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Ksp with significant anion hydrolysis

  1. Apr 5, 2014 #1

    Qube

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    When calculating a Ksp for a salt with a significantly basic anion, e.g. silver cyanide, AgCN, given the experimental molar solubility value (s), should one write the Ksp expression as:

    [itex]K_{sp} = [Ag^{+}][CN^{-}]^{x}[/itex], where x is the fraction of the concentration cyanide ion at equilibrium with respect to s ...

    or as

    [itex]K_{sp} = [Ag^{+}][CN^{-}]^{x}[HCN]^{1-x}[/itex]?

    I'm thinking it's the latter because Ksp is defined as the ion concentration product and obviously when silver cyanide dissolves in water solution we get not only cyanide ion in water but also hydrocyanic acid. The first expression would be incomplete. The complete dissolution equation of AgCN is:

    [itex]AgCN \leftrightharpoons Ag^{+} + (1/9)CN^{-} + (8/9)HCN[/itex]

    The coefficients represent the hydrolysis of the cyanide ion factored in with the ionization of hydrocyanic acid to make more cyanide ion. Hydrolysis of the cyanide ion with molar solubility given by Wikipedia to be s = 0.00023 M is about 91%. However, about 1.7% of the HCN which is formed through cyanide ion hydrolysis ionizes to form cyanide ion again. So call "net" hydrolysis about 88%. There is approximately 1 cyanide ion to every 9 hydrocyanic acid molecules in solution. So should the Ksp therefore be:

    [itex]K_{sp} = [Ag^{+}][CN^{-}]^{1/9}[HCN]^{8/9}[/itex]?

    Which, because of mass balance:

    [itex] s = [Ag^{+}] = [CN^{-}] + [HCN][/itex]

    is equivalent to:

    [itex]K_{sp} = [s/9]^{1/9}[8s/9]^{8/9}[/itex]?
     
    Last edited: Apr 5, 2014
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  3. Apr 6, 2014 #2

    Borek

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    As you mentioned, Ksp is ion concentration product, so it just [Ag+][CN-].

    Now, equilibrium in the solution is described by several equations, starting with three equilibria:

    [tex]K_{sp} = [Ag^+][CN^-][/tex]

    [tex]K_w = [H^+][OH^-][/tex]

    [tex]K_a = \frac {[H^+][CN^-]}{[HCN]}[/tex]

    ending with mass balance and charge balance:

    [tex][Ag^+] = [HCN] + [CN^-][/tex]

    [tex][Ag^+] + [H^+] = [CN^-] + [OH^-][/tex]

    (actually there is more to it, as Ag+ gets complexed by CN-, creating Ag(CN)2-, but let's not go there).

    Molar solubility equals either [Ag+] or [HCN]+[CN-].
     
  4. Apr 6, 2014 #3

    Qube

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    Thanks. But that will give me a new Ksp which is dramatically (more than just a few magnitudes) different from the Ksp of AgCN in literature. Why the big difference?
     
  5. Apr 6, 2014 #4

    Borek

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    No, it won't.

    Hard to comment not knowing where you got this information from.
     
  6. Apr 6, 2014 #5

    Qube

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    These are the values I've been using:

    [itex]K_{a} HCN = 6.2*10^-10[/itex]

    As we can see, the conjugate acid of the cyanide ion, [itex]NC^{-}[/itex], is fairly weak. So we can expect the cyanide ion itself to be a rather good base. At at such a low molar solubility value, we expect hydrolysis of the cyanide ion to be rather extensive. I get about 88% after factoring in ionization of the hydrocyanic acid.

    so [itex]K_{b} NC- = Kw / K_{a} HCN[/itex]

    The s value for silver cyanide was taken from Wikipedia.

    [itex]s = 0.000023 g/100 mL[/itex]

    http://en.wikipedia.org/wiki/Silver_cyanide

    And this value can be readily found (values vary somewhat across sources but all of the sources have the molar solubility value pegged at [itex]10^{-16}[/itex] to [itex]10^{-17}[/itex]. I'm using the value below:

    [itex]Ksp AgCN = 2.2 * 10^{-16}[/itex].

    So:

    Revised Ksp (with consideration of anion hydrolysis):

    [itex]K_{sp} = [Ag^{+}][NC^{-}] = [s/9]^{1/9} = 9.2 * 10^{-4}[/itex].

    Plus it's clear that …

    [itex]K_{sp} = [Ag^{+}][NC^{-}] = [s/9]^{1/9}[/itex] is probably dramatically different from [itex]K_{sp} = [Ag^{+}][NC^{-}] = [/itex].
     
    Last edited: Apr 6, 2014
  7. Apr 6, 2014 #6

    Borek

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    You have calculated something you call "revised Ksp", it has nothing to do with Ksp as is defined in a standard way, and you are surprised its value is different from the standard one?

    Ksp is defined, calculated and determined using equilibrium concentrations of ions, not using solubility.
     
  8. Apr 6, 2014 #7

    Qube

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    Yes, the equilibrium concentration of ions is what I plugged into my Ksp; I used the measured molar solubility value from Wikipedia as the empirical value on the EQ concentration of ions. Because we know that in solution

    [itex]s = [Ag^{+}] + [Ag(CN)_{2}^{-}] = [NC^{-}] + [HCN][/itex]

    And that [itex]K_{sp} = [Ag^{+}]_{equilibrium}[NC^{-}]_{equilibrium}[/itex]

    And [itex][Ag^{+}]_{equilibrium} =!![NC^{-}]_{equilibrium}[/itex]

    And from Wikipedia we have the molar solubility of AgCN. We can take it as a decent measure of [itex][Ag^{+}][/itex].

    [itex]s = [Ag^{+}]_{equilibrium} = [NC^{-}]_{equilibrium} + [HCN]_{equilibrium}[/itex]

    And [itex][NC^{-}]_{equilibrium}[/itex] is easily ascertained using the [itex]K_{b}[/itex] of the cyanide ion.

    But we can still go from solubility to equilibrium concentration of ions, right? As shown above?



    I think the issue we have here is that we are lacking a good, formal definition of the solubility product. Does the solubility product really exclude molecular products of salt disassociation? Just about every textbook I've consulted has no say on this specific matter, but then just about every textbook ignores anion hydrolysis, and for those which do brave into the messy chemistry of anion hydrolysis, do not say anything about this issue.
     
    Last edited: Apr 6, 2014
  9. Apr 6, 2014 #8

    Borek

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    We don't lack anything, you just made it up (probably because you are mixing solubility and Ksp). I told what the definition is, and apparently all books you consulted defined it the same way. They don't speak about other ions, because other ions are NOT part of Ksp.

    Yes. They are necessary when you want to calculate solubility, but they are not required for Ksp. Ksp is NOT a measure of solubility, it is an equilibrium constant of the dissolution reaction. When the dissolution is followed by other reaction, they increase solubility, but they are not part of Ksp.
     
  10. Apr 6, 2014 #9

    Qube

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    What do you mean other ions are not part of Ksp? What if I have calcium carbonate; would I not factor in the equilibrium concentrations of not only carbonate ion but also hydrogen carbonate ion?


    Did I calculate Ksp correctly - was the revised Ksp calculated correctly and do you see where I'm going with it?
     
  11. Apr 6, 2014 #10

    Borek

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    I mean exactly what I said. For calcium carbonate

    [tex]K_{sp} = [Ca^{2+}] [CO_3^{2-}][/tex]

    That's by unambiguous definition that you are stubbornly refusing to accept.

    No, you would not. Hydrogen carbonate doesn't matter for Ksp, it matters for solubility - but solubility and Ksp are different things.

    Let's get back to AgCN. Let's say we keep pH buffered at pH that equals pKa of HCN - if so, [CN-]=[HCN].

    [tex]K_{sp} = [Ag^+][CN^-][/tex]

    mass balance is

    [tex][Ag^+] = [HCN] + [CN^-][/tex]

    as we know [CN-]=[HCN], we can write

    [tex][Ag^+] = 2[CN^-][/tex]

    (that works only for this particular pH!)

    We have two equations in two unknowns now. Solve for [Ag+]

    [tex][Ag^+] = \frac {\sqrt{K_{sp}}}{2}[/tex]

    and you will get the AgCN solubility in this particular pH. No need to redefine Ksp, calculated solubility takes presence of both HCN and CN- into account.
     
  12. Apr 6, 2014 #11

    Qube

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    Right, and for that system,

    [itex][Ca^{2+}]_{equilibrium} =![CO_3^{2-}]_{equilibrium}[/itex]

    unless the system is buffered at a certain pH, of course, and so we cannot say

    [itex]K_{sp} = [Ca^{2+}] [CO_3^{2-}] = [/itex], where s is the equilibrium concentration of each ion, or square root of Ksp, or molar solubility.

    Right?

    This is all fine, but we don't have a 1:1 base to conjugate acid buffer solution of silver cyanide. So we cannot define cyanide ion concentration as half the silver ion concentration. Now what? Does my method apply? I don't see why not ... after all I am using the equilibrium concentrations of each ion.
     
  13. Apr 7, 2014 #12

    Borek

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    I am starting to think you are misunderstanding one of the concepts used, I just don't know which one yet.

    Solubility - it is the amount of the substance that dissolves. It doesn't tell anything about equilibrium concentrations (although it puts some restrictions on their values).

    Equilibrium concentration - concentration of a specific ion observed in the solution, once it is at equilibrium. Doesn't matter where the ion comes from.

    Ksp - is expressed using equilibrium concentrations.



    Wrong. Molar solubility is NOT the equilibrium concentration. Furthermore, neither is guaranteed to be equal to square root of Ksp. You already wrote that [Ca2+]≠[CO32-], which precludes this result.

    If so, we do what I explained much earlier - we describe whole system with a full set of equations, describing every mass balance, charge balance, and every equilibrium, and we solve the whole system.
     
  14. Apr 7, 2014 #13

    Qube

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    Well, I wrote molar solubility is the concentration of calcium ion. And equal to the sum of all carbonate derived species.

    I'll have to think about this further.

    Thank you for the insight, I probably did foul up somewhere and I just have to see where.
     
  15. Apr 16, 2014 #14

    Qube

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    Okay, this is what my teacher did, and indeed he did use molar solubility to arrive at Ksp. "Offending" parts in orange boxes. Borek, can you clarify?

    http://i.minus.com/jFjOBJDj83U4M.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  16. Apr 16, 2014 #15

    Borek

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    Sorry, I have serious problems following what he wrote.

    First part - about molar solubility - looks perfectly correct.

    He is right you can't neglect hydrolysis of PO43-. It doesn't stop equation

    [tex]K_{sp} = [Ag^+]^3[PO_4^{3-}][/tex]

    from being right, you just can't plug concentration calculated from the solubility data directly into this definition.

    Molar solubility s is

    [tex]s = \frac {[Ag^+]}3 = [H_3PO_4] + [H_2PO_4^-] + [HPO_4^{2-}] + [PO_4^{3-}][/tex]

    so to properly calculate Ksp from the solubility you should calculate what part of the phosphate is in the form of PO43-. Not trivial, but doable (assuming you know all dissociation constants).

    If I read his print correctly that's exactly what he did, just neglecting further hydrolysis steps.
     
  17. Apr 16, 2014 #16

    Qube

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    Okay, I did what you suggested, and I got 1.65*10^-13. I plugged in the amount of phosphate ion at equilibrium after considering phosphate ion hydrolysis (which is ~99% using molar solubility of silver phosphate as the initial concentration of phosphate ion).

    This answer is several orders of magnitude off from 1) reference values and 2) several more magnitudes off from his calculated value of Ksp for silver phosphate. Here is the remainder of his work in which he actually finds a new Ksp.

    http://i.minus.com/j4fBLdAe9gEB5.png [Broken]

    Also why did he drop the phosphate term from the new Ksp? True, hydrolysis is significant, but still, shouldn't that term be included? The amount of phosphate remaining as he calculated is in the order of something times ten to the negative eighth power; including that term would most definitely have a big impact on the Ksp.
     
    Last edited by a moderator: May 6, 2017
  18. Apr 17, 2014 #17

    Borek

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    No way is it correct. We start with molar solubility in the 10- range, [Ag+]3 alone is 10-15, and that's before even taking PO43- into consideration.

    I disagree with his statement "this is Ksp". This is dissolution equilibrium, but not the solubility product. Ksp has a well defined meaning, mentioned many times in this very thread.
     
    Last edited by a moderator: May 6, 2017
  19. Apr 17, 2014 #18

    Qube

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    Good point. I think I forgot to cube the silver ion concentration. Cubing it and account for the amount of phosphate left upon hydrolysis gives me a Ksp of 4.0 * 10^-23, which not that far off from my prof's calculated value.

    [itex]K_{sp} = [Ag^{+}]^{3}[PO_{4}^{3-}] = [1.55 x 10^{-5}]^{3}[1.55 x 10^{-5} - 1.5489 x 10^{-5}] = 4.0 x 10^{-23}[/itex]


    What do you think the correct Ksp for silver phosphate is? This webpage gives it as something to the negative 17th power.

    http://www.ktf-split.hr/periodni/en/abc/kpt.html [Broken]

    Wikipedia gives a similar value.

    http://en.wikipedia.org/wiki/Silver_phosphate

    Using your definition of Ksp and my teacher's definition of Ksp we both get something to the negative 23rd power. The values however still differ significantly.

    So who's correct?

    Did the literature really not take into account hydrolysis of the phosphate ion?
     
    Last edited by a moderator: May 6, 2017
  20. Apr 17, 2014 #19

    Qube

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    If I understand you correctly, you are saying that Ksp is the ion concentration product, where the ions are the constituent ions of the solid salt. Because hydroxide ion is NOT a constituent ion of silver phosphate, it should be dropped from the Ksp (but can be included in the dissociation equilibrium).
     
  21. Apr 17, 2014 #20

    Borek

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    TBH - no idea. I see pKsp around 17 here (in my reference books) as well, and it clearly doesn't make sense.

    These values are often repeated and copied from older sources, so it is enough that someone at some moment printed wrong value and it is repeated since then. The best approach would be to search the literature, but I have no access.
     
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