Problem 11
Question
Sodium Carbonate cannot be used in place of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) for the identification of \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}\) and \(\mathrm{Sr}^{2+}\) ions (in group V) during mixture analysis because : (a) \(\mathrm{Mg}^{2+}\) ions will also be precipitated. (b) Concentration of \(\mathrm{CO}_{3}^{2-}\) ions is very low. (c) Sodium ions will react with acid radicals. (d) \(\mathrm{Na}^{+}\)ions will interfere with the detection of \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}\) ions.
Step-by-Step Solution
Verified Answer
(a) Sodium Carbonate precipitates Mg²⁺ ions.
1Step 1: Understanding the Role of Ammonium Carbonate
In group V cation analysis, \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+},\) and \(\mathrm{Sr}^{2+}\) ions are usually identified by precipitating them as carbonates using ammonium carbonate. Ammonium carbonate provides carbonate ions \(\left(\mathrm{CO}_{3}^{2-}\right)\) which precipitate these metal ions as their respective carbonates: \(\mathrm{CaCO}_{3}, \mathrm{BaCO}_{3}, \text{and} \mathrm{SrCO}_{3}\).\. In this reaction, ammonium ions \(\left(\mathrm{NH}_4^{+}\right)\) help maintain a neutral pH, preventing other metal ions from precipitating.
2Step 2: Evaluating Sodium Carbonate as a Substitute
Sodium carbonate \(\left(\mathrm{Na}_2\mathrm{CO}_3\right)\) can also provide carbonate ions for precipitating \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}, \text{and} \mathrm{Sr}^{2+}\) ions. However, its use poses complications. The sodium ions \(\left(\mathrm{Na}^{+}\right)\) in sodium carbonate can react with other ions in the solution, potentially leading to incorrect conclusions.
3Step 3: Effect on Magnesium Ions
Importantly, sodium carbonate could also precipitate magnesium ions \(\left(\mathrm{Mg}^{2+}\right)\) present in the mixture because magnesium carbonate \(\left(\mathrm{MgCO}_{3}\right)\) is also insoluble in water. This does not happen with ammonium carbonate under similar conditions because \(\mathrm{NH}_4^{+}\) ions prevent the precipitation of magnesium carbonate by competing with \(\mathrm{Mg}^{2+}\).
4Step 4: Selecting the Correct Reason
The presence of sodium carbonate would cause difficulties due to its effect of precipitating \(\mathrm{Mg}^{2+}\) ions along with the group V ions which is not desirable. Therefore, \(\mathrm{Mg}^{2+}\) precipitation by sodium carbonate would interfere with the correct identification of \(\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}, \text{and} \mathrm{Sr}^{2+}\), validating option (a) as the correct answer.
Key Concepts
Cation IdentificationGroup V CationsCarbonate PrecipitationInterference in Ion Detection
Cation Identification
Cation identification is a crucial step in qualitative analysis, which involves detecting and identifying the presence of cations in a mixture. These positively charged ions can belong to various groups and require specific methods for accurate detection. In qualitative analysis, identifying cations like \(\mathrm{Ca}^{2+},\mathrm{Ba}^{2+},\mathrm{Sr}^{2+}\) is important due to their chemical properties and reactions.
One common method of identifying cations is through precipitation reactions. These reactions help isolate specific cations from a mixture based on the formation of insoluble compounds. During analysis, reagents are carefully chosen to ensure that only the target cations precipitate, leaving others in the solution. This precision aids in differentiating between similar cations, making the identification process both accurate and efficient.
One common method of identifying cations is through precipitation reactions. These reactions help isolate specific cations from a mixture based on the formation of insoluble compounds. During analysis, reagents are carefully chosen to ensure that only the target cations precipitate, leaving others in the solution. This precision aids in differentiating between similar cations, making the identification process both accurate and efficient.
Group V Cations
Group V cations, which include \(\mathrm{Ca}^{2+},\mathrm{Ba}^{2+},\mathrm{Sr}^{2+}\), play a significant role in qualitative analysis due to their distinct reaction patterns. These cations are usually detected by precipitating them as carbonates using ammonium carbonate. This method is particularly effective because these metal ions form insoluble carbonate salts, visible as precipitates.
- Ammonium carbonate ensures that other ions, like magnesium, do not precipitate, as \(\mathrm{NH}_4^{+}\) ions maintain a neutral pH.
- The selective precipitation of Group V cations helps in accurately identifying them even in complex mixtures with multiple ions present.
Carbonate Precipitation
Carbonate precipitation is a fundamental technique used to isolate cations like \(\mathrm{Ca}^{2+},\mathrm{Ba}^{2+},\mathrm{Sr}^{2+}\). This process involves the formation of insoluble carbonate salts when carbonate ions react with these metal ions.
To achieve this, ammonium carbonate is often preferred as it effectively provides the carbonate ions necessary for precipitation without causing unwanted reactions with \(\mathrm{Mg}^{2+}\) ions.
To achieve this, ammonium carbonate is often preferred as it effectively provides the carbonate ions necessary for precipitation without causing unwanted reactions with \(\mathrm{Mg}^{2+}\) ions.
- Ammonium carbonate's decomposition releases \(\mathrm{NH}_4^{+}\) and carbonate ions, which facilitate the precipitation process.
- The controlled pH due to \(\mathrm{NH}_4^{+}\) also ensures specificity, preventing neutral magnesium carbonate from forming.
Interference in Ion Detection
Interference in ion detection occurs when unintended reactions cause inaccurate identification or measurement of ions. When using sodium carbonate instead of ammonium carbonate, sodium ions in the solution can lead to several problems.
- Sodium ions (\(\mathrm{Na}^{+}\)) can react with other ions, leading to unexpected precipitates and complicating the analysis.
- Additionally, \(\mathrm{Mg}^{2+}\) ions might precipitate as \(\mathrm{MgCO}_3\), which is undesirable if you are specifically testing for Group V cations.
Other exercises in this chapter
Problem 10
The cation that will not be precipitated by \(\mathrm{H}_{2} \mathrm{~S}\) in the presence of dil. \(\mathrm{HCl}\) is : (a) \(\mathrm{Pb}^{2+}\) (b) \(\mathrm{
View solution Problem 10
An organic compound ' \(A\) ' is oxidized with \(\mathrm{Na}_{2} \mathrm{O}_{2}\) followed by boiling with \(\mathrm{HNO}_{3}\). The resultant solution is then
View solution Problem 11
Which one of the following is likely to give a precipitate with \(\mathrm{AgNO}_{3}\) solution ? \(\quad\) (a) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}\) (b) \
View solution Problem 12
Upon treatment with ammoniacal \(\mathrm{H}_{2} \mathrm{~S}\), the metal ion that precipitates as a sulfide is (a) \(\mathrm{Fe}(\mathrm{III})\) (b) Al(III) (c)
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