# Solubility and Gibbs free energy Qs

1. Apr 9, 2010

### LeilaNami

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
ΔG0=-RTlnK
Kp=Kc(RT)Δn(gas)

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
Kc=.958
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. Apr 9, 2010

### 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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