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In more powerful toothpastes, like Crest Pro-Health, stannous fluoride is used in place of sodium fluoride.

According to Wikipedia:

Stannous fluoride converts the calcium mineral apatite into fluorapatite, which makes tooth enamel more resistant to bacteria generated acid attacks. In toothpastes containing calcium minerals, sodium fluoride becomes ineffective over time, while stannous fluoride remains effective in strengthening tooth enamel

Is the stannous fluoride more effective in this reaction simply due to having two fluoride ions per mole, or is there something more complicated in the mechanism of the reaction in which the tin spectator ion is beneficial?

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  • $\begingroup$ I took a look at the paper that Wiki links to, but the abstract doesn't tell anything about SnF2 and in my opinion just adds to the confusion. Do you know anyone who could provide you with the full article? $\endgroup$
    – Antimon
    Commented Oct 17, 2012 at 15:48
  • $\begingroup$ @Antimon I will try to get a hold of it in the next couple of days. Thanks for checking it out anyway. I get the feeling the actual results may be in another (cited) study in the paper. $\endgroup$
    – jonsca
    Commented Oct 17, 2012 at 23:12

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Found an answer to this question in a book called 'Pediatric Dentistry'.

Apparently, when Stannous Fluoride reacts with dental enamel (hydroxyapatite), both the fluoride and the tin react to form Stannous Trifluorophosphate $(\ce{Sn3F3PO4})$ -- This is highly resistant to acid attack.

At high concentrations of stannous fluoride (I don't know what counts as "high", the book doesn't say), apparently the following compound is formed as well: Trifluorostannate -- This is also highly acid resistant.

In addition, Calcium Fluoride is also produced, which then further reacts with hydroxyapatite forming Fluorhydroxyapatite (which is also rather acid resistant, although it seems that the stannous trifluorophosphate is more resistant.)

Sodium fluoride only produces Fluorhydroxyapatite as its reaction.

It appears that Stannous Fluoride is a superior agent because it produces 1 or 2 additional compounds (that become part of the surface enamel) that are more acid resistant than just the one that Sodium Fluoride produces. These other compounds (which become a part of the surface of the tooth) also have an anti-microbial effect, thus inhibiting the growth of the very bacteria which are directly responsible for producing the majority of the acid found in dental decay.

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$\ce{SnF2}$ appears to have a predominantly covalent structure while $\ce{NaF}$ is ionic. So I think $\ce{SnF2}$ might release fluoride ions slower (more reluctantly) than $\ce{NaF}$.

Since toothpaste contains $\ce{Ca}$ containing minerals as well (as mentioned in the Wikipedia page), highly insoluble $\ce{CaF2}$ is formed easily when more fluoride ions are present. This is bad since $\ce{CaF2}$ will be inactive.

So since $\ce{SnF2}$ doesn't release fluoride ions as much (unless it comes into contact with water – it undergoes hydrolysis fast. Again this is good since more fluoride will be available only while you are brushing your teeth.), it has a longer shelf life than $\ce{NaF}$ I guess.

This also means that it doesn't matter what fluoride you use as long as its covalent. However I suspect tin is used because it happens to be one of the covalent fluorides relatively harmless to us (see this).

This is all just a guess so take it with a grain of salt... I don't have many sources to back it up (except Wikipedia of course!). Do point out any mistakes.

EDIT

As Georg has mentioned in the comments, toothpastes have a significant amount of water. Apparently toothpastes containing stannous fluoride contain a strong complexing agent which prevents oxidation of the $\ce{Sn^2+}$ ion (I don't have access to the paper but I read the abstract here).

Also the abstract mentions that $\ce{Sn^2+}$ has a antibacterial effect (most probably due to it being a heavy metal ion) and this is another reason why it is used (in contrast to $\ce{Na+}$ which has almost no antibacterial effects).

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  • $\begingroup$ So You think toothpastes are free of water? Think about stability of SnII with respect to oxidation and write a new, hoprfully less nonsensy answer. $\endgroup$
    – Georg
    Commented Apr 16, 2013 at 10:28
  • $\begingroup$ @Georg I agree that I didn't consider that toothpastes have a significant amount of water. I have already mentioned that it was just a guess. There is nothing more to be said... $\endgroup$
    – kaliaden
    Commented Apr 16, 2013 at 12:40
  • $\begingroup$ I have edited in why $SnF_2$ is stable even though there is water in the toothpaste. $\endgroup$
    – kaliaden
    Commented Apr 16, 2013 at 13:04
  • $\begingroup$ I'll accept this for now, but if you run into any sources supporting your position, please feel free to incorporate them in. And, thanks! $\endgroup$
    – jonsca
    Commented Apr 21, 2013 at 23:48

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