Synthetic Inorganic Chemistry

DIFFEEENT BEHAVIOR OF OXIDES AND PEROXIDES 161

sodium peroxide cool completely, and then drop it into the

dilute acid. Test the solution with titanium sulphate. The

characteristic test will be obtained.

(d) Prepare another sample of sodium peroxide in the

same way. Scrape it off the porcelain into a dry test tube.

Add 5 cc. of water. Note the effervescence and that the

gas inflames a glowing splinter. Boil the solution until

effervescence ceases. Acidify with sulphuric acid and test

with titanium sulphate. The test is negative.

In (d) an amount of oxygen is evolved corresponding to one-
half the oxygen of the Na 2 O 2. (See also discussion of Experi-
ment 11, p. 70.) This leaves in the solution the oxide Na 2 O or
rather NaOH the product of the reaction of Na20 with water.

Na 2 O 2 + H 2 O -»2 NaOH + JO,
On acidification we get a simple neutralization;
2NaOH + H 2 SO 4 ^ Na 2 SO 4 + 2H 2 O
In (c) the peroxide has not been given a chance to decompose
before the treatment with acid. Since hydrogen peroxide was
found as a product, the reaction is probably

Na 2 O 2 + H 2 SO 4 -> Na 2 SO 4 + H 2 O 2

Evaporation of the solution yields crystals of Na 2 SO 4 -10H 2 O
identical in every respect to the sodium sulphate that can be
obtained from (d), thus confirming the above equations.
In the same way barium oxide BaO neutralizes an acid, with the
formation of water:
BaO + 2HN0-, -»Ba (NO 3 ), + H 2 O
and barium peroxide yields hydrogen peroxide:
BaO 2 + 2HNO 3 ->Ba(NO 3 )2 + H 2 O 2
It is obvious that the reaction of the peroxide with an acid is a
metathesis in which the O 2 radical is concerned just as the 0 is
concerned in the neutralization of an ordinary oxide. It is fur-
thermore obvious from the formulas H2O2 and Na 2 O 2 and BaO 2 ,
if we ascribe the ordinary valence to hydrogen, sodium, and barium,
that the valence of the O 2 radical is 2.
Thus the peroxides are not really simple binary compounds in