A solution contains `1.4 xx 10^(-3)M AgNO_(3)`. What concentration of `KC1` will be required to initiate the precipitation of `AgC1 ? K_(sp) AgC1 = 2.8 xx 10^(-10)`

Explanation

To find the concentration of KCl required to initiate the precipitation of AgCl from a solution containing 1.4 x 10^(-3) M AgNO3, we can follow these steps:

### Step 1: Understand the Dissociation of Compounds
AgNO3 dissociates in solution to give Ag⁺ and NO3⁻ ions:
\[ \text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^- \]

### Step 2: Determine the Concentration of Ag⁺
The concentration of Ag⁺ ions in the solution is equal to the concentration of AgNO3:
\[ [\text{Ag}^+] = 1.4 \times 10^{-3} \, M \]

### Step 3: Write the Dissociation Equation for KCl
KCl dissociates in solution to give K⁺ and Cl⁻ ions:
\[ \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \]

### Step 4: Define the Concentration of Cl⁻
Let the concentration of KCl be \( x \, M \). Therefore, the concentration of Cl⁻ ions will also be:
\[ [\text{Cl}^-] = x \, M \]

### Step 5: Write the Expression for Ionic Product (IP)
The ionic product (IP) for AgCl can be expressed as:
\[ \text{IP} = [\text{Ag}^+][\text{Cl}^-] \]

### Step 6: Set Up the Equation for Precipitation
For precipitation to occur, the ionic product must equal the solubility product (Ksp) of AgCl:
\[ \text{IP} = K_{sp} \]
\[ [\text{Ag}^+][\text{Cl}^-] = K_{sp} \]

Substituting the known values:
\[ (1.4 \times 10^{-3})(x) = 2.8 \times 10^{-10} \]

### Step 7: Solve for x
Now, we can solve for \( x \):
\[ x = \frac{2.8 \times 10^{-10}}{1.4 \times 10^{-3}} \]

Calculating the right-hand side:
\[ x = 2.0 \times 10^{-7} \, M \]

### Step 8: Conclusion
Thus, the concentration of KCl required to initiate the precipitation of AgCl is:
\[ \text{Concentration of KCl} = 2.0 \times 10^{-7} \, M \]

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