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vera_ewing

  • one year ago

At equilibrium, the concentration of calcium ions in a calcium chromate (CaCrO4) solution is 1 x 10^–4 mol/L at 25°C. In the same solution, the concentration of chromate ions is also 1 x 10^–4 mol/L. What is the Ksp of calcium chromate? Show your work.

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  1. vera_ewing
    • one year ago
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    @aaronq Please help!

  2. welshfella
    • one year ago
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    sorry I've forgotten this stuff

  3. aaronq
    • one year ago
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    Write the dissociation, then it's corresponding equilibrium equation (which is the Ksp equation). Plug in your values and solve

  4. vera_ewing
    • one year ago
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    What's the dissociation and corresponding equilibrium equation?

  5. aaronq
    • one year ago
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    the dissociation of CaCrO4 (in water)

  6. aaronq
    • one year ago
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    and by that i mean a chemical equation

  7. vera_ewing
    • one year ago
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    Okay, so what does the chemical equation look like? Can you set it up and then I solve?

  8. aaronq
    • one year ago
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    nah, you have to try it

  9. vera_ewing
    • one year ago
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    I don't know how :( Can you maybe explain it?

  10. aaronq
    • one year ago
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    CaCrO4 is an ionic compound, right? what ions make up CaCrO4?

  11. vera_ewing
    • one year ago
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    Not sure. It's covalent and ionic though, right?

  12. aaronq
    • one year ago
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    it's ionic only. You should be able to identify the ions here, if you can't look the compound up on wikipedia and read a little about it

  13. welshfella
    • one year ago
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    what i do remember is that Ca CrO4 disassociates in water to [Ca++] and [Cr2O4 --]

  14. welshfella
    • one year ago
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    and sp means solubility product so i guess Ksp is the product of Ca++ ions and CrO4-- ions

  15. vera_ewing
    • one year ago
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    Yep, I figured it out :) thanks

  16. welshfella
    • one year ago
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    so that 10-4 8 10^-4 = 10^-8

  17. welshfella
    • one year ago
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    oh Ok wd

  18. Photon336
    • one year ago
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    CaCrO4 ----> Ca^2+ + CrO4^2- Ksp = [Ca^2+][CrO4^2-] Ksp = [x][x] Ksp = concentration of ions dissolved, like any K it's an equilibrium value, and it is temperature dependent. I'm assuming that the equilibrium is already reached and that the solution is saturated. B.c from my understanding that's what Ksp is. (Qsp = Ksp) Their concentrations are equal for both ions in the problem. So I think it's (1x10^-4)(1x10^-4) = Ksp. 10^-8

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