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PH calculation

  1. May 1, 2006 #1
    This problem is driving me nuts. :tongue2:

    The polyprotic acid, H2SO4 has a complete disassociation at the first step and its pKa for the second step is 1.9. When CaSO4 dissolves ina strong acidic solution of HCl, the final concentration of Ca2+ is 0.03 M and [SO42-] is 0.008 M. What is the pH of the solution?

    Okay, so I did the equilibrium for the first reaction. Doesn't tell me much except that I see it produces hydronium and sulfate at the end of the second step...

    H2SO4 + H2O -> H3O+ (aq) + HSO4- (aq)

    then

    HSO4- (aq) + H2O (l) -> H3O+ (aq) + SO42- (aq)

    ...the second reaction seems to produce sulfate as well, but I can't quite figure out how to tie them together, mostly because I can't figure out how the second equilibrium reaction works. Working from

    CaSO4 + H2O -> Ca2+ + SO42-

    This seems too simple... do I need to add HCl and H2O on the left? If so then I get a chloride ion on the right that I don't have a concentration for, ditto for the hydronium... of course, since I'm just adding stuff to both sides, it doesn't matter, does it. Still, it seems like I'm missing something I'd need to find Kb.

    Anyway, I'm presuming that if I could figure out this second reaction I could work out Kb and thus pKb, then pOH, pH, success. Although that leaves out the given pH... ugh. Stuck.
     
    Last edited: May 1, 2006
  2. jcsd
  3. May 1, 2006 #2
    If I am not mistaken, [tex] CaSO_{4} [/tex] is a salt that is insoluble in water. However, in the presence of strong acid, it dissolves, and the reaction should be as follows:-
    [tex] CaSO_{4} + 2HCl --> H_{2}SO_{4} + CaCl_{2} [/tex]. Now, as sulphuric acid is a product of the reaction, you can use the information provided about it to help you calculate the pH!

    NOTE:
    1) Assume no HCl is present after the reaction is complete, so the acidity of the final product comes from sulphuric acid alone.
    2) Since sulphuric acid dissociates twice, my guess is you need to sum up the concentrations of [tex] H^+ [/tex] produced in each dissociation to determine the total concentration of [tex] H^+ [/tex] produced. You can directly sum up the concentrations as the volume of the final product is constant.
     
    Last edited: May 1, 2006
  4. May 1, 2006 #3
    Whew, I do believe I get it now. Thank you so much. If there's a problem like this on my final today, you may well have just saved my college career :-)
     
  5. May 1, 2006 #4
    omg chemistry. you like this stuff? masochists xD all respect to you tho =)
     
  6. May 1, 2006 #5

    Borek

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    Staff: Mentor

    You can't ignore HCl, as it is almost solely responsible for the solution pH. Take a look at the following equations - they contain the key to the solution:

    [SO42- ] + [HSO4- ] = [Ca2+ ]

    Ka = [SO42- ][H+ ]/[HSO4- ]

    There are two unknowns only - [H+ ] being one of them.

    Best,
    Borek
     
    Last edited by a moderator: Aug 13, 2013
  7. May 1, 2006 #6
    Queries

    Hi Borek!

    May I know why we cannot ignore HCl? I thought it had completely reacted with the [tex] CaSO_{4} [/tex]? Also, not much information is provided about it, so I thought we need not consider it...

    Now that I think about it, do we even need to use the [tex] K_{a} [/tex] value? After all, isn't [tex] [H^+]_{total} = [H^+]_{1st\ dissociation} + [H^+]_{2nd\ dissociation} [/tex]?

    And when we analyze the question further, isn't this equivalent to [tex] [H^+]_{total} = 0.03 + [SO_{4}^{2-}]_{final} [/tex]?
    Thanks for your help!
     
    Last edited: May 1, 2006
  8. May 2, 2006 #7

    Borek

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    Staff: Mentor

    In short - you are not asked what is pH in effect of hydrolyzis, but what pH will force known situation is the solution.

    Situation in the solution is such that in low pH (forced by HCl) SO42- gets protonated shifting CaSO4 dissolution to the right. This low pH is due to the HCl presence, not due to CaSO4 dissolution.

    This value is not given here, but pKso for CaSO4 is 4.62, so saturated solution is about 0.0043M - about seven times less concentrated than 0.03M given in the question. Something pushes this equilibrium - and it is a presence of the H+ from other source than hydrolyzis.

    pH of saturated CaSO4 solution is almost perfectly neutral (7.06 when calculated using BATE).

    Best,
    Borek
     
    Last edited by a moderator: Aug 13, 2013
  9. May 2, 2006 #8
    Query

    So this reaction
    [tex] CaSO_{4} + 2HCl --> H_{2}SO_{4} + CaCl_{2} [/tex] does not occur?
     
  10. May 2, 2006 #9

    Borek

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    Staff: Mentor

    H2SO4 is too strong acid for HSO4- to get protonated in measurable quantities. So

    CaSO4 + H+ -> Ca2+ + HSO4-

    is what is really happening in the solution.

    Best,
    Borek
     
    Last edited by a moderator: Aug 13, 2013
  11. May 2, 2006 #10

    GCT

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    Science Advisor
    Homework Helper

    When CaSO4 dissolves in water, the solvated SO4 2- reacts as a base, you can determine the Kb from the Ka for HSO4 -. For this problem, you probably don't need to consider the latter compounds reactivity as a base since it indicates that sulfuric acid dissociates completely.

    From the solubility constant of CaSO4, you can determine the concentration of SO4 2- in an equilibrium saturated solution. You can then compare this concentration with explicitly given concentration due to the presence of HCl. That is the deviancy will indicated how much of the SO4 2- had reacted further. The setup for this equation, however, is slightly complicating, but I'll let you get started on that.

    How much SO4 2- reacted pertains directly to how much HSO4 - was formed, its equilibrium concentration. You can use this equilibrium concentration as well as the provided Kb, or Ka if you prefer, to determine the pH. The equilibrium constant equation is sufficient for the determination of pH, you don't need to worry about HCl. That's the nice thing about these equations, is that you can determine the context of the solution simply by observing the dynamics of HSO4 -, it acts somewhat as an indicator.
     
  12. May 2, 2006 #11

    Borek

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    Staff: Mentor

    Solubility constant was not given in the question. Equilibrium concentrations were.

    Best,
    Borek
     
    Last edited by a moderator: Aug 13, 2013
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