3.The solubility product for silver (I) chloride is 1.6x10^-10 while that for sliver (I) iodide is 8.5x10^-17. Suppose 5g of AgCl is put in a liter of 0.1M sodium iodide solution. If a solid is eventually isolated, what is it, and how did it form?
This is a good thinking question but there are so many variables that are not listed. Which solid is more soluble? So which should predominate IF GIVEN ENOUGH TIME.
AgCl is insoluble / dissolves really slowly, but I'm guessing from the question that it means the latter. It should form AgI given enough time.
To determine what solid is eventually isolated and how it forms in this scenario, we need to consider the solubility of the compounds involved and the concept of precipitation.
First, let's look at the solubility product (Ksp) values provided:
- The solubility product for silver (I) chloride (AgCl) is 1.6x10^-10.
- The solubility product for silver (I) iodide (AgI) is 8.5x10^-17.
Ksp represents the equilibrium constant for the dissolution of a sparingly soluble salt in water. When the product of the concentrations of the ions in a solution exceeds the Ksp value, precipitation occurs, resulting in the formation of a solid.
Now, let's consider the reaction occurring between AgCl and sodium iodide (NaI) in the given scenario:
AgCl (s) + NaI (aq) -> AgI (s) + NaCl (aq)
We are given that 5g of AgCl is put in a liter of 0.1M sodium iodide solution. To determine if a solid is formed, we need to compare the ion product (IP) with the Ksp value for both AgCl and AgI.
For AgCl:
IP(Cl-) = [Cl-] * [Ag+]
The concentration of chloride ions is determined by the dissociation of sodium chloride (NaCl) from the sodium iodide solution. Assuming 100% dissociation of NaCl, the concentration of chloride ions is equal to the concentration of NaCl, which is 0.1M.
Therefore, IP(Cl-) = 0.1M * [Ag+]
For AgI:
IP(I-) = [I-] * [Ag+]
The concentration of iodide ions is determined by the 0.1M sodium iodide solution. Therefore, IP(I-) = 0.1M * [Ag+]
Now, let's compare the IP to the Ksp values:
For AgCl:
IP(Cl-) = 0.1M * [Ag+]
IP(Cl-) = 0.1M * [Ag+] > Ksp(AgCl) = 1.6x10^-10
Since the ion product for AgCl exceeds the Ksp value, precipitation of AgCl will occur. A solid of silver (I) chloride (AgCl) will form.
For AgI:
IP(I-) = 0.1M * [Ag+]
IP(I-) = 0.1M * [Ag+] < Ksp(AgI) = 8.5x10^-17
Since the ion product for AgI is less than the Ksp value, AgI will remain dissolved in the solution. No precipitation of AgI will occur.
In summary, when 5g of AgCl is put in a liter of 0.1M sodium iodide solution, a solid of silver (I) chloride (AgCl) will form while silver (I) iodide (AgI) remains dissolved in the solution.