Chapter 34: Problem 82
Three separate samples of a solution of single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. \(\mathrm{NaCl}\) and one formed a black precipitate with \(\mathrm{H}_{2} \mathrm{~S} .\) The salt could be (a) \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{MnSO}_{4}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (d) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\)
Short Answer
Step by step solution
Understand Precipitate Formation
Evaluate Reactions with Ammonia
Analyze Black Precipitate Formation
Final Analysis
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Precipitate Formation
- Silver ions (\( \mathrm{Ag^+} \)) and chloride ions (\( \mathrm{Cl^-} \)) come together to form \( \mathrm{AgCl} \).
- The \( \mathrm{AgCl} \) is insoluble, hence it precipitates out as a white solid.
Ammonia Reactions
- Silver nitrate (\( \mathrm{AgNO_3} \)), upon reacting with ammonia, forms a complex ion. Excess \( \mathrm{NH_3} \) reacts with the silver to slowly dissolve \( \mathrm{Ag_2O} \) back into the solution, allowing for complex ions.
- Mercury(II) nitrate (\( \mathrm{Hg(NO_3)_2} \)) also forms a white precipitate when added to ammonia due to the formation of coordination complexes.
Sulfide Precipitation
- This reaction occurs because metal sulfides are typically highly insoluble in water.
- The formation of such precipitates is pivotal in qualitative analysis for detecting the presence of specific metal ions.
Lead Nitrate Reactions
- When reacting with sodium chloride (\( \mathrm{NaCl} \)), it forms a white precipitate of lead chloride (\( \mathrm{PbCl_2} \)), identifying the presence of lead ions.
- In reaction with hydrogen sulfide (\( \mathrm{H_2S} \)), it results in a black precipitate (\( \mathrm{PbS} \)), which is highly characteristic and useful for identification of lead.