Certainly! Let's address the chemical equation and determine if a reaction occurs:
When we mix sodium sulfate [tex]\( \text{Na}_2\text{SO}_4 \)[/tex] with barium iodide [tex]\( \text{BaI}_2 \)[/tex], we need to consider the possible products and whether a precipitate forms. The ions in solution are:
[tex]\( \text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \)[/tex]
[tex]\( \text{BaI}_2 \rightarrow \text{Ba}^{2+} + 2 \text{I}^- \)[/tex]
Upon mixing, the ions can combine to form:
1. Sodium iodide [tex]\( \text{NaI} \)[/tex]
2. Barium sulfate [tex]\( \text{BaSO}_4 \)[/tex]
Now we check the solubility rules:
1. Sodium iodide [tex]\( \text{NaI} \)[/tex] is soluble in water.
2. Barium sulfate [tex]\( \text{BaSO}_4 \)[/tex] is insoluble in water and will form a precipitate.
So, by combining these ions:
[tex]\[ \text{Na}_2\text{SO}_4 (aq) + \text{BaI}_2 (aq) \rightarrow 2 \text{NaI} (aq) + \text{BaSO}_4 (s) \][/tex]
Thus, a reaction occurs and a precipitate of barium sulfate is formed:
[tex]\[ \text{BaSO}_4 (s) \][/tex]
So the balanced chemical equation for the reaction is:
[tex]\[ \text{Na}_2\text{SO}_4 (aq) + \text{BaI}_2 (aq) \rightarrow 2 \text{NaI} (aq) + \text{BaSO}_4 (s) \][/tex]
This confirms that a reaction does occur and results in the formation of a precipitate.
