What is the solubility of silver iodide in a solution containing 0.20 M sodium iodide? The K_sp value for silver iodide is 1.5 $\times$ 10^-16.

Question

Quizplus · Accepted Answer

To determine the solubility of silver iodide in the presence of sodium iodide, we need to use the common ion effect. The addition of sodium iodide provides a source of iodide ions, which will decrease the solubility of silver iodide. The equilibrium expression for the dissolution of silver iodide is:

AgI(s) ⇌ Ag+(aq) + I-(aq)

The solubility product constant (Ksp) for silver iodide is given as 1.5 × 10^-16. According to the principle of equilibrium, the product of the concentrations of the ions must equal the Ksp:

Ksp = [Ag+] [I-]

If we assume that x is the solubility of silver iodide in the presence of sodium iodide, then the concentration of iodide ions in the solution will be 0.20 M (from the sodium iodide). Therefore, the equilibrium expression becomes:

Ksp = [Ag+] (0.20)

[Ag+] = Ksp/0.20 = (1.5 × 10^-16)/0.20 = 7.5 × 10^-17 M

So, the solubility of silver iodide in the presence of 0.20 M sodium iodide is 7.5 × 10^-17 M, which is closest to option A.

What Is the Solubility of Silver Iodide in a Solution ×\times×

What Is the Solubility of Silver Iodide in a Solution $\times$