Ba2+(aq)+2I−(aq)+2Na+(aq)+SO2−4(aq)→BaSO4(s)+2I−(aq)+2Na+(aq)
B
a
2+
+S
O
2−
→BaSo
The equation you provided is a chemical equation that represents a double displacement reaction between the ions in the given aqueous solutions. Specifically, it shows the reaction between barium(II) ions (Ba2+), iodide ions (I-), sodium ions (Na+), and sulfate ions (SO4 2-).
To understand the reaction, let's break it down:
Ba2+(aq) + 2I-(aq) + 2Na+(aq) + SO4 2-(aq) → BaSO4(s) + 2I-(aq) + 2Na+(aq)
On the reactant side (left side of the arrow), we have:
- Ba2+(aq): Barium ions in aqueous solution.
- 2I-(aq): Two iodide ions in aqueous solution.
- 2Na+(aq): Two sodium ions in aqueous solution.
- SO4 2-(aq): Sulfate ions in aqueous solution.
On the product side (right side of the arrow), we have:
- BaSO4(s): Barium sulfate, which is a solid precipitate formed during the reaction.
- 2I-(aq): Two iodide ions, which remain in the aqueous solution.
- 2Na+(aq): Two sodium ions, which also remain in the aqueous solution.
In this reaction, the barium ions (Ba2+) and the sulfate ions (SO4 2-) combine to form the insoluble compound barium sulfate (BaSO4), which precipitates out of the solution as a solid.
As for the ions that are not involved in the precipitation reaction (iodide ions and sodium ions), they remain in the aqueous solution before and after the reaction. They are called spectator ions because they do not actively participate in the chemical reaction.
To balance the equation, you need to ensure that the number of atoms on both sides of the equation is the same. However, the equation you provided is already balanced.
Hence, the balanced equation for the reaction is:
Ba2+(aq) + 2I-(aq) + 2Na+(aq) + SO4 2-(aq) → BaSO4(s) + 2I-(aq) + 2Na+(aq)
I hope this explanation helps you understand the given chemical equation and the reaction it represents!