How do you calculate the solubility product of calcium fluoride?
Hello everyone,
As already mentioned, this is about calculating a solubility product, and I'm having some difficulties with it. I understand the calculation of solubility, but I'm less familiar with how to calculate the solubility product. Here are my notes.
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You have ions: Ca2+ and two Fluoride F ions.
You calculated the quantity of CaF2. In the solution are however Ca2+ / F- /F-
So KL = x3
If there is no 4 KL=[Ca2+] x (2 [Ca2+])^2.
> KL = x3
Rather not, at least when x represents an obvious concentration. Because of c(F-)= 2 c(Ca2+)
I thank you very much, that confused me!
You have a fat mistake in the formula. The solubility product should be:
K(AB2)=[A]*[B]2
with the unit mol3/L3 for K
Here find the formulas (even if study marter is the last red).
I thank you so I have to pull the 3rd root because of mol^3? But if I use the formula from studysmarter, then I still have 4.41*10^-8…… or do I have to count 0.00021* 0.00021^2? I don't know exactly where my mistake is.
To supplement the other answers, the factor before the fluoride and the exponent after the fluoride have different meanings. The factor in front of it only expresses the stochelometric ratio in pure calcium fluoride, the ions occur in a ratio of 1 to 2. This, however, is not a must the fluoride could occur, for example, in other concentrations. If desired, the concentration of the fluoride is increased by three times the calcium has thus now for the first fluoride that it is necessary to have the triple chance of a "contact" for the second one that it is required the chance also three times, ie, overall it has multiplied.
Thus, if the concentration of calcium x is then the concentration of fluoride 2x and the solubility product is consequently 4x^3
I ask the real chemists for this compelling comparison of forbearance with a mechanism has nothing to do with that of course
How would you have thought that the ^3 disappeared from the unit?
Of course you have to reckon [Ca2+]*[F-]^2, I wrote that. yes, third root.
These are absolute foundations that you also find in the Gerdes or Jander-Blasius.
That's what made me confused, why if that's F2 to double again? How to make sure that the formula is ^2, but not yet a doubling so *2
Have you looked over the link? AK value is given below for CaF2. You need to drag the root when you are recurring K->c.
0.00021mol/L* 0.00042mol/L^2 = K.
I still don't understand it completely! :/ I write the invoice in detail: 0.00021mol/L* 0.00021mol/L^2= 9.261*10^-12mol^3/L^3 then I draw the root and this gives 3.043*10-6mol^3/L^3… so I don't know where from
4*10^-11
Fluoride ion concentration is NOT the same as calcium, but twice. Then you also come to K=4E-11 mol3/L3.
I still thank you. That means the result would be expected and the third. Root etc 3.0431*10^-6 mol/l