A sample of AgCI was treated with `5.00mL` of `1.5M` `Na_(2)CO_(3)` solubility to give `Ag_(2)CO_(3)` . The remaining solution contained `0.0026g of CI^(-)` per litre. Calculate the solubility product of AgCI. `(K_(SP)for Ag_(2)CO_(3)=8.2xx10^(-12))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of Na2CO3 used
We know the volume and molarity of Na2CO3:
- Volume (V) = 5.00 mL = 0.005 L
- Molarity (M) = 1.5 M

Using the formula for moles:
\[ \text{Moles of Na}_2\text{CO}_3 = M \times V = 1.5 \, \text{mol/L} \times 0.005 \, \text{L} = 0.0075 \, \text{mol} \, \text{or} \, 7.5 \, \text{mmol} \]

### Step 2: Determine the moles of Cl⁻ remaining in the solution
We are given that the remaining solution contains 0.0026 g of Cl⁻ per liter. First, we convert grams to moles:
- Molar mass of Cl = 35.5 g/mol

Calculating the moles of Cl⁻:
\[ \text{Moles of Cl}^- = \frac{0.0026 \, \text{g}}{35.5 \, \text{g/mol}} = 7.30 \times 10^{-5} \, \text{mol/L} \]

### Step 3: Calculate the concentration of Cl⁻ in the solution
Since the concentration is given per liter, we can directly use the calculated moles:
\[ \text{Concentration of Cl}^- = 7.30 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Calculate the concentration of carbonate ions (CO₃²⁻)
Since Na2CO3 dissociates into 2 Na⁺ and CO₃²⁻, the concentration of CO₃²⁻ will be equal to the initial moles of Na2CO3 divided by the total volume of the solution.

The total volume after mixing is:
\[ \text{Total Volume} = 5 \, \text{mL} + \text{Volume of AgCl solution} \]

Assuming the volume of AgCl solution is negligible compared to the added Na2CO3, we can approximate:
\[ \text{Concentration of CO}_3^{2-} = \frac{7.5 \, \text{mmol}}{0.005 \, \text{L}} = 1.5 \, \text{M} \]

### Step 5: Write the equilibrium expression for Ag2CO3
The solubility product (Ksp) expression for Ag2CO3 is:
\[ K_{sp} = [Ag^+]^2[CO_3^{2-}] \]

Given:
\[ K_{sp} = 8.2 \times 10^{-12} \]

### Step 6: Calculate the concentration of Ag⁺
Let the concentration of Ag⁺ be \( x \). We can substitute the known values into the Ksp expression:
\[ 8.2 \times 10^{-12} = x^2 \times 1.5 \]

Solving for \( x \):
\[ x^2 = \frac{8.2 \times 10^{-12}}{1.5} \]
\[ x^2 = 5.47 \times 10^{-12} \]
\[ x = \sqrt{5.47 \times 10^{-12}} \approx 2.34 \times 10^{-6} \, \text{mol/L} \]

### Step 7: Calculate Ksp for AgCl
The Ksp for AgCl can be expressed as:
\[ K_{sp} = [Ag^+][Cl^-] \]

Substituting the values:
\[ K_{sp} = (2.34 \times 10^{-6}) \times (7.30 \times 10^{-5}) \]

Calculating:
\[ K_{sp} = 1.71 \times 10^{-10} \]

### Final Answer:
The solubility product of AgCl is:
\[ K_{sp} = 1.71 \times 10^{-10} \]

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