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 solve the problem, we need to determine the concentration of NaCl required to initiate the precipitation of AgCl from a solution containing 2.3 x 10^(-3) M AgNO3, given that the solubility product (Ksp) of AgCl is 2.8 x 10^(-10).

### Step-by-Step Solution:

1. **Write the Dissociation Reaction:**
The dissociation of AgNO3 and NaCl in water can be represented as:
\[ \text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^- \]
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]

When AgNO3 and NaCl are mixed, AgCl precipitates according to the reaction:
\[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl (s)} \]

2. **Identify the Known Values:**
- Concentration of AgNO3 = 2.3 x 10^(-3) M
- Ksp of AgCl = 2.8 x 10^(-10)

3. **Set Up the Expression for Ksp:**
The solubility product expression for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

4. **Substitute Known Values:**
Let the concentration of NaCl required to initiate precipitation be \( S \) (which is equal to the concentration of Cl^- ions). The concentration of Ag^+ ions from AgNO3 is given as 2.3 x 10^(-3) M. Therefore, we can write:
\[ K_{sp} = (2.3 \times 10^{-3})(S) \]

5. **Equate to Ksp:**
Substitute the value of Ksp into the equation:
\[ 2.8 \times 10^{-10} = (2.3 \times 10^{-3})(S) \]

6. **Solve for S:**
Rearranging the equation to solve for \( S \):
\[ S = \frac{2.8 \times 10^{-10}}{2.3 \times 10^{-3}} \]

Now, calculate \( S \):
\[ S = \frac{2.8}{2.3} \times 10^{-10 + 3} = \frac{2.8}{2.3} \times 10^{-7} \]
\[ S \approx 1.22 \times 10^{-7} \, \text{M} \]

7. **Conclusion:**
The concentration of NaCl required to initiate the precipitation of AgCl is approximately \( 1.22 \times 10^{-7} \, \text{M} \).