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Chemistry Homework Help => CBSE XI and XII Chemistry => Topic started by: Pranjal Singh on February 04, 2022, 02:18:54 AM

Title: Quantitative analysis
Post by: Pranjal Singh on February 04, 2022, 02:18:54 AM
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Title: Re: Quantitative analysis
Post by: chenbeier on February 04, 2022, 07:12:22 AM
Answer 3 is Chromium. In reducing conditions the ion is green Cr3+. Oxidation guides to Chromate CrO4 2- , what forms a yellow precipitate of PbCrO4
Title: Re: Quantitative analysis
Post by: uma on February 04, 2022, 07:13:55 PM
Quote from: chenbeier on February 04, 2022, 07:12:22 AM
Answer 3 is Chromium. In reducing conditions the ion is green Cr3+. Oxidation guides to Chromate CrO4 2- , what forms a yellow precipitate of PbCrO4
A good answer
Title: Re: Quantitative analysis
Post by: jakecunningham on February 22, 2022, 09:25:02 AM
Third option is the correct one.
Explanation:
The green colour is produced in the borax bead test of a chromium (III) salt. This green colour is due to Cr(BO2)3
Na2B4O7 . 10H2O         Na2B4O7       NaBO2 + B2O3
NaBO2 + Cr2O3         Cr(BO2)  (green)
And When chromium ion is oxidised by hydrogen peroxide in the basic solution, yellow chromate ion is formed by the following reaction:
2Cr3+ + 3H2O2 + 10OH- ⇋       2CrO42- + 8H20          (CrO42-  is yellow)
When it is treatment with Pb2+   It forms precipitate of PbCrO4 which is also of yellow colour.

Pb2+ + CrO42-   ⇋   PbCrO4(s)  (Yellow)