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I tried to prepare $\ce{Sn(NO3)2}$ using tin(II) chloride and nitric acid according to the reaction

$$\ce{SnCl2 + 2 HNO3 -> Sn(NO3)2 + 2 HCl},$$

but stannous chloride appears to be poorly soluble (turbid solution). What is the correct method to prepare $\ce{Sn(NO3)2}$?

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    $\begingroup$ SnCl2 is pretty well soluble. This reaction is no good for other reasons. $\endgroup$
    – Ivan Neretin
    Commented Feb 7, 2017 at 5:33
  • $\begingroup$ I stay away from all nitrates and cyanides, at all costs. I wanna stay legit in the eyes of everybody. $\endgroup$
    – Dehbop
    Commented Feb 7, 2017 at 6:14
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    $\begingroup$ I don't think that $\ce{Sn(NO3)2}$ exists as a salt. You can evidently make it in solution by dissolving $\ce{Sn(OH)2}$ in dilute nitric acid. $\endgroup$
    – MaxW
    Commented Feb 7, 2017 at 6:31
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    $\begingroup$ Related: chemistry.stackexchange.com/questions/8748/make-tin-nitrate $\endgroup$
    – Ivan Neretin
    Commented Feb 7, 2017 at 7:25

1 Answer 1

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Stannous Nitrate is obtained in white leaflets having the composition $\ce{Sn(NO3)2.20H2O}$ by cooling to $\pu{-20 °C}$. a solution of stannous oxide in nitric acid of density 1.2. It is also formed by the action of highly diluted nitric acid on the metal, some of the acid thus suffering maximum reduction to ammonia: 

$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$

According to C. H. H. Walker, tin dissolves in nitric acid forming stannous and stannic nitrates, the relative proportion of these two salts produced depending on the temperature and the strength of the acid; moreover, the yellowish white precipitate which separates when the somewhat concentrated acid is employed is said to be a hydrated, ill-defined, stannic nitrate.

Ref.: http://tin.atomistry.com/stannous_nitrate.html

You can also check the post linked by Ivan.

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  • $\begingroup$ All of this suggests that you get mixture of compounds every time, IMO there should be a better way. $\endgroup$
    – Mithoron
    Commented Feb 7, 2017 at 14:45

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