Calculate the solubility product of AgCl if the solubility of the salt in saturated solution at 298 K is 0.00198 g `L^(-1)`.

Explanation

To calculate the solubility product (Ksp) of AgCl given its solubility in a saturated solution at 298 K, we can follow these steps:

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

### Step 2: Define the solubility (S)
Let the solubility of AgCl in moles per liter be represented as S. When AgCl dissolves, it produces one mole of Ag⁺ and one mole of Cl⁻ for every mole of AgCl that dissolves. Therefore, at equilibrium:
- [Ag⁺] = S
- [Cl⁻] = S

### Step 3: Write the expression for Ksp
The solubility product (Ksp) is given by the expression:
Ksp = [Ag⁺][Cl⁻] = S × S = S²

### Step 4: Convert solubility from grams per liter to moles per liter
The solubility of AgCl is given as 0.00198 g/L. To convert this to moles per liter, we need the molar mass of AgCl.

- Molar mass of Ag = 108 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Now, convert the solubility:
\[ S = \frac{0.00198 \text{ g/L}}{143.5 \text{ g/mol}} \]

### Step 5: Calculate the solubility in moles per liter
Calculating the above expression:
\[ S = \frac{0.00198}{143.5} \approx 1.38 \times 10^{-5} \text{ mol/L} \]

### Step 6: Calculate Ksp
Now, substitute the value of S into the Ksp expression:
\[ Ksp = S² = (1.38 \times 10^{-5})² \]

### Step 7: Perform the calculation
Calculating the square:
\[ Ksp = 1.90 \times 10^{-10} \]

### Final Answer
The solubility product (Ksp) of AgCl at 298 K is approximately:
\[ Ksp = 1.90 \times 10^{-10} \]

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