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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.

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  1. Photon336
    • one year ago
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    Can someone tell me whether this is the correct logic towards approaching this problem? \[CaF _{2} --> Ca ^{2+} +2F ^{-}\] \[Ksp = [x][2x]^{2}\]

  2. Rushwr
    • one year ago
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    yeah that's right I guess !

  3. Rushwr
    • one year ago
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    I'm not sure if it is approaching or not but its correct

  4. Rushwr
    • one year ago
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    What's the volume of CaCl2 in the addition of HF?

  5. Photon336
    • one year ago
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    They did not give it

  6. Photon336
    • one year ago
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    \[10.0 mL/1000*CaCl _{2} * (0.01) M = 1*10^-4 mol CaCl _{2} \] \[20.0mL/1000*(0.01mol/L) HF = 2x10\]

  7. Photon336
    • one year ago
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    2x10^-4 mol of HF

  8. Photon336
    • one year ago
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    need to figure out the concentrations of both ions

  9. Photon336
    • one year ago
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    but they said that the volume was diluted to 1L.

  10. Photon336
    • one year ago
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    HF --> 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

  11. Photon336
    • one year ago
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    \[ Ksp=[x][2x]2\] [1x10^-4M][2(2x10^-4)]^2 = Qsp (1.6x10^-7)(1x10^-4) = 1.6x10^-11

  12. Photon336
    • one year ago
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    \[Qsp = 1.6x10^-11 \]

  13. Photon336
    • one year ago
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    Ksp = 4.1x10^-11

  14. Rushwr
    • one year ago
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    So its gonna be a no

  15. Photon336
    • one year ago
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    \[Qsp < Ksp \] so no it won't.

  16. Photon336
    • one year ago
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    Yeah, 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.

  17. Photon336
    • one year ago
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    1. 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.

  18. Photon336
    • one year ago
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    You 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.

  19. Photon336
    • one year ago
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    But 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.

  20. Photon336
    • one year ago
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    Qsp is the [ ] of ions at any point while Ksp is the equilibrium value.

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