K s ​ p of AgCl is 1×10 −10 . Its solubility in 0.1MKNO 3 ​ will be:

Explanation

To find the solubility of AgCl in a 0.1 M KNO₃ solution, we will follow these steps:

### Step 1: Write the dissociation equation for AgCl
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

### Step 2: Define the solubility product (Ksp)
The solubility product constant (Ksp) for AgCl is given by the equation:
Ksp = [Ag⁺][Cl⁻]

### Step 3: Set up the expression for Ksp
Let the solubility of AgCl in the solution be 's'. Therefore, at equilibrium:
- [Ag⁺] = s
- [Cl⁻] = s

Thus, we can express Ksp as:
Ksp = s × s = s²

### Step 4: Substitute the given Ksp value
We know that Ksp of AgCl is 1 × 10⁻¹⁰. Therefore:
s² = 1 × 10⁻¹⁰

### Step 5: Solve for 's'
To find 's', we take the square root of both sides:
s = √(1 × 10⁻¹⁰) = 10⁻⁵

### Step 6: Consider the effect of KNO₃
When we add KNO₃ to the solution, it dissociates into K⁺ and NO₃⁻ ions. However, KNO₃ does not provide any common ions (Ag⁺ or Cl⁻) that would affect the solubility of AgCl. Therefore, the solubility of AgCl remains unchanged.

### Final Answer
The solubility of AgCl in 0.1 M KNO₃ is still 10⁻⁵ M.

---