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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.
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