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Equilibrium Constant of a Salt knowing the solubility
Calculate the equilibrium constant in $\ce\\M ^ 3$ of the heterogeneous system in aqueous solution:
$\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$
Knowing that the solubility of the salt is $15.9 \ce{mg/L}$ .
- equilibrium
- aqueous-solution
- $\begingroup$ Ok but i can't post a photo of my efforts. I tried three times with differents methods but they gave the same result that is wrong. $\endgroup$ – Benjamin Commented Dec 29, 2014 at 13:50
- $\begingroup$ No problem. The point is, we want to keep the community clean from 'users' who only want to get the answer and in other words, want to make the community do their homework. The thing you should do is to show that you have understood the concepts behind solving the problem. e.g.: Say that first I should find the M for $\ce{CaF2}$ and then I should use the formula of measuring the equilibrium constant which is .... $\endgroup$ – M.A.R. Commented Dec 29, 2014 at 13:55
- $\begingroup$ @MARamezani Yes i did like this: Solubility= radq^3 of Kps/4 Because Kps=2s^2 *s (in this case because the salt gives 2 moles of F and 1 of Ca) I found then the result of the Keq, and it is 2.9*10^-7 (this is one of the method, the other one was using molarity for the salt and...) The problem is that three methods gave me this result and the book says that the correct answer is another. $\endgroup$ – Benjamin Commented Dec 29, 2014 at 14:04
The molar solubility $s$ equals:
$$s= \frac{15.9\times 10^{-3}}{78}= 0.2038\times10^{-3} \ce{mol/L}$$
According to the equilibrium: $\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$ , the equilibrium constant is:
$$K_{sp}=4s^3$$
- By substituting the value of $s$ in the above equation, we find: $$K_{sp}=0.034\times10^{-9}$$
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