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Homework Help: Products or Reactants ?

  1. Aug 16, 2009 #1
    1. The problem statement, all variables and given/known data
    When the following substances are introduced into water do the resulting equilibria favour products or reactants?

    a) NH4F

    These are other questions like this one:

    3. The attempt at a solution

    The only way i can determine if products or reactants are favoured is if i do a Keq calculation. If Keq > 1 then its products. If Keq < 1 then its reactants.

    [tex]NH_{4}F \rightleftharpoons NH4^{+} + F^{-}[/tex]

    This is again a hydrolysis reaction because i have to determine which ion is the spectator ion. It says in my book that:
    "Spectator ions can be thought of as being the result of dissolving a strong base or a strong acid. The positive ion (cation) is considered to result from a base, BOH, and the negative ion (anion) is considered to result from an acid, HA"

    If thats correct then the cation (NH4+) came from NH3 and the anion (F-) came from HF so that the entire equilibria is:

    [tex]NH_{3} + HF \rightleftharpoons NH_{4}F + H_{2}O[/tex]

    But neither NH3 or HF are strong acids or bases. I'm stuck. How can I proceed through with the rest of this question ? I still don't understand Hydrolysis that im stuck here in the very beginning of this question. PLEASE HELP.

    The answer btw is:
    reactants; Keq = 8.5 x 10-7
  2. jcsd
  3. Aug 16, 2009 #2
    Firstly, [tex]NH_{4}F \rightleftharpoons NH4^{+} + F^{-}[/tex] is wrong. An ionic salt dissociates completely in aqueous media.
    Construct the individual hydrolysis equations for the ions:

    [tex]NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{3} + H_{3}O^{+}[/tex]

    [tex]F^{-} + H_{2}O \rightleftharpoons HF + OH^{-}[/tex]

    So, the equilibria equation is actually [tex]NH_{4}^{+} + F^{-} \rightleftharpoons NH_{3} + HF[/tex], which yields
    [tex]K_{C} = \frac{[NH_{3}][HF]}{[NH_{4}^{+}][F^{-}]}[/tex]
    From the Ka and Kb values of the hydrolysis of these ions, the ratios can be computed.
  4. Aug 16, 2009 #3


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    Staff: Mentor

    You can't ignore water.

  5. Aug 16, 2009 #4
    both ions come from weak acid/base is that the reason why you included the hydrolysis of both ions ?

    WoW... ur right i got 8.787878 x 10-7 is it close enough to the answer in a couple post prior?
  6. Aug 16, 2009 #5
    I have two other problems which I COMPLETELY DO NOT UNDERSTAND. I'm serious.

    c) H3PO4 and Na2HPO4
    d) H2O2 and kHS
    e) MgO
    f) H2S and LiNO2

    starting with c):

    I separated H3PO4 and Na2HPO4 into ions:
    [tex]H_{3}PO_{4} \rightarrow H_{2}PO_{4}^{-} + H^{+}[/tex]
    [tex]Na_{2}HPO_{4} \rightarrow 2Na^{+} + HPO_{4}^{2-}[/tex]

    and combining the anions:
    [tex]H_{2}PO_{4}^{-} + HPO_{4}^{2-} \rightleftharpoons HPO_{4}^{2-} + H_{2}PO_{4}^{-}[/tex]

    [tex]Keq = \frac{K_{a} \mbox{(reactant acid)}}{K_{a} \mbox{(product acid)}} = \frac{ \mbox{6.3 x 10^{-8}}}{ \mbox{6.3 x 10^{-8}}}[/tex]

    There is something wrong with that, what am I doing wrong ?

    starting with d):

    [tex]H_{2}O_{2} \rightarrow HO_{2}^{-} + H^{+}[/tex]
    [tex]KHS \rightarrow K^{+} + HS^{-}[/tex]

    [tex]HO_{2}^{-} + HS^{-} \rightleftharpoons O_{2}^{2-} + H_{2}S[/tex]
    [tex]HO_{2}^{-} + HS^{-} \rightleftharpoons H_{2}O_{2} + S^{2-}[/tex]

    This is so confusing :confused:

    starting with e):

    [tex]MgO \rightarrow Mg^{2+} + O^{2-}[/tex]

    [tex]Mg^{2+} + 2H_{2}O \rightleftharpoons Mg(OH)_{2} + 2H^{+}[/tex]

    [tex]O^{2-} + 2H_{2}O \rightleftharpoons H_{2}O + 2OH^{-}[/tex]
    -------------------------------------- (You told me not to ignore water so here it is)
    [tex]Mg^{2+} + O^{2-} + 4H_{2}O \rightleftharpoons Mg(OH)_{2} + H_{2}O + 2H_{2}O[/tex]

    Simplifies down to:
    [tex]Mg^{2+} + O^{2-} + H_{2}O \rightleftharpoons Mg(OH)_{2}[/tex]

    Now, HOW can you tell which one are acids/bases.

    Starting with f):

    [tex]H_{2}S \rightarrow HS^{-} + H^{+}[/tex]
    [tex]LiNO_{2} \rightarrow Li^{+} + NO^{2-}[/tex]

    HS- Ka = 1.3 x 10-13
    Kb = 1 x 10-7
    Kb > Ka therefore HS- will react as a base in solution

    Li+ came from a strong base (LiOH) so Li+ will not hydrolyze but NO2- will. NO2- will act as a base in solution.

    BUT HOW CAN THEY BOTH BE BASES? One of them has to be an acid and one of them a base. I know that NO2- will be a base because it doesnt have another hydrogen to ionize. So I know NO2- is going to be a base and HS- will be an acid when the anions are mixed.

    [tex]HS^{-} + NO^{2-} \rightleftharpoons HNO_{2} + S^{-2}[/tex]

    [tex]K_{eq} = \frac{K_{a} \mbox{(reactant acid)}}{K_{a} \mbox{(product acid)}}= \frac{ \mbox{1.3 x 10^{-13}}}{ \mbox{5.1 x 10^{-4}}}[/tex]

    [tex] = 2.5 x 10^{-10}[/tex] (which is WRONG!)

    What can I do now ?

    The answers:

    c: products; Keq = 1.1 x 105
    d: reactants; Keq = 2.4 x 10-5
    e: products; Keq > 10^22 (acid = H2O)
    f: reactants; Keq = 2.0 x 10-4

    Im in need of a lot of clarification.
    Last edited: Aug 16, 2009
  7. Aug 17, 2009 #6
    Sorry don't have much time atm, so will only answer c) for now. Will look at rest later.
    The equilibria in (c) should be individually:
    [tex]H_{3}PO_{4} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + H_{3}O^{+}[/tex]
    [tex]HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + OH^{-}[/tex]

    [tex]H_{3}PO_{4} + HPO_{4}^{2-} \rightleftharpoons 2H_{2}PO_{4}^{-}[/tex],
    (Phosphoric acid is a weak acid and only dissociates partially.)
    which yields
    [tex]K_{eq} = \frac{[H_{2}PO_{4}^{-}]^{2}}{[H_{3}PO_{4}][HPO_{4}^{2-}]} = \frac{K_{a}(H_{3}PO_{4}) K_{b}(HPO_{4}^{2-})}{10^{-14}}[/tex]

    Qualitatively, this is an acid-base reaction, and thus the position of the equilibrium will lie on the side of the products, as the depletion of the H+ and OH- ions drives the individual equilibria towards the formation of products.
    Last edited: Aug 17, 2009
  8. Aug 17, 2009 #7
    How did you get 10-14 for the denominator ? and why? I know its Ka for H2O but why ?
  9. Aug 17, 2009 #8
    and also what method are you using? The method in my worksheet is kinda different. I have scanned the sheet for you so you can see that different method.
  10. Aug 17, 2009 #9
    [H_{3}O^{+}][OH^{-}] = 10^{-14}[/tex]
    But from your equations [tex][H_{3}PO_{4}] = [H_{3}O^{+}][/tex] and [tex] [HPO_{4}^{2-}] = [OH^{-}][/tex], not forgetting you started with only the reactants. Thus,
    [tex][H_{3}PO_{4}][HPO_{4}^{2-}] = 10^{-14}[/tex]
  11. Aug 18, 2009 #10
    I figured out all the other questions. I did another way and got the same answer. The only one giving me trouble is e) MgO

    To see my method:

    [tex]H_{3}PO_{4} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + H_{3}O^{+}[/tex]

    [tex]Na_{2}HPO_{4} \rightarrow 2Na^{+} + HPO_{4}^{2-}[/tex]

    [tex]HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{2-} + OH^{-}[/tex]

    [tex]H_{3}PO_{4} + HPO_{4}^{2-} \rightleftharpoons H_{2}PO_{4}^{-} + H_{2}PO_{4}^{-}[/tex] (Note that i didnt include [tex]2H_{2}PO_{4}^{-}[/tex] just so I could see the distinction between which species reacted as acids/bases)

    K_{eq} = \frac{K_{a} \mbox{(product acid)}}{K_{a} \mbox{(reactant acid)}} = \frac{7.1 \mbox{x} 10^{-3}}{6.3 \mbox{x} 10^{-8}} = \mbox{112698.4127 which is approx. 1.1 x 10^{5}}

    Now, for that troublesome e) MgO

    I know that its a salt and in my worksheet it said all salts are soluble and 100% ionized so,

    [tex]MgO \rightleftharpoons Mg^{2+} + O^{2-}[/tex]

    [tex]Mg^{2+} + 4H_{2}O \rightleftharpoons \underbrace{Mg(OH)_{2}}_{ \mbox{That is my first guess}} + 2H_{3}O^{+}[/tex]

    [tex]O^{2-} + H_{2}O \rightleftharpoons OH^{-} + OH^{-}[/tex]

    When combined:
    [tex]Mg^{2+} + O^{2-} \rightleftharpoons Mg(OH)_{2} + OH^{-}[/tex]

    Now wheres [tex]K_{a} \mbox{(reactant acid)}[/tex] the "reactants" dont have any ionizable hydrogen to behave as an acid. I am stuck with this one. Please help.
  12. Aug 18, 2009 #11
    You forgot the water again, methinks. (look at the equations you stated more carefully) H2O is the acid in this case.
    Last edited: Aug 18, 2009
  13. Aug 18, 2009 #12
    What do you mean?
  14. Aug 18, 2009 #13
    When you combined your equations, your water disappears.
  15. Aug 18, 2009 #14
    No it doesnt. This is what I get.

    [tex]Mg^{2+} + O^{2-} + H_{2}O \rightleftharpoons Mg(OH)_{2}[/tex]
    Last edited: Aug 18, 2009
  16. Aug 18, 2009 #15
    Exactly. You had left out the H20 in post #10. So in this case H2O is your "acid".
  17. Aug 18, 2009 #16
    The key equilibria should be:
    [tex]O^{2-} + H_{2}O \rightleftharpoons 2OH^{-}[/tex] (almost a direct arrow!)
    [tex]Mg^{2+} + 2OH^{-} \rightleftharpoons Mg(OH)_{2}[/tex]
    The second one is essentially ignorable due to the relatively low Ksp of magnesium hydroxide, and given that we expect a very very high concentration of hydroxide ions.
  18. Aug 18, 2009 #17
    what about [tex]K_{a} \mbox{(product)}[/tex] in post #14 I know now thanks to you that [tex]K_{a} \mbox{(reactant)}[/tex] is [tex]H_{2}O[/tex] in post #14

    All these problems came from these worksheets:
    Last edited: Aug 18, 2009
  19. Aug 18, 2009 #18
    Two last sheets giving a total of 5 pages.

    I just hope someone can explain these problems to me so i can understand them faster.
  20. Aug 19, 2009 #19
    There's no [tex]K_{a} \mbox{(product)}[/tex] for the second equilibria, which is not a dissociation of an acid or a base, but a solubility equilibria. So, you can treat the second equilibria as negligible as the high concentration of OH- from equilibria one would drive the second equilibria to the extent that most of the Mg2+ would be precipitated out as the product.
  21. Aug 19, 2009 #20
    (d) Hydrogen peroxide is a weak acid, while the hydrogen sulfite ion is a weak base.
    [tex]H_{2}O_{2} + HS^{-} \rightleftharpoons HO_{2}^{-} + H_{2}S[/tex]

    (f) Once again, hydrogen sulfite is a weak acid, and thus you do not dissociate it and work with the conjugate base. Hence,
    [tex]H_{2}S + NO^{2-} \rightleftharpoons HS^{-} + HNO_{2}[/tex]

    Key takeaway: Leave the weak acids and bases as they are and don't dissociate them completely to give the conjugate bases/acids and construct your equilibria from there -> the equilibria starts with the weak acids/bases, which only partially dissociate in water.
  22. Aug 19, 2009 #21
    Can you explain a little bit more clearer ?
  23. Aug 19, 2009 #22
    Can you explain a little bit more clearer ? Also, how do you arrive at an answer such as in the worksheet with Keq > 1022 ??? how did they get such a big number and from where ?
  24. Aug 20, 2009 #23
    You still haven't explained to me WHY and HOW in the worksheet there is that answer. You are not helping me when you just explain what YOU think. I want you to actually take a look in page 5 of those scanned pages i took the time to scan for you so you could help me out. If you still do not want to help me I'll take my post somewhere else where somebody else is better at explaining things than you guys. I was very patient with you BUT how much patience can I take..
  25. Aug 20, 2009 #24


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    I tried to take a look at the scan #5, but after clicking once to open and once more to bring up a window with an enlarged version, neither of the two forms of the image were readable.
  26. Aug 20, 2009 #25
    Make sure guys, that you download adobe reader *THE LATEST ONE* from adobe.com. If you still can't see it in ur browser window then download the picture and save it to wherever you want and then open it up with whatever default program ur computer uses to view images.

    I also know that Mg(OH)2 is a somewhat strong base but not completely strong. Does it have a Kb ?
    Last edited: Aug 20, 2009
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