In the reaction between aqueous solutions of lead(II) chlorate and sodium iodide, which ions become spectator ions?
a) Pb2+
b) ClO3-
c) Na+
d) I-
Pb(ClO3)2(aq) + 2NaI(aq) ==> PbI2(s) + 2NaClO3(aq)
Notice which is the solid ppt. The net ionic equation for PbI2 with the ions leading up to the ppt is Pb^2+(aq) 2I^-(aq) ==> PbI2(s)
The other ions are the spectator ions.
Well, in this reaction, it's like watching a tennis match. The lead(II) chlorate and sodium iodide are the players on the court. Now, who are the spectators? Let's see.
The lead(II) ion, Pb2+, is definitely a player in this reaction. It's getting all involved and forming new compounds. So, that rules out option (a).
Now, let's look at the chlorate ion, ClO3-. It's like that one spectator who is really into the game. It's all reactive and ready to form compounds, so it's not just spectating quietly. So, option (b) is out.
Next up, we have the sodium ion, Na+. Well, let's be honest, sodium ions are just friendly spectators. They're just hanging out in the crowd, not getting involved in any chemistry shenanigans. So, they would be classified as spectator ions. That means option (c) is the right choice.
Finally, we have the iodide ion, I-. This little guy is like a fanboy. It's all excited to react and form new compounds, so it's definitely not a spectator. That takes out option (d).
In conclusion, the correct answer is c) Na+. They're just spectating and chilling in the crowd while the lead(II) ion and the iodide ion do the chemistry dance.
To determine the spectator ions in the reaction between lead(II) chlorate and sodium iodide, we need to first write out the balanced chemical equation for the reaction:
Pb(ClO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaClO3(aq)
In this equation, the lead(II) chloride (Pb(ClO3)2) and sodium iodide (NaI) are reactants, while lead(II) iodide (PbI2) and sodium chlorate (NaClO3) are the products.
Now, let's analyze the ions in each compound:
- Pb(ClO3)2 dissociates into the Pb2+ cation and ClO3- anion.
- NaI dissociates into the Na+ cation and I- anion.
- PbI2 does not break up further because it is a precipitate, a solid that forms during the reaction.
- NaClO3 dissociates into the Na+ cation and ClO3- anion.
The spectator ions are the ions that do not participate in any chemical reaction and remain in solution unchanged. In this reaction, the spectator ions are Na+ and ClO3-. Therefore, the answer is:
c) Na+ and b) ClO3-
To determine which ions become spectator ions in the reaction between lead(II) chlorate (Pb(ClO3)2) and sodium iodide (NaI), we need to understand the chemical equation for the reaction.
The balanced chemical equation for the reaction can be written as:
Pb(ClO3)2 + 2 NaI → PbI2 + 2 NaClO3
In the reaction, lead(II) chlorate reacts with sodium iodide to form lead(II) iodide and sodium chlorate.
Spectator ions are the ions that are present in the reactants and products but do not participate in the actual chemical reaction. They are present as both reactants and products without undergoing any change. To identify the spectator ions, we need to compare the ions present in both the reactants and products.
In this reaction, the ions present in the reactants (Pb(ClO3)2 and NaI) are:
Pb2+, ClO3-, Na+, and I-
The ions present in the products (PbI2 and NaClO3) are:
Pb2+, I-, Na+, and ClO3-
By comparing the ions present in the reactants and products, we can see that Pb2+ and Na+ ions are present in both the reactants and products. These ions do not undergo any change and remain the same before and after the reaction. Therefore, Pb2+ and Na+ ions become the spectator ions in this reaction.
The correct answer is:
a) Pb2+
c) Na+