294 ELEMENTS OF GROUP IV
to produce the maximum precipitate and again the amount
necessary to redissolve the precipitate.
Repeat using 2 cc. of IN Pb(NO 3 ) 2 instead of SnCl 2 and
note that a very much larger volume of the NaOH is necessary
to redissolve the precipitate.Pb(OH) 2 is much more basic than Sn(OH) 2 ; it is correspond-
ingly more weakly acidic, as is shown by the greater excess of base
required to convert it to the soluble salt.
H 2 Sn0 2 + 2Na0H -» Na 2 Sn0 2 + 2H 2 O
H 2 Pb0 2 + 2Na0H -»• Na 2 Pb0 2 + 2H 2 OSodium stannite and sodium plumbite both hydrolyze easily but
the latter much more so, consequently the greater amount of base
to overcome its tendency to hydrolyze.
- Stannic Acid. Heat 0.5 gram of tin in a casserole with
a little 16 N HN0 3. Note that red gases are evolved, that
the metal disintegrates, and that a white powder insoluble
in the nitric acid, and later insoluble in water, is formed.
Concentrated nitric acid oxidizes tin to the dioxide which, in a
hydrated form, usually called meta-stannic acid (approximately
H 2 Sn0 3 ), is left as the white insoluble residue.
- Thio-Salts of Tin. Perform Experiment 1 under
Preparation 43. Stannous sulphide does not dissolve in
Na 2 S solution. Addition of sulphur causes it to dissolve.
Addition of HC1 to the solution produces a yellow precipitate
and an evolution of hydrogen sulphide.
Sulphur and oxygen are interchangeable in sulphides and oxides.
Metal oxides (basic) and non-metal oxides (acidic) combine to
form salts. Likewise metal sulphides may combine with sulphides
of weakly metallic or non-metallic elements to form salts, the so-
called thio-salts, or sulpho-salts. Thio-salts of a few of the ele-
ments, notably tin, are very well denned. Stannous sulphide does
not form a thio-salt; but addition of sulphur converts it to stannic
sulphide which does form a soluble thio-salt with the sulphide of
an alkali metal.
Na 2 S + SnS 2 -»• Na 2 SnS 3