Determine the solubility of AgCl in 0.1 M `BaCl_2 [K_(sp)" for "AgCl=1 xx 10^(-10)]`

Answer

To determine the solubility of AgCl in a 0.1 M BaCl₂ solution, we will follow these steps:

### Step 1: Write the dissociation equations
AgCl dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

Barium chloride dissociates as:
\[ \text{BaCl}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Identify the concentrations of ions
In a 0.1 M BaCl₂ solution, the concentration of Ba²⁺ ions is 0.1 M. The dissociation of BaCl₂ produces 2 Cl⁻ ions for every formula unit, so the concentration of Cl⁻ ions from BaCl₂ is:
\[ \text{[Cl}^-] = 2 \times 0.1 \, \text{M} = 0.2 \, \text{M} \]

### Step 3: Set up the expression for Ksp
The solubility product constant (Ksp) expression for AgCl is:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given that \( K_{sp} \) for AgCl is \( 1 \times 10^{-10} \).

### Step 4: Define the solubility
Let the solubility of AgCl in this solution be \( S \). At equilibrium, the concentrations of the ions will be:
- \([\text{Ag}^+] = S\)
- \([\text{Cl}^-] = 0.2 + S\)

However, since \( S \) will be very small compared to 0.2 M, we can approximate:
\[ [\text{Cl}^-] \approx 0.2 \, \text{M} \]

### Step 5: Substitute into the Ksp expression
Now we can substitute these values into the Ksp expression:
\[ K_{sp} = S \times 0.2 \]
Setting this equal to the given Ksp:
\[ 1 \times 10^{-10} = S \times 0.2 \]

### Step 6: Solve for S
To find \( S \):
\[ S = \frac{1 \times 10^{-10}}{0.2} \]
\[ S = 5 \times 10^{-10} \, \text{M} \]

### Conclusion
The solubility of AgCl in a 0.1 M BaCl₂ solution is \( 5 \times 10^{-10} \, \text{M} \).

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