why is lead 2 iodide produced from the reaction of lead 2 nitrate and potassium iodide?
The driving force is that PbI2 is a solid that is insoluble in the aqueous solution present with Pb(NO3)2 and KI
Lead(II) iodide (PbI2) is produced from the reaction between lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI) due to a double displacement reaction, also known as a metathesis reaction or a precipitation reaction.
In this reaction, the lead(II) ion (Pb2+) from lead(II) nitrate reacts with the iodide ion (I-) from potassium iodide. The product, lead(II) iodide, is insoluble and thus precipitates out of the solution.
To understand the reaction further, let's break it down step by step:
1. Pb(NO3)2 (lead(II) nitrate) dissociates into Pb2+ and NO3- ions.
Pb(NO3)2 -> Pb^2+ + 2NO3^-
2. Similarly, KI (potassium iodide) dissociates into K+ and I- ions.
KI -> K+ + I-
3. The Pb2+ ion from lead(II) nitrate combines with the I- ion from potassium iodide.
Pb^2+ + 2I- -> PbI2 (lead(II) iodide)
4. The resulting product, PbI2, is a yellow solid that precipitates out of the solution because it is insoluble in water.
In summary, the reaction between lead(II) nitrate and potassium iodide forms lead(II) iodide because of the exchange of ions. The reaction is represented by the following equation:
Pb(NO3)2 + 2KI -> PbI2 + 2KNO3
It's important to note that writing and balancing chemical equations requires understanding the valence charges and solubility rules of ions. These rules can be learned through studying chemistry or referring to reliable sources such as textbooks or online educational resources.