Notices
Results 1 to 11 of 11

Thread: Acids

  1. #1 Acids 
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    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


    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Ph.D.
    Join Date
    Feb 2005
    Location
    Dublin, Ireland
    Posts
    945
    i think you do count the water, as far as i know.

    best bet. do both

    which one comes out closest?


    Stumble on through life.
    Feel free to correct any false information, which unknown to me, may be included in my posts. (also - let this be a disclaimer)
    Reply With Quote  
     

  4. #3  
    Him
    Him is offline
    Forum Sophomore Him's Avatar
    Join Date
    Aug 2005
    Posts
    181
    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.
    he who forgets...will be destined to remember (Nothing Man - Pearl Jam)
    Reply With Quote  
     

  5. #4  
    Forum Sophomore biohazard87's Avatar
    Join Date
    Nov 2005
    Location
    With your mom
    Posts
    181
    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.
    Noodles happen when you kiss a stranger in the alps.

    Mi padre tiene un impala.
    Reply With Quote  
     

  6. #5  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    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?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  7. #6  
    Forum Professor
    Join Date
    May 2005
    Posts
    1,893
    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.
    Reply With Quote  
     

  8. #7  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    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.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  9. #8  
    Forum Professor
    Join Date
    May 2005
    Posts
    1,893
    Quote 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.
    Reply With Quote  
     

  10. #9  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    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?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  11. #10  
    Forum Professor
    Join Date
    May 2005
    Posts
    1,893
    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.
    Reply With Quote  
     

  12. #11  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    :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
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •