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Will NaF + CaCO3 precipitate much CaF2?
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Alan Jia
Will NaF + CaCO3 precipitate much CaF2?
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.
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.
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.More
When you say "And the F- ions from the NaF will just drive the Ca++ concentration down even lower". Isnt 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...?More
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++. Thats why I suggested adding CO2 - to raise Ca++.More
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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
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.
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
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.
@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.More
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? Ill work on a concrete example and cite your answer assuming my expansion is correct.More
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.
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.
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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.
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.
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