Use a table of soluble and insoluble ionic compounds to predict the precipitate that will form in the reaction of Na3(PO4) + CsBr
1.Only NaBr will precipitate
2.Only Cs3(PO4) will precipitate
3.Both NaBr and Cs3(PO4) will precipitate
4.There will be no precipitate
The Correct Answer and Explanation is :
To determine whether a precipitate will form when Na₃PO₄ reacts with CsBr, we need to analyze the solubility of the potential products. This reaction involves the exchange of ions between sodium phosphate (Na₃PO₄) and cesium bromide (CsBr).
The reaction can be written as:
[
\text{Na}_3\text{PO}_4 (aq) + \text{CsBr} (aq) \rightarrow \text{NaBr} (aq) + \text{Cs}_3\text{PO}_4 (s)
]
Step 1: Identifying Products
The two potential products in this reaction are sodium bromide (NaBr) and cesium phosphate (Cs₃PO₄).
- Sodium bromide (NaBr): Sodium (Na⁺) and bromide (Br⁻) are both ions that typically form soluble compounds. According to solubility rules, most salts of alkali metals (like Na⁺) are soluble in water, and bromides (Br⁻) are also typically soluble, with a few exceptions (e.g., silver bromide, AgBr). Since sodium and bromide are both from groups that form soluble compounds, NaBr is soluble in water and will not form a precipitate.
- Cesium phosphate (Cs₃PO₄): Cesium (Cs⁺) is an alkali metal, and phosphate (PO₄³⁻) compounds are generally insoluble, with exceptions primarily for alkali metal phosphates. However, cesium phosphate (Cs₃PO₄) is one of the few phosphate compounds that is soluble because cesium is an alkali metal, and alkali metal phosphates are generally soluble. Thus, Cs₃PO₄ will dissolve in water and will not form a precipitate.
Step 2: Conclusion
Since both NaBr and Cs₃PO₄ are soluble in water, no precipitate will form in this reaction.
Thus, the correct answer is:
4. There will be no precipitate.
This conclusion aligns with the solubility rules for ionic compounds: alkali metal halides (like NaBr) are soluble, and alkali metal phosphates (like Cs₃PO₄) are also soluble. Therefore, no insoluble products are produced in this reaction, and the solution remains clear.