To Dr.Bob

The problem you helped me with: A liter of a solution saturated at 25 deg C with calcium oxalate, CaC2O4 is evaporated to dryness, giving a .0061 gm residue of CaC2)4. Calculate the concentrations of the inons, and the molar solubility and the solubility product constant for this salt at 25 deg C. Show the balanced reactions.

Second part of problem: Answer the above question, using a solution of 0.15M CaCl2 as the solvent instead of pure water.

Thx to your help I'm a step closer to understanding those types of problems. Here goes...
I first write the "reversible reaction" because it does not dissociate to completion (correct?.
1) CaC2O4 >>> Ca^2+ + C2O4^2-
<<<

2. Now I find the first part of the question : the concentration of the ions.
.0061 g CaC2O4/L * 1molCaC2O4/128.1g
= 4.76E-5M/L CaC2O4
Because the ions are 1:1 ration,Ca^2 = C2O4- >> 4.76E-5M (first part solved)
Second part of problem: the molar solubility (or how much is dissolved)
Actually, I may have answered it already: it is 4.76E-5M/L
On to the third part of question: Ksp
[Ca^2+][C2O4^2-]= 2.27E-9 ( a constant)

Am I ok so far before I move on to the secon part of the problem? If so...

We used the same dissociation equation above. It is:
CaC2O4>>>(reversible)Ca^2+ + C2O4^2-

From here, I lose it a bit, though I can see the logic you applied to solving it. For example: you said let y = solubility of CaC2O4
Because Ca2+ is also in CaCl2, then it is .15M Since we must use the same Ksp as above then Ksp is 2.27E-9
Then how do I proceed to calculate the concentrations of ions, the molar solubility and the new Ksp since we are using a different solvent: 0.15 M CaCl2. I'm close to understanding the problem. The different solvent is confusing me a bit.
By the way, you mentioned (y + 0.15)y. But if we neglect the inside y, I don't see that we have a quadratic formula. It simply becomes 0.15y and solve for y.

This is what you wrote:
Ksp = (Ca^2+)(C2O4^2-) = ??
(Y + 0.15)(Y) = ??

Is the first part correct. What can I do to solve the second part of problem? Thank you Dr.Bob


You ALMOST have it but not quite. Let me do this in two posts so as to keep one post for being too long. Tne first one. No, CaC2O4 DOES dissociate completely. Hpwever, it doesn't dissolve completey; therefore, it is correct to write a double arrow and show the solid on the left and squeous ions on the right like so.
CaC2O4(s) <---> Ca^+2(aq) + C2O4^-2(aq)
I omitted the aq part in my first post yesterday.
The remainder of your reasoning is ok for #1 and the answers you have are correct.


For #2, there are several points.
Actually, there isn't a different solvent which is another reason I think the problem is not stated very well. Water is still the solvent; it just happens that the solvent has some CaCl2 dissolved in it and that affects the solubility of CaC2O4 because the Ca from CaCl2 provides a common ion to CaC2O4. According to LeChatelier's Principle, the addition of Ca^+2 (from whatever source) will shift the equilibrium to the left and that means less Ca^+2 and less C2O4^-2 and more solid CaC2O4. The net effect is that the solubility of CaC2O4 is decreased.

I usually write two equations to show what is going on.
CaC2O4(s)<===>Ca^+2(aq) + C2O4^-2(aq) AND
CaCl2 ===>Ca^+2 + 2Cl^-(aq)

We want the solubility of CaC2O4 under these new conditions so we call it y mols/L. For every one mole of CaC2O4 that dissolves, the (Ca^+2) = y mols/L and the (C2O4^-2) = y mols/L.

For CaCl2, the problem tells us that it is 0.15M; therefore, (Ca^+2) = 0.15 and (Cl^-) = 0.30M (or mols/L). Note that CaCl2 is VERY soluble and no Ksp is given for it since it completely dissolves. Ksp only works for sparingly soluble electrolytes. CaC2O4, AgCl, AgBr, AgI, Hg2Cl2, CuS, BaSO4, etc etc.are sparingly soluble. Most text list a set of Ksp values in the back of the book. If it is a complete table the ones not listed are soluble.

I have the feeling from your post that although you say we use Ksp from the first part that you don't really believe it is a constant for you ask, "then how do I proceed to calculate the concentrations of ions, the molar solubility and the new Ksp since ------." First, I think you have the concentrations answered. (Ca^+2) is y Molar from CaC2O4 and 0.15M from CaCl2; (C2O4^-2) is y M; (Cl^-) is 0.30M and (CaC2O4) = y M.

The next point is that Ksp is Ksp is Ksp. There is no NEW Ksp. It wouldn't be a constant, would it, if it changed? (The constant IS temperature dependent; i.e., it will change with temperature but it will not change due to concentrations.) That was my second point yesterday when I stated that I thought the problem was stated poorly. Anyway, we go with the Ksp, plug in the concentrations above, and calculate y.
Ksp = (y + 0.15)(y) = 2.27 x 10^-9 and solve for y.

About ignoring the "inside" y. You are absolutely correct that we get away from the quadratic if we say y + 0.15 = 0.15. We can do that IF (and ONLY IF) y is small. The EASY way is to make that assumption, in order to avoid the quadratic, then compare the answer at the end and SEE if y actually is small compared to 0.15. So we would have, if we make the assumption [that y + 0.15 = 0.15], then (0.15)(y) = 2.27 x 10^-9 and y = 1.51 x 10^-8M. Three things are important here. The first is that do you see that the solubility of CaC2O4 has decreased just as we said it would because of the presence of the common ion and the effect of LeChatelier's Principle? The solubility of CaC2O4 in a saturated solution was what? 4.76 x 10^-5 M. Adding 0.15M CaCl2 to it reduces the solubility to 1.51 x 10^-8 M. Pretty dramatic--huh? The second point is how LeChatelier's Principle works because of the common ion. (By the way, adding C2O4^-2 (for example we could have added Na2C2O4) would be a common ion too and it works the same way). The third thing is that now we compare y with y + 0.15 to see if our assumption is correct. If we add 1.51 x 10^-8M to 0.15M we get essentially 0.15 M so our assumption is ok and we avoided solving a quadratic. I know this is lengthy but I hope you get through this. This is such an iomportant concept in chemistry and you will see applications of LeChatelier's Principle thoughtout life (even in events that are not chemical events). Please repost and tell me in detail if something is still confusing.

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asked by Anonymous
  1. what up?!

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    posted by alan
  2. iecsnw jxgardy jknshtx rnbjdakp pyqtivr ubxwny qclkndr

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    posted by ylker hzoduwe
  3. to solve this for the Molar concentration of the Ions then 1.51 ee -8 M + .15 molar becomes the concentration of 2 Ca 2+ and 1.15.ee -8 + .30 M becomes the concentration of 2 Cl-. Ksp the we solve ofr KSP using .15 x .15 ^2?

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    posted by Claudia

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