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Question: A Black ppt is obtained on boiling white ppt. of silver at \(T > {300^o}C\). Identify the compound o...

A Black ppt is obtained on boiling white ppt. of silver at T>300oCT > {300^o}C. Identify the compound of silver having white ppt.
A. Ag2SO4A{g_2}S{O_4}
B. Ag2CO3A{g_2}C{O_3}
C. AgClAgCl
D. None of these

Explanation

Solution

Qualitative chemical analysis is the method for determination of the chemical composition of an unknown sample and it involves the specific tests for the presence of certain elements and if the sample is a complex mixture, then a systematic analysis must be made in order to identify all constituents present in the sample.

Complete answer:
As per given question, we need to identify the silver precipitate on the basis of the observations made while performing specific tests. So, the general properties of precipitate of silver formed while performing specific tests are as follows:

PrecipitateColourKsp{K_{sp}}Solubility (molL−1mol{L^{ - 1}})
Ag2CO3A{g_2}C{O_3}White8.1×10−128.1 \times {10^{ - 12}}1.2×10−41.2 \times {10^{ - 4}} (Most soluble)
AgOHAgOHBrown6.8×10−96.8 \times {10^{ - 9}}8.2×10−58.2 \times {10^{ - 5}}
AgClAgClWhite1.8×10−101.8 \times {10^{ - 10}}1.3×10−51.3 \times {10^{ - 5}}
AgBrAgBrYellow7.7×10−137.7 \times {10^{ - 13}}8.8×10−78.8 \times {10^{ - 7}}
AgIAgIYellow8.3×10−168.3 \times {10^{ - 16}}1.2×10−81.2 \times {10^{ - 8}} (Least soluble)

Now, according to the question we need to find the chemical formula of white precipitate. So as per our observations, it may be Ag2CO3A{g_2}C{O_3} or AgClAgCl because both are white precipitates. We need to perform further tests as mentioned in the question in order to identify the white precipitate as follows:
On initial heating of both the compounds, following products are obtained:
Ag2CO3→ΔAg2O(brown)+CO2A{g_2}C{O_3}\xrightarrow{\Delta }A{g_2}O({\text{brown}}) + C{O_2}
2AgCl→Δ2Ag(black) + Cl22AgCl\xrightarrow{\Delta }2Ag({\text{black) + }}C{l_2}
Hence, on heating of silver carbonate, silver oxide which is a brown precipitate as mentioned in the question. On excess heating of silver carbonate, the reaction takes place as follows:
2Ag2O→Δ4Ag(black)+O22A{g_2}O\xrightarrow{\Delta }4Ag({\text{black}}) + {O_2}
A black ppt. of silver metal is obtained on excess heating of silver carbonate. Therefore, we can conclude that the compound of silver having white ppt is silver carbonate with chemical formula Ag2CO3A{g_2}C{O_3}.
Thus, option (B) is the correct answer.

Note:
Remember that on heating of silver chloride in the presence of light it directly converts into silver metal and releases chlorine gas to show black-white effect which is most commonly used in photographic films. Thus, silver chloride is widely used in the manufacture of photographic emulsions.