Answer :
To solve the problem and determine which product will precipitate out, we need to consider the solubility rules provided and the nature of the reactants in the given chemical reaction:
[tex]\[ K_2SO_4(\text{aq}) + Ba(NO_3)_2(\text{aq}) \rightarrow ? \][/tex]
The reactants are potassium sulfate ([tex]\(K_2SO_4\)[/tex]) and barium nitrate ([tex]\(Ba(NO_3)_2\)[/tex]), both in aqueous form. When these compounds react, they undergo a double displacement reaction:
[tex]\[ K_2SO_4(\text{aq}) + Ba(NO_3)_2(\text{aq}) \rightarrow KNO_3(\text{aq}) + BaSO_4(\text{s}) \][/tex]
Let's break it down in detail:
1. Formation of Potassium Nitrate [tex]\(KNO_3\)[/tex]:
- According to Rule 2, nitrates [tex]\((NO_3^-)\)[/tex] are always soluble. Therefore, potassium nitrate [tex]\(KNO_3\)[/tex] will dissolve in water and remain in an aqueous form.
2. Formation of Barium Sulfate [tex]\(BaSO_4\)[/tex]:
- According to Rule 4, sulfates [tex]\((SO_4^{2-})\)[/tex] are generally soluble except when they are combined with barium [tex]\((Ba^{2+})\)[/tex], calcium [tex]\((Ca^{2+})\)[/tex], strontium [tex]\((Sr^{2+})\)[/tex], or lead [tex]\((Pb^{2+})\)[/tex]. In this case, the sulfate is combining with barium:
[tex]\[ Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \][/tex]
- Barium sulfate [tex]\(BaSO_4\)[/tex] is insoluble in water, meaning it will form a precipitate.
After analyzing the reaction, the precipitate that forms is barium sulfate [tex]\(BaSO_4\)[/tex].
Let's match this to the answer choices:
- A. [tex]\(KNO_3\)[/tex] – This compound is soluble and does not precipitate.
- B. [tex]\(Ba(NO_3)_2\)[/tex] – This is a reactant, not a product, and it remains in solution.
- C. [tex]\(K_2SO_4\)[/tex] – This is a reactant, not a product.
- D. [tex]\(BaSO_4\)[/tex] – This is the correct answer: the product that precipitates out.
Therefore, the correct answer is:
[tex]\[ \boxed{D} \][/tex]
[tex]\[ K_2SO_4(\text{aq}) + Ba(NO_3)_2(\text{aq}) \rightarrow ? \][/tex]
The reactants are potassium sulfate ([tex]\(K_2SO_4\)[/tex]) and barium nitrate ([tex]\(Ba(NO_3)_2\)[/tex]), both in aqueous form. When these compounds react, they undergo a double displacement reaction:
[tex]\[ K_2SO_4(\text{aq}) + Ba(NO_3)_2(\text{aq}) \rightarrow KNO_3(\text{aq}) + BaSO_4(\text{s}) \][/tex]
Let's break it down in detail:
1. Formation of Potassium Nitrate [tex]\(KNO_3\)[/tex]:
- According to Rule 2, nitrates [tex]\((NO_3^-)\)[/tex] are always soluble. Therefore, potassium nitrate [tex]\(KNO_3\)[/tex] will dissolve in water and remain in an aqueous form.
2. Formation of Barium Sulfate [tex]\(BaSO_4\)[/tex]:
- According to Rule 4, sulfates [tex]\((SO_4^{2-})\)[/tex] are generally soluble except when they are combined with barium [tex]\((Ba^{2+})\)[/tex], calcium [tex]\((Ca^{2+})\)[/tex], strontium [tex]\((Sr^{2+})\)[/tex], or lead [tex]\((Pb^{2+})\)[/tex]. In this case, the sulfate is combining with barium:
[tex]\[ Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \][/tex]
- Barium sulfate [tex]\(BaSO_4\)[/tex] is insoluble in water, meaning it will form a precipitate.
After analyzing the reaction, the precipitate that forms is barium sulfate [tex]\(BaSO_4\)[/tex].
Let's match this to the answer choices:
- A. [tex]\(KNO_3\)[/tex] – This compound is soluble and does not precipitate.
- B. [tex]\(Ba(NO_3)_2\)[/tex] – This is a reactant, not a product, and it remains in solution.
- C. [tex]\(K_2SO_4\)[/tex] – This is a reactant, not a product.
- D. [tex]\(BaSO_4\)[/tex] – This is the correct answer: the product that precipitates out.
Therefore, the correct answer is:
[tex]\[ \boxed{D} \][/tex]