Chemistry Help (on behalf of someone else)

[rizwaan]

1) CaF2 is added to a 0.1 M Ca(NO3)2 solution. At what concentration of F- will CaF2 begin to precipitate? Ksp of CaF2 is 4e-11??

2) The solubility of CaF2 (ksp = 4 e-11) in a 0.1 M solution of Ca(NO3)2 is approximately?

Need help finding the solutions for these. I'm posting this for someone else actually who's not that good with using forums. Please be as detailed as you can.

thank you very much in your advance for your help!

 

[Borek]

You were told at chemical forums that nobody is going to do the question for you. You will do better service to your friend showing her/him how to use forums so that we can guide her/him through the question.

 

[rizwaan]

Hi I'm very sorry but I copied and pasted the wrong info.

This one has more of an explanation. Let me know if you need anymore info with this.

1) Given the reactions and thermodynamic data below, calculate the enthalphy of formation for C6H5OH kcal/mol.

(kcal)
C6H5OH + 7 O2 --> 6 CO 2 + 3 H2O 729.8
C + O2 ---> CO2 -94.4
2 H2 = O2 ---> 136.8

I used Hess's law and flipped the first equation and got -729.8 kcal and multiplied the second one by 6 to get 566.4 and added everything to get -26.6 kcal.
But this is not the right answer...apparently is +41.7?

2)
The balanced equation below is for a nonsponstaneous reaction (enthalpy of formation = 131 kj/mol and
entropy of formation = 134 J/(mol x k) ). Assuming that the change in enthalpy and entropy do not vary with temperature, at what temperature will the reaction become spontaneous?

C(s) + H2O(l) --->CO(g) + H2(g)

To solve this I used delat G = delta H - T(delta S).
I got 978 K and subtracted it by 273 to get 704 C. However the answer is 1022 C?

Thank you very much for your help.

 

[Borek]

I used Hess's law and flipped the first equation and got -729.8 kcal and multiplied the second one by 6 to get 566.4 and added everything

And you have ignored third equation? Show what you got (what reaction equation) after doing what you wrote you did.

 

[DeeNos]

1) To determine the concentration of F- at which CaF2 will begin to precipitate, we need to use the solubility product constant (Ksp) of CaF2. Ksp is the equilibrium constant for the dissolution of a sparingly soluble salt, and it is defined as the product of the concentrations of the ions raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. In this case, the balanced equation for the dissolution of CaF2 is CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq). Therefore, the Ksp expression for CaF2 is Ksp = [Ca2+][F-]^2.

To determine the concentration of F- at which CaF2 will begin to precipitate, we need to set up an ICE (initial, change, equilibrium) table and use the Ksp expression to calculate the equilibrium concentration of F-. The initial concentration of Ca2+ is 0.1 M, and since CaF2 is a strong electrolyte, it dissociates completely into Ca2+ and F-. Therefore, the initial concentration of F- is also 0.1 M.

ICE Table:

CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
I 0 M 0.1 M 0.1 M
C -x +x +2x
E 0 M (0.1+x) M (0.1+2x) M

Using the Ksp expression, we can set up the equilibrium expression as Ksp = (0.1+2x)(x)^2 = 4e-11. Solving for x, we get x = 6.32e-6 M. This means that at a concentration of 6.32e-6 M F-, CaF2 will begin to precipitate.

2) To determine the solubility of CaF2 in a 0.1 M solution of Ca(NO3)2, we need to use the same approach as above. Since Ca(NO3)2 is a strong electrolyte, it dissociates completely into Ca2+ and 2NO3-. Therefore, the initial concentration of Ca2+ is 0.1 M and the initial concentration of F- is 2(0.1) = 0.2 M.

ICE Table:

CaF