2) Identify the Unknown (# 1) a) Addition of pH 0.5 H 2 S resulted in formation

ID: 967624 • Letter: 2

Question

2) Identify the Unknown (# 1)

a) Addition of pH 0.5 H2S resulted in formation of a precipitate.

b) Addition of 3M KOH caused this precipitate to dissolve (gave a solution)

c) Addition of 6M HCl followed by pH 0.5 H2S resulted in a yellow precipitate.

What can you conclude from these results___________________________

3) Unknown # 2

a) A test tube may contain Bi2S3 and/or CuS.

b) 6M HNO3 is added to this test tube, mixed and then centrifuged.

c) To the resulting supernatant liquid 6M NH3 is added and the tube centrifuged again.

e) A white precipitate and colorless supernatant liquid result.

What do these observations tell you___________________________

This is inorganic salt analysis for basic radicals

when H2S is added in pH 0.5 , it shows group radicals higher than II group

addition of KOH dissolved the ppts it shows III group is absent

again adding HCl and H2S there were ppts (yellow colored) it showed absence of v group radicals as they ppt only in basic medium

so VI th group radical is present

a) it showed presence of Bi /cu sulphides which is possible if in second group add H2S

b) on adding HNO3 ppt dissolved and adding NH3 sol white ppt it shows presence of Bi as Bi(OH)3 is white in colour

Had it been Cu(OH)2 ..it would have been red/green

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