I am dissolving tin in nitric acid. What reactions will take place depending on the conentration of $\rm{HNO}_3$ and temperature?
Particularly, I am interested when the metastannic acid $\rm{H_2SnO_3}$ is likely to be produced, according to
$\rm{Sn + 4 HNO_3 → H_2O + H_2SnO_3 + 4 NO_2}$
P. S. References to books or scientific articles are welcome.
It is a common belief that tin dissolves in dilute nitric acid to produce tin(II) nitrate. The reactions are addressed in a previous answer of mine. However, employing a more hot and concentrated nitric acid yields an unstable compound called hydrated stannic oxide also known as β-stannic acid. The compound also has a α-form which is later converted to β form. The relevant information can be found here:
[...]tin dissolves slowly, in very dilute nitric acid to produce stannous nitrate (but) when hot and more concentrated acid is employed, is very unstable and quickly decomposes, yielding the form of hydrated stannic oxide known as β-stannic acid. Probably α-stannic acid is first produced [...] which then passes into the β-form. The product is, therefore, liable to contain both α- and β-forms, and the pure β-acid is obtained by dissolving this product in sodium hydroxide solution, and then adding excess of concentrated soda, which precipitates sodium p-stannate whilst the α-salt remains in solution. Pure β-stannic acid, or β-metastannic acid, is then obtained by decomposing the sodium salt with acid. It is also formed by the hydrolysis of its sodium salt at 60° C., and by boiling a dilute solution of stannic chloride, with or without the addition of nitric acid.[...]
[...]Gelatinous, precipitated β-stannic acid has the empirical composition $\ce{SnO2.4H2O}$, when air-dried $\ce{SnO2.2H2O}$, and when dried in a vacuum, $\ce{SnO2.H2O}$; [...]
