What will be the solubility of AgCl in 0.05 M NaCl aqueous solution if solubility product of AgCl is `1.5xx10^(-10)` ?

Options

A. `3xx10^(-9)" mol L"^(-1)`
B. `0.05" mol L"^(-1)`
C. `1.5xx10^(-5)" mol L"^(-1)`
D. `3xx10^(9)" mol L"^(-1)`

Explanation

To find the solubility of AgCl in a 0.05 M NaCl solution, we can follow these steps:

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

NaCl also dissociates in water:
\[ \text{NaCl (s)} \rightleftharpoons \text{Na}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Identify the concentrations
From the dissociation of NaCl, we know that in a 0.05 M NaCl solution, the concentration of Cl⁻ ions is:
\[ [\text{Cl}^-] = 0.05 \, \text{M} \]

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product (Ksp) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given \( K_{sp} = 1.5 \times 10^{-10} \).

### Step 4: Substitute known values into the Ksp expression
Let the solubility of AgCl in the solution be \( s \) (which is the concentration of Ag⁺). Therefore:
\[ [\text{Ag}^+] = s \]
\[ [\text{Cl}^-] = 0.05 \, \text{M} \]

Now substituting these into the Ksp expression:
\[ K_{sp} = s \times 0.05 \]

### Step 5: Solve for s
We can rearrange the equation to solve for \( s \):
\[ s = \frac{K_{sp}}{[\text{Cl}^-]} \]
Substituting the known values:
\[ s = \frac{1.5 \times 10^{-10}}{0.05} \]

### Step 6: Calculate the solubility
Calculating \( s \):
\[ s = \frac{1.5 \times 10^{-10}}{0.05} = 3.0 \times 10^{-9} \, \text{M} \]

### Conclusion
Thus, the solubility of AgCl in a 0.05 M NaCl solution is:
\[ \boxed{3.0 \times 10^{-9} \, \text{M}} \]

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