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# oxalate concentration for calcium oxalate to precipitate

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7. The Ksp for calcium oxalate (CaC2O4) is 2.7 x 109. If the free Ca+2 concentration were 1mM, what would need to be the oxalate concentration for calcium oxalate to precipitate?

8. What would be the voltage of a Zn|Zn+2||Cu+2|Cu cell, once the concentration of Zn+2 went to 1.5M and the CU+2 concentration diminished to 0.5M?

https://brainmass.com/chemistry/oxidation-reduction-and-electrochemistry/oxalate-concentration-calcium-oxalate-precipitate-133459

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7. The Ksp for calcium oxalate (CaC2O4) is 2.7 x 109. If the free Ca+2 concentration were 1mM, what would need to be the oxalate concentration for calcium oxalate to precipitate?
CaC2O4Â¬ --ïƒ  Ca2+ + (C2O4)2-
If the reaction quotient is greater than the Ksp, precipitation occurs.

We know the concentration of Ca2+ is 1 mM = 0.001 M, so

So the oxalate concentration needs to be greater than for calcium oxalate to precipitate.
I believe the Ksp should be 2.7 x 10-9 since calcium oxalate is only soluble.

8. What would be the voltage of a Znâ”‚Zn+2â•‘Cu+2â”‚Cu cell, once the concentration of Zn+2 went to 1.5M and the CU+2 concentration diminished to 0.5M?
The voltage of any electrochemical cell is a function of the molar concentrations of the compounds involved in the cell, as described by the Nernst equation:
aA(s) + bB+(aq) ---> cC(s) + dD+(aq)
Ecell = EÂ° - (0.06/n) log {[D+]d/[B+]b}

For Znâ”‚Zn2+â•‘Cu2+â”‚Cu cell,
Zn(s) + Cu2+(aq) ---> Cu(s) + Zn2+(aq)
n = 2, then the voltage equation becomes

Now we need to figure out E0, which is the standard electrode potential.
Cu2+(aq) + 2e- ---> Cu(s) EÂ° reduction = +0.34 V
Zn(s) ---> Zn2+(aq) + 2e- EÂ° oxidation = -0.76 V
So
Now we know the two ion concentrations,

So voltage of the cell is 1.0857 V with the given concentrations.

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