1
$\begingroup$

If I mix sodium fluoride, calcium carbonate powder, and water, under what conditions (if any) would there be an equilibrium?

I.e. assume we start out with fully dissociated $\ce{Na+(aq) + F-(aq)}$, because it's fairly soluble. But when I add the $\ce{CaCO3}$, it's only slightly soluble, so assume an excess of solid $\ce{CaCO3}$ and a small amount dissolved. Will that be stable with $\ce{NaF}$ concentration much greater than the saturated concentrations of $\ce{CaCO3}$, or will $\ce{CaF2}$ start to precipitate, and will it continue until exhaustion of a reactant, or reach equilibrium?

I'm wondering if it's the relative solubilities of $\ce{CaCO3}$ and $\ce{CaF2}$ that will drive the equilibrium (if that's what it is) one way or the other. I can't work out how to apply the theory of an equilibrium constant because of the excess solid reactant and corresponding possible precipitate messes up how concentrations balance out in a normal equilibrium… And I'm wondering if the absolute concentration of $\ce{NaF}$ effects things as well.

Any pointers appreciated. Or even some jargon to say what this kind of reaction is called.

$\endgroup$
  • 3
    $\begingroup$ I'm voting to close this question as off-topic because this is a personal/medical question since you're try to formulate a toothpaste. To give any kind of advice on a personal care product is just letting the camels nose under the tent. $\endgroup$ – MaxW Jan 11 at 8:48
  • 1
    $\begingroup$ Removed post script. Now it's a general question $\endgroup$ – Neil Stephens Jan 11 at 8:57
  • $\begingroup$ I may be wrong but its not possible to make CaF2 because CaF2 is soluble and CaCO3 is not . $\endgroup$ – aryan bansal Jan 11 at 14:10
  • 1
    $\begingroup$ No, they're both slightly soluble $\endgroup$ – Neil Stephens Jan 11 at 20:46
  • $\begingroup$ Updated question to be as general as possible so that it is on-topic. $\endgroup$ – Neil Stephens Jan 12 at 2:53
2
$\begingroup$

Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate.

The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's well below the solubility of NaF.

However, the presence of additional carbonate could be used to reduce the calcium concentration. Similarly, any addition of acid will convert carbonate to bicarbonate and increase the maximum calcium concentration.

UPDATE: In the above answer, I assumed a basic familiarity with solubility products. For those not familiar with those, here's more detail.

The key quantitative measure of solubility of ionic compounds is the solubility product, usually indicated as $K_{sp}$, which is the product of the concentrations of the separate ions.

For $\ce{CaCO3}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{CO3^2-}]$. Reported values vary somewhat, but are typically around $2\times 10^{-8}$.

Likewise, for $\ce{CaF2}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{F-}][\ce{F-}]$, and the reported values are around $4\times 10^{-11}$. Thus, the condition for keeping fluoride in solution is

$[\ce{Ca^2+}][\ce{F-}][\ce{F-}]<4\times 10^{-11}$.

If $\ce{CaCO3}$ is dissolved in distilled water to maximum solubility, the calcium ion concentration is $\sqrt{K_{sp}}\approx 0.15$ mM, which is the basis of the calculation above.

$\endgroup$
  • $\begingroup$ Attempt to expand your explanation: CaF2 will start to precipitate as long as the concentration of fluoride ions is > ~0.5mM and calcium ions > 0.25mM. Since there is an excess of CaCO3, the precipitation will continue until either the fluoride ions are reduced to 0.5mM or enough carbonate ions build up to prevent more than 0.25mM of calcium dissolving...? But the 'common ion effect' not only works with the carbonate. Fluoride ions will lower the solubility of CaF2, so that has to be accounted for as well? I'll work on a concrete example and cite your answer assuming my expansion is correct. $\endgroup$ – Neil Stephens Jan 11 at 21:19
  • $\begingroup$ One has to consider calcium concentration not originated from CaCO3 dissolution, in the case the solvent contains any. $\endgroup$ – Poutnik Jan 12 at 6:38
  • 1
    $\begingroup$ @Poutnik - I interpreted the question to mean that CaCO3 is the only other source of calcium, and I specified "distilled water" in the answer to make clear that I am assuming no other source of calcium. $\endgroup$ – Andrew Jan 12 at 14:28
  • $\begingroup$ I understand, but both cases are worthy to consider $\endgroup$ – Poutnik Jan 12 at 14:30
  • $\begingroup$ I think the question was pretty clear that CaCO3 is the only source of calcium $\endgroup$ – Neil Stephens Jan 14 at 7:33
1
$\begingroup$

Consider three compositions:

A. 2NaF + CaCO3 B. CaF2 + Na2CO3, and

C. NaF + 0.5 CaCO3 + 0.5 CaF2 + 0.5 Na2CO3.

Using data from the CRC Handbook (62nd ed), the heats of formation of A and B are respectively 560.47 and 560.6 kcal, so there is little driving force to make a reaction go to completion. Note that A should be near neutral pH, but B and C would be about 12 pH.

The difficulty with trying to make A become B is that CaCO3 is so insoluble that there aren't enough Ca++ ions to make a new crystal of CaF2 rapidly. And the F- ions from the NaF will just drive the Ca++ concentration down even lower - and it doesn't get better as the reaction proceeds, because the CO3-- ion produced is just as effective at keeping the Ca++ level down.

But if you make up system C, which is just 50% reacted and followed its pH over time, you might be able to tell if the reaction goes forward or backward, or more likely, just remains the same. The reaction just doesn't have enough dissolved Ca++ to react in a reasonable time.

But an acid would increase the concentration of Ca++ ions. CO2 does the least change to the system; then the system becomes D:

D. 2NaF + CO2 + H2O + CaCO3 --> CaF2 + 2NaHCO3.

The left-hand side of the equation now has a total heat of formation of 727.47 kcal and the right-hand side has 743.3 kcal, a change of 2.1%. Not much, but maybe that would force the system at a more reasonable rate to the desired reactants. NaHCO3 gives a pH of about 8.5. Fascinating that Ca(HCO3)2 is more soluble than NaHCO3! The CO2 that becomes entrapped in the NaHCO3 is a reactant, not a catalyst. The CO2 could be applied in excess (pressure) so that it is partially a catalyst, dissolving the CaCO3 and CaF2 until you dry the mix, at which time the CO2 volatilizes.

$\endgroup$
  • $\begingroup$ When you say "And the F- ions from the NaF will just drive the Ca++ concentration down even lower". Isn't that actually a driver for the CaF2 to form? Because them being in solution reduces the solubility of CaF2, but not CaCO3, so the F- does push down Ca++, but more should dissolve from CaCO3 until CO3-- builds up...? $\endgroup$ – Neil Stephens Jan 12 at 20:41
  • $\begingroup$ Yes, F- from NaF drives the formation of CaF2. But with CaCO3 alone in H20, or with CaF2 alone in water, the Ca++ concentration is low; adding NaF (case A) or Na2CO3 (case B) drives the Ca++ concentration VERY low because the concentration of F- from NaF is way greater than CO3-- from CaCO3 and the concentration of CO3-- from Na2CO3 is way greater than F- from CaF2. What you say could be the best explanation if the reaction goes rapidly, but I think the rate would be very slow because of the low concentration of Ca++. That's why I suggested adding CO2 - to raise Ca++. $\endgroup$ – James Gaidis Jan 12 at 22:02
  • 1
    $\begingroup$ Another reason the reaction might slow down, in addition to low concentrations of Ca++, is that F- might coat the CaCO3 rather than precipitate as a separate crystal of its own. That would stifle dissolution of CaCO3. The same could happen with CO3-- on CaF2. $\endgroup$ – James Gaidis Jan 14 at 4:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.