so no. of moles is given). How do I manage with calculations?

If for instance I have 0.1 mol A dissolved in 0.5 L water, then do I say that [A] = (0.1) / (0.5) = 0.2 M and go on with pH calculations, or do I simply take the no. of moles, disregarding volume of water (or solution in general) and say [A] = 0.1 M?

The first makes more sense to me but I'm not sure.  9. Originally Posted by Stranger
Yes, I understand this. My only problem now is, when I dissolve a solid or gas in the solution (so no. of moles is given). How do I manage with calculations?

If for instance I have 0.1 mol A dissolved in 0.5 L water, then do I say that [A] = (0.1) / (0.5) = 0.2 M and go on with pH calculations.
This is correct. Find the concentration of your weak acid in the solution, then use the Ka (or pKa, or whatever they give you) to calculate the H+ concentration from there.  10. Ok, but one last question. What confuses me is: when I add say 0.1M HCl to water, I ignore the volume of water because water's contributions to provide H+ is very weak. So I say pH = - log 0.1 = 1, right?

My problem now is, wouldn't the volume of water affect the molarity of the acid I added? decreasing it if volume of water >1L, and increasing it if volume of water < 1, like it did when we dissolved a solid or a gas?  11. If you have 0.1M HCl, then the pH of that HCl solution is 1. If you poor some more water into your HCl solution you have now changed the molarity, so the pH will change. You will need to figure out the new molarity of the HCl solution and recalculate the pH using that new molarity.  12. :x :x :x :x :x

I finally figured it out. I misunderstood this stupid sentence from the begining:

We should first note that molar concentrations are independent of solution volume. That is, [H3O] in 0.015 M HCl is the same whether we are describing 1L, 10L or 100mL of solution.
Ugh... I read it quickly and I thought he meant it didn't matter whether we put 0.015M in 1L, 10L or 100mL water, pH would be the same because (and I got this fragment from somewhere else) water do not contribute to pH.

All got mixed... waste of time... and now I have to correct all the problems I answered... mmm, on the bright side there is no contradiction now. Well, thanks very much. I'll try to be more careful next time  Bookmarks
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