A solution contains `2.3 xx 10^(-3) M Ag NO_(3)`. What concentration of NaCl will be required to initiate the precipitation of AgCl.
The solubility product of AgCl is `2.8 xx 10^(-10)` ?

Answer

To determine the concentration of NaCl required to initiate the precipitation of AgCl from a solution containing 2.3 x 10^(-3) M AgNO3, we can follow these steps:

### Step 1: Understand the Reaction
The precipitation of AgCl occurs when AgNO3 reacts with NaCl:
\[ \text{AgNO}_3 (aq) + \text{NaCl} (aq) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \]

### Step 2: Write the Expression for the Solubility Product (Ksp)
The solubility product (Ksp) for AgCl is given as:
\[ K_{sp} = [Ag^+][Cl^-] \]
Where \([Ag^+]\) and \([Cl^-]\) are the molar concentrations of silver and chloride ions at equilibrium.

### Step 3: Identify Given Values
- The concentration of AgNO3 is:
\[ [Ag^+] = 2.3 \times 10^{-3} \, M \]
- The solubility product \( K_{sp} \) of AgCl is:
\[ K_{sp} = 2.8 \times 10^{-10} \]

### Step 4: Set Up the Equation
To initiate precipitation, the ionic product must equal the solubility product:
\[ K_{sp} = [Ag^+][Cl^-] \]
Substituting the known values:
\[ 2.8 \times 10^{-10} = (2.3 \times 10^{-3})[Cl^-] \]

### Step 5: Solve for [Cl^-]
Let \( [Cl^-] = S \) (the concentration of NaCl needed):
\[ 2.8 \times 10^{-10} = (2.3 \times 10^{-3})S \]
Now, rearranging to solve for \( S \):
\[ S = \frac{2.8 \times 10^{-10}}{2.3 \times 10^{-3}} \]

### Step 6: Calculate S
Calculating the value:
\[ S = \frac{2.8 \times 10^{-10}}{2.3 \times 10^{-3}} \approx 1.22 \times 10^{-7} \, M \]

### Conclusion
The concentration of NaCl required to initiate the precipitation of AgCl is approximately:
\[ S \approx 1.22 \times 10^{-7} \, M \]

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