Does the mixing of copper (I) nitrate with sodium iodide produce a precipitate? If so, give the molecular equation.
To determine if a precipitate is formed when copper (I) nitrate is mixed with sodium iodide, we need to evaluate the solubility rules.
Solubility rules are guidelines that determine the solubility of different compounds in water. According to the rules:
1. Most nitrate (NO3-) compounds are soluble, including copper (I) nitrate (CuNO3).
2. Most iodide (I-) compounds are soluble, including sodium iodide (NaI).
Based on these rules, both copper (I) nitrate and sodium iodide are soluble in water.
When two soluble compounds are combined, no precipitate forms because the resulting species remain in solution as individual ions.
The molecular equation for the reaction between copper (I) nitrate and sodium iodide can be written as follows:
CuNO3 + 2NaI → CuI2 + 2NaNO3
However, it should be noted that this equation does not represent the actual chemical species present in solution. In reality, copper (I) nitrate dissociates into copper (I) cations (Cu+) and nitrate anions (NO3-), while sodium iodide dissociates into sodium cations (Na+) and iodide anions (I-). The resulting ionic equation more accurately represents the reaction:
Cu+ + 2I- → CuI2
Although no precipitate is formed, a color change can be observed. Copper (I) iodide, which is present in the reaction, has a characteristic yellow color.
Yes, the mixing of copper (I) nitrate with sodium iodide does indeed produce a precipitate. The molecular equation for this reaction is as follows:
CuNO3 + 2NaI -> CuI2 + 2NaNO3
In this equation, copper (I) nitrate reacts with sodium iodide to form copper (II) iodide (precipitate) and sodium nitrate.