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Homework Help: Solubility and Gibbs free energy Qs

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data
    There are a problems I wasn't sure about on my test. Trying to get answers for ones I had to guess on.

    1. What is the maximum [Mg] that can be achieved in an aqueous solution of Mg(OH)2 with a pH=12.50? Ksp (Mg(OH)2)=5.6e-12

    2. Determine the max solubility of CaF2 that can dissolve in a 0.10 M HF solution. Given Ksp (CaF2)=1.5e-10 and Ka (HF)= 3.5e-4

    3. Consider the reaction below:
    N2(g) + 3H2(g) -> 2NH3
    The standard free energy change for this reaction is ΔG0 = -33.3 kJ at 25oC. What is the reaction mixture consisting of 0.50 atm N2, 0.50 atm H2 and 10.0 atm NH3?

    4. For the following, determine Kp at 25C.
    NH4HS (s) <--> H2S (g) + NH3 (g)
    ΔH0=83.47 kJ/mol and ΔG0=17.5 kJ/mol

    2. Relevant equations
    Ksp = [products]n
    Ka = [products]n/[reactants]m

    3. The attempt at a solution

    1. So I have no idea how to do this one with a certain pH and don't have a clue even where to start. I know how to do this without a given pH however.

    2. I think this uses the common ion effect.
    HF + H2O <--> F- + H3O
    Using the Ka given I found that [F]= 5.74e-3 M
    Since this is a common ion, it pushes the reaction back towards CaF2 and so technically, the amount dissolved is the same as [Ca].
    CaF2 -> Ca + F2
    So Ksp = [2S]2
    1.5e-10 = [Ca][(5.74e-3)+2S]2
    [Ca]=??? I'm not getting one of the answer choices

    3.Kp=P(NH3)2/[P(N2) x (P(H2))3]
    Kp= 1.6e3
    1.6e3 = Kc[(0.0821)(298)]-2
    ΔG = -33.3 + (8.314e-3)(298)ln(.958) = ??? Not getting one of the answer choices

    4. Not getting anywhere close to one of the answers. Pretty much tried to solve it the same as #3 but in reverse I guess.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 9, 2010 #2


    User Avatar

    Staff: Mentor

    1. If pH is given you can calculate concentration of [OH-], rest should be obvious - this is a common ion effect.

    2. What are possible answers? My bet is that you can ignore F- from the CaF2 dissolution, I tried with and without and the difference was around 0.2%.

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