the solubility product of AgI at 25 degreeC is 1.0x10^-16 mol^2L^-2.the solubility of AgI in10^-4 N solution of KI at 25degreeC is approximately(in mol L^-1)

To determine the solubility of silver iodide (AgI) in a 10^-4 N solution of potassium iodide (KI), we need to consider the common ion effect and how it influences the solubility of sparingly soluble salts. The solubility product (Ksp) of AgI at 25°C is given as 1.0 x 10^-16 mol²/L². Let's break down the steps to find the solubility of AgI in the presence of KI.

Understanding the Common Ion Effect

The common ion effect states that the solubility of a salt decreases in a solution that already contains one of its constituent ions. In this case, KI dissociates in solution to provide potassium ions (K⁺) and iodide ions (I^-). The presence of I^- ions from KI will affect the solubility of AgI.

Setting Up the Equilibrium Expression

The dissolution of AgI can be represented by the following equilibrium:

• AgI (s) ⇌ Ag⁺ (aq) + I^- (aq)

The solubility product expression for this equilibrium is:

• Ksp = [Ag⁺][I^-]

Calculating the Concentration of Iodide Ions

In a 10^-4 N solution of KI, the concentration of I^- ions is 10^-4 mol/L. Let’s denote the solubility of AgI in this solution as 's' (in mol/L). Therefore, at equilibrium, the concentrations will be:

• [Ag⁺] = s
• [I^-] = 10^-4 mol/L

Substituting into the Ksp Expression

Now we can substitute these values into the Ksp expression:

• Ksp = (s)(10^-4)

Since we know Ksp for AgI is 1.0 x 10^-16, we can set up the equation:

• 1.0 x 10^-16 = s × 10^-4

Solving for Solubility 's'

To find 's', we rearrange the equation:

• s = (1.0 x 10^-16) / (10^-4)

Calculating this gives:

• s = 1.0 x 10^-12 mol/L

Final Result

The solubility of AgI in a 10^-4 N solution of KI at 25°C is approximately 1.0 x 10^-12 mol/L. This illustrates how the presence of a common ion (I^-) significantly reduces the solubility of AgI compared to its solubility in pure water.