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Photon336
 one year ago
The solubility constant of calcium fluoride CaF2 is 4.0x10^11 will 10.0mL
Will of 0.01M cacl2 and 20.0 mL of 0.01M HF to added sufficient water to make a 1.0L solution. Would CaF2 precipitate because of this?
Yes
No
Yes if temperature is increased
There is insufficient data.
Photon336
 one year ago
The solubility constant of calcium fluoride CaF2 is 4.0x10^11 will 10.0mL Will of 0.01M cacl2 and 20.0 mL of 0.01M HF to added sufficient water to make a 1.0L solution. Would CaF2 precipitate because of this? Yes No Yes if temperature is increased There is insufficient data.

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Can someone tell me whether this is the correct logic towards approaching this problem? \[CaF _{2} > Ca ^{2+} +2F ^{}\] \[Ksp = [x][2x]^{2}\]

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0yeah that's right I guess !

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure if it is approaching or not but its correct

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.0What's the volume of CaCl2 in the addition of HF?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1They did not give it

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[10.0 mL/1000*CaCl _{2} * (0.01) M = 1*10^4 mol CaCl _{2} \] \[20.0mL/1000*(0.01mol/L) HF = 2x10\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1need to figure out the concentrations of both ions

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1but they said that the volume was diluted to 1L.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1HF > H + F CaCl2 > Ca + 2Cl 1 mol of Ca ions. 1 mol of F ions from those equations. i'm guessing that this is the next step. so we do 1x10^4 mol/1L = 10^4 M Ca ions 2x10^4 mol/1L = 2x10^4 M F ions

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[ Ksp=[x][2x]2\] [1x10^4M][2(2x10^4)]^2 = Qsp (1.6x10^7)(1x10^4) = 1.6x10^11

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[Qsp = 1.6x10^11 \]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[Qsp < Ksp \] so no it won't.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, this was a long problem but I think what you had to do was to write out the reactions and keep track of how many ions you had total for each.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.11. you multiply the Molarity of each by the volume to get the number of moles. 2. you find the new concentration of each compound (new volume) for 1L 3. then you figure out how many total moles of ions you have, in this case for each of them it's one. 4. after that you set up the Ksp for CaF2 which was [x][2x]^2 5. plug in and solve.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1You also realize that Qsp < Ksp which means that no precipitate will form. if you raised the temperature you'll change the value of your Ksp and if you don't add any more of those compounds nothing will happen as a result. it can't be yes. Ksp is an equilibrium value, so with all equilibrium values they will be temperature dependent.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1But the general idea behind this question was that you didn't really have enough of both substance to really cause precipitate to form. for that Qsp >Ksp for that to happen.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1Qsp is the [ ] of ions at any point while Ksp is the equilibrium value.
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