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Homework Help: Bad Chemistry, Correct Math?

  1. Mar 21, 2014 #1


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    1. The problem statement, all variables and given/known data

    Find the pH of [itex]1.0 x 10^{-7} M [/itex] [itex]H_{2}S[/itex] solution.

    2. Relevant equations

    We can take three approaches to this problem, one which is mathematically and chemically correct, one mathematically correct but not necessarily chemically correct, and one which is unequivolcally wrong. All involve the [itex]K_{a}[/itex] of [itex]H_{2}S[/itex]:

    [itex]K_{a}= 1.1x10^{-7} = [H_{3}O^{+}][[HS^{-}]/[H_{2}S][/itex]

    Approach 1:

    Use charge balance and mass balance. Start from the below relationships and substitute into the Ka expression.

    [itex][H_{3}O^{+}] = [HO^{-}] + [HS^{-}][/itex]

    [itex][H_{2}S] + [HS^{-}] = 1.1x10^{-7} M[/itex]

    Approach 2:

    Realize there are two significant sources of hydronium ion - one source being water since the solution is so dilute - the acid ionization constant approaches the initial molarity of the acid - and the actual acid itself.

    Use both the acid ionization constant and the auto ionization of water (Kw) to find pH. Use the acid ionization constant below with x = not to hydronium ion molarity but instead hydronium ion contribution from the acid.

    Approach 3 (wrong but what a typical chemistry student would do):

    Follow through on the below formula, with [itex]x = [H_{3}O^{+}][/itex].

    [itex]K_{a} = x^{2}/M_{i}-x [/itex]

    3. The attempt at a solution

    Now, I just showed my chem teacher this method, and although it yields the right answer, he told me it was unequivocally wrong.

    I see what he's saying. We should only plug in the equilibrium molarities into our Ka expression. We can think what we want, but ignore the chemistry, and we run the peril of being wrong. We can delineate between hydronium ion contribution between the acid and water all we want, but the one thing we should be plugging into the Ka expression is neither of the aforementioned but rather the sum or equilibrium concentration of the aforementioned.

    What do you think?
    Last edited: Mar 21, 2014
  2. jcsd
  3. Mar 21, 2014 #2


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    Please show - step by step - how you are trying to solve, I have problems following just the description.
  4. Mar 21, 2014 #3


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    You have overestimated us! :smile: You have given us the hints and left us to do the calculations. We like to do it the other way round! :biggrin:

    Since you seem to have done the calculations why not give them?

    Because of the way itex works your X has become [tex]x [/tex]

    I understand you to be saying that the Ka of H2S is 1.1X10-7. As I didn't know at first what you meant and didn't trust it (because you have made a mistake in your 3rd equation :tongue2:) I googled it http://www.sciencegeek.net/tables/ka.shtml and found 9.1X10-8 which is close enough, there are probably several different values around.

    You are right if what you are saying is that usual approximations are tricky to use here because of the extreme dilution making the protons originating from H2S of comparable concentration to those in pure water. I think you'd have to take into account everything present and do a full calculation if you want to calculate pH to 2 decimal places.

    For rough calculations a chemist [PLAIN]https://imageshack.com/a/img84/3696/grindm0.gif [Broken] [Broken]sneer might say there are at most 10-7M protons that could original from H2S which additional to 10-7M in water, total cannot be more than 2X10-7M, -> pH not lower than 6.7. A more refined but still chemist's [PLAIN]https://imageshack.com/a/img84/3696/grindm0.gif [Broken] [Broken] handwavy circular argument would be to say that looks about half dissociated - [HS-]/[H2S] around unity fits with Ka/[H+] around unity at near neutrality (10-7M H+) and 10-7 Ka. So we'd have about 0.5X10-7M protons originating from H2S plus about 10-7M from water, 1.5X10-7M total and a pH around 6.8.

    How close is this to the exact value you have calculated? :smile:
    Last edited by a moderator: May 6, 2017
  5. Mar 22, 2014 #4


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  6. Mar 22, 2014 #5


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    Honestly, I still have problems understanding what it is all about. I mean - I don't get why you do approximations that are obviously questionable, instead of just going straight ahead and solving full system of equations. Yes, it will yield a third degree polynomial, which is a little bit more complicated to solve, but it gives an answer that is correct without any doubts.

    Last edited: Mar 22, 2014
  7. Mar 22, 2014 #6
    Why not just use an ICE table with [H+] = 10-7 initially and then 10-7 + x. In other words use the common ion effect + ICE table but don't neglect the initial concentration of acid from water auto-hydrolysis. You get a polynomial without any assumptions. Not sure what the limitations of that approach are.
    Last edited: Mar 22, 2014
  8. Mar 22, 2014 #7


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    How are you going to use ICE table for two separate equilibria?

    I am not telling it can't be done somehow, I just don't see it.
  9. Mar 22, 2014 #8
    I was shown by a grad student a few years ago and never thought much about it. The initial hydronium concentration is taken to be 10-7 and the final hydronium concentration is 10-7 + x.

    The resulting Ka expression is (x)(10-7 + x) / ([HA]o - x) which becomes a second order polynomial.

    I'm on my dads tablet and it is tough to type. I can show the full calculation later.


    EDIT: Here's my recollection of the calculation for a weak and dilute acid in water. This is how I was taught to work with ICE tables by using the common ion effect. In the most basic case you neglect the hydronium due to auto-hydrolysis of water but not if we are working with dilute acid such that [HA]° ~ 10-7 M.

    HA + H2O → H3O+ + A-

    All units are M, where applicable.

    Initially: [HA]=[HA]°; H3O+=10-7; [A-]=0.

    Extent of rxn: x.

    End: [HA]=[HA]° - x; H3O+=10-7 + x; [A-]=x.

    Ka = ( [H3O+] [A-] ) / [HA] = ( (10-7 + x) x ) / ( [HA]° - x )
    Last edited: Mar 22, 2014
  10. Mar 22, 2014 #9
    Let x represent the concentration of HS, and let 10-7 - y represent the concentration of OH. Then H = 10-7-y + x, and H2S = 10-7 - x. So, from the equilibria:
    (10-7-y+x)(x)=1.1 x 10-7(10-7 - x)

    (10-7-y + x)(10-7 - y)=10-14

    Let z = 10-7-y

    x(z+x)=1.1 x 10-7(10-7 - x)

    z(z + x)=10-14

    Let x = 10-7X

    X(Z+X)=1.1 (1 - X)


  11. Mar 23, 2014 #10


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    You are ignoring water dissociation. 10-7 is a correct initial concentration of H+, but any changes in pH will shift water autodissociation consuming part of H+, and 10-7 will be no longer valid.
  12. Mar 23, 2014 #11


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    Sorry to say that, but you did a lot to make following your derivation difficult to follow.

    what is y?

    So now z is concentration of OH-? Why another variable, one which hides non-defined variable introduced earlier?

    I don't see how is it going to help find pH.
  13. Mar 23, 2014 #12


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    This is becoming a "reinvent the wheel" thread.

    I am not going to repeat whole (and full) derivation I posted here: http://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution, this is just the final equation 6.9:

    [tex][H^+]^3 + K_a[H^+]^2 - (C_aK_a+K_w)[H^+] - K_aK_w = 0[/tex]

    where Ca is the analytical concentration of the weak acid.

    You can try to simplify the polynomial for specific cases. There is an additional discussion here: http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base that shows which approximations work for which acids and which concentrations. Scroll down to the plot - white area is where typical approximations cease to work and you have to use above equation (but in most of the area solution is simply neutral).
  14. Mar 23, 2014 #13
    Well I feel lied to, and a bit silly for not ever really thinking the calculation through. I was actually thinking along similar lines last night as I was writing it up but decided to get input anyway.

    Ancora imparo, thanks Borek!
  15. Mar 23, 2014 #14
    y is the amount of H and OH that react with one another.

    Ths substitution just seemed to simplify the equations.
    I can substitute Z in either of the equations and solve for X.
  16. Mar 23, 2014 #15


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    OK, I finally see what you did. But unless I am missing something again, this approach yields 3rd degree polynomial in either z or x, doesn't it?
  17. Mar 23, 2014 #16
    Yes. I just wanted to show a different way of setting up the problem (in case this approach might work better for someone). Also, I thought that I had found a simple way of solving the equations, but later I found I had made a mistake in a subsequent algebraic step.

  18. Mar 23, 2014 #17


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    OK, so that's equivalent to the equation I posted.

    Actually your

    H = 10-7 - y + x = OH + x

    is just another way of writing

    [H+] = [OH-] + [HS-]

    which is nothing else but charge balance.
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