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Questions and Answers
Which of the following cations does NOT belong to Group I cations?
Which of the following cations does NOT belong to Group I cations?
- Bismuth (Bi^{3+}) (correct)
- Mercurous (Hg_2^{2+})
- Silver (Ag^{+})
- Lead (Pb^{2+})
What is the role of HCl in the precipitation of Group I cations?
What is the role of HCl in the precipitation of Group I cations?
- To form soluble salts of the cations
- To prevent the formation of insoluble salt of bismuth (correct)
- To oxidize the cations
- To increase the solubility of the precipitates
Why should a slight excess of HCl be added during the analysis?
Why should a slight excess of HCl be added during the analysis?
- To fully dissolve all the cations
- To obtain complete precipitation and prevent peptization of the insoluble chlorides (correct)
- To increase the temperature of the solution
- To ensure the chlorides remain soluble
What is the formula of the precipitate formed by the addition of cold dilute HCl in Group I cations?
What is the formula of the precipitate formed by the addition of cold dilute HCl in Group I cations?
Which of the following chemical equations represents the formation of lead chloride precipitate?
Which of the following chemical equations represents the formation of lead chloride precipitate?
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Study Notes
Group I Cations (Silver Group)
- The group includes lead, silver, and mercurous, with their respective ions: $Pb^{2+}$, $Ag^{+}$, and $Hg_2^{2+}$
Steps of Analysis
- Precipitation is the first step of analysis, using cold dilute $HCl$ as the group reagent
- The group reagent $HCl$ precipitates the ions as chlorides, with the following formulas: $AgCl$, $PbCl_2$, and $Hg_2Cl_2$, resulting in a white precipitate
- $HCl$ is preferred over $NH_4Cl$ because it prevents the formation of insoluble bismuth oxychloride precipitate
- A slight excess of $HCl$ should be added to ensure complete precipitation and prevent peptization of the insoluble chlorides
- The chemical equation for the precipitation of $PbCl_2$ is: $Pb^{2+} + 2Cl^- + H_2O \rightarrow PbCl_2 \downarrow$
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