Calculate solubility of fluorite in water

[hexa]

Hello,

I posted this already in the chemistry forum, not noticing there's a homeowrk help forum here. Sorry for that. I'm still 2 more weeks abroad and the person with my on this field work didn't take this course so she's also not a big help for me. Other people who have to take this exam are also currently abroad at other location, so asking them doesn't work either. And nobody has the key to my appartment so sending the book was also no option. Still no big problems with other topics though.

Here's the original posting:
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I'm currently abroad and have to learn for a chemistry exam which will take place the day after I'm back home. Unfortunately I've forgotten to take along a book and now have difficulties solving one type of problem, also as the local library doesn't have books in a language I understand and there's just one computer with very expensive internet connection here.

One example problem:

Chlorite Mg5Al2SiO3O10(OH)8 dissolves incongruently.

Write down the reaction assuming that H4SiO4^0 is part of that reaction.

If ground water at 25 degrees celsius with a molarity of 10^-3.38 is in equilibrium with chlorite and the ph is 8.5, what is the concetration of H4SiO4^0 in mg/l. Assume that a=m

Another problem:

Calculate the solubility of fluorite (CaF2) in water at 10, 20 and 30 degrees celsius and express the answer of Ca^2+ in mg/L (gamma = 1).

There are more, rather similar problems in previous exams. I just chose those two as representative examples. Please can someone at least explain to me how to solve them? If I had that book with me it would not be a problem to look it up myself but at the moment I'm only guessing around. No problems yet with other topics as they are explained nicely in the book I actually took along.

thanks a lot,
hexa

 

[GCT]

those can't be the original form of the problems, if you want to get an answer post the exact problem.
for instance,

Calculate the solubility of fluorite (CaF2) in water at 10, 20 and 30 degrees celsius and express the answer of Ca^2+ in mg/L (gamma = 1).

assuming they actually want you to calculate the solubility, they would have at least given you some data to work with.

 

[thepatientmental]

Hello hexa,

No worries about posting in the wrong forum, we are happy to help with any homework questions here! Let's tackle the first problem about calculating the solubility of fluorite (CaF2) in water.

To solve this problem, we need to use the solubility product constant (Ksp) for CaF2, which is 3.45 x 10^-11 at 25 degrees Celsius. This value represents the equilibrium constant for the dissociation of CaF2 into Ca^2+ and 2F^- ions in water.

To calculate the solubility, we can use the following equation:

Ksp = [Ca^2+][F^-]^2

Where [Ca^2+] represents the concentration of Ca^2+ ions in solution and [F^-] represents the concentration of F^- ions in solution.

Since we are given the temperature (10, 20, and 30 degrees Celsius) and asked to express the concentration in mg/L (which is equivalent to ppm), we can use the following conversions:

10 degrees Celsius = 283.15 K
20 degrees Celsius = 293.15 K
30 degrees Celsius = 303.15 K

Now, let's plug in the values and solve for [Ca^2+].

At 10 degrees Celsius:
3.45 x 10^-11 = [Ca^2+][F^-]^2
3.45 x 10^-11 = [Ca^2+](2[F^-])^2
3.45 x 10^-11 = [Ca^2+][F^-]^2
3.45 x 10^-11 = [Ca^2+] x (2[F^-])^2
3.45 x 10^-11 = [Ca^2+] x 4[F^-]^2
3.45 x 10^-11 = [Ca^2+] x 4 x (10^-3.38)^2
3.45 x 10^-11 = [Ca^2+] x 4 x 10^-6.76
3.45 x 10^-11 = [Ca^2+] x 4 x 10^-6.76
3.45 x 10^-11 = [Ca^2+] x 4 x 10^-6.76
3.45 x 10^-11 = [Ca^2+] x 4 x 10