The solublity product of AgI at 25*C is 1*10^(16) mol^2 lt^(-2)

the solublity of AgI in 10^(-4)normal solution of KI at 25*C will be.

plese explain with step & cause.

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 solubility. The solubility product (K_sp) of AgI is given as 1 x 10^-16 mol² L^-2 at 25°C. Let's break this down step by step.

Understanding the Solubility Product

The solubility product constant (K_sp) is an equilibrium constant that applies to the dissolution of sparingly soluble salts. For AgI, the dissolution can be represented as:

AgI (s) ⇌ Ag⁺ (aq) + I^- (aq)

The K_sp expression for this equilibrium is:

K_sp = [Ag⁺][I^-]

Calculating Solubility in Pure Water

In pure water, if we let the solubility of AgI be 's' mol/L, then at equilibrium:

• [Ag⁺] = s
• [I^-] = s

Thus, we can write:

K_sp = s × s = s²

Substituting the value of K_sp:

1 x 10^-16 = s²

Solving for 's':

s = √(1 x 10^-16) = 1 x 10^-8 mol/L

Introducing the Common Ion Effect

When KI is added to the solution, it dissociates completely into K⁺ and I^- ions. The presence of I^- ions from KI will shift the equilibrium to the left, reducing the solubility of AgI due to the common ion effect.

In a 10^-4 N solution of KI, the concentration of I^- ions is 10^-4 mol/L. Let’s denote the new solubility of AgI in this solution as 's'_new.

Setting Up the New Equilibrium

At equilibrium, we have:

• [Ag⁺] = s'_new
• [I^-] = 10^-4 + s'_new ≈ 10^-4 (since s'_new will be very small compared to 10^-4)

Now, we can write the K_sp expression again:

K_sp = [Ag⁺][I^-] = s'_new × (10^-4)

Substituting the known K_sp value:

1 x 10^-16 = s'_new × (10^-4)

Solving for the New Solubility

Rearranging the equation gives:

s'_new = (1 x 10^-16) / (10^-4) = 1 x 10^-12 mol/L

Final Thoughts

The solubility of AgI in a 10^-4 N solution of KI at 25°C is therefore 1 x 10^-12 mol/L. This significant decrease in solubility compared to pure water (1 x 10^-8 mol/L) illustrates the common ion effect, where the presence of a common ion (I^-) reduces the solubility of the salt.