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The solubility product of a salt having same molecular formula as $\ce{Ca_3(PO_4)_2}$ is $1.08\cdot10^{-28}$. Then find the number of moles of ions present in the saturated solution.

My approach: I found out the solubility of the salt i.e.: $x=10^{-6}M$

This means $x$ moles of salt are present in a litre. As 1 mole of the salt yields 5 moles of ions I thought the answer would be $5x$.

Help required.

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  • $\begingroup$ Did the given answer not match your answer? $\endgroup$ – Satwik Pasani May 25 '14 at 13:49
  • $\begingroup$ What is the solubility product equal to? $\endgroup$ – LDC3 May 25 '14 at 13:51
  • $\begingroup$ I don't know the answer. It is a MCQ question. The options did not match my answer. $\endgroup$ – Rudstar May 25 '14 at 14:29
  • $\begingroup$ Perhaps you have to consider hydrolysis of the phosphate anion. $\endgroup$ – Dissenter May 25 '14 at 16:25
  • $\begingroup$ @Rudstar Your approach looks correct to me. It would be helpful if you provided the actual question and the answer options. For example, was $\ce{K_sp}$ given to you or did you look it up? What salt was used in the problem? Did they want the answer in moles/L or gm/L? $\endgroup$ – ron May 25 '14 at 16:58
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For your salt, calcium phosphate, or $\ce{Ca_3(PO_4)_2}$, this is the correct dissolution equilibrium:

$\ce{Ca_3(PO_4)_2 \leftrightharpoons 3Ca^{2+} + 2PO_{4}^{3-}}$

Upon dissolution of one unit of calcium phosphate, we get five units of ions total, true, but the solubility product is defined as follows (from the IUPAC Gold Book):

The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.

For our purposes, we will use concentrations instead of activities. Concentrations are a rough gauge of activities in very dilute solutions (such as this one involving the semisoluble salt, calcium phosphate).

Also since you mentioned this was a multiple-choice test, I highly doubt that the following hydrolysis reaction will be relevant, even though at such low concentrations of phosphate anion, the phosphate anion should be nearly completely hydrolyzed.

$\ce{PO_{4}^{3-} + H_2O ->HOPO_3^{2-} + HO^-}$

Back to the problem. Upon dissolution of one unit of calcium phosphate, we should get 3 units of calcium ion and 2 units of phosphate ion.

Therefore, based on the IUPAC definition:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}$

Taking $\ce{x}$ to be one unit of calcium ion:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}=[3x]^3[2x]^2$

Solving the above yields $x = 10^{-6}$.

Therefore, with respect to your original question, the number of moles of calcium phosphate present initially can be number greater than 0 moles. For the amount of calcium phosphate present at equilibrium, then yes, $5x$ appears to be the right answer.

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  • $\begingroup$ Doesn't that just give the same answer for x that Rudstar reported above? $\endgroup$ – ron May 25 '14 at 16:51
  • $\begingroup$ I'm fairly sure it shouldn't because it appears he set $K_{sp}$ equal to $5x$. But then I haven't run through the calculations myself. $\endgroup$ – Dissenter May 25 '14 at 16:52
  • $\begingroup$ Oops, I see that I misinterpreted the question because I didn't read the title. Sorry! $\endgroup$ – Dissenter May 25 '14 at 16:53

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