Question

Solubility of \(\mathrm{AgCl}\) in \(10^{-2} \text{M} \mathrm{NaCl}\) solution is (\( K_{sp_{\mathrm{AgCl}}} = 1.8 \times 10^{-10} \)):

• A. \( 10^{-2} \text{M} \)
• B. \( 1.8 \times 10^{-12} \text{M} \)
• C. \( 1.8 \times 10^{-8} \text{M} \)
• D. \( 1.8 \times 10^{-2} \text{M} \)

Hint:

For solubility with a common ion:
- Write the \( K_{sp} \) expression.
- Account for the common ion concentration from other sources.
- Use the approximation \( S \ll [\text{common ion}] \) for sparingly soluble salts.
- Verify the approximation by checking the \( K_{sp} \).

Answer 1

The Correct Option is C

The solubility of \(\mathrm{AgCl}\) is affected by the presence of \(\mathrm{NaCl}\), which provides a common ion (\(\mathrm{Cl}^-\)). The solubility product constant (\( K_{sp} \)) for \(\mathrm{AgCl}\) is: \[ \mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{Cl}^-(aq), \quad K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] = 1.8 \times 10^{-10} \] -
Step 1: Define solubility
Let the solubility of \(\mathrm{AgCl}\) in the \(\mathrm{NaCl}\) solution be \( S \) (in \(\text{mol L}^{-1}\)). Dissolving \(\mathrm{AgCl}\) contributes:
- \([\mathrm{Ag}^+] = S\) - \([\mathrm{Cl}^-] = S + [\mathrm{Cl}^-]_{\text{from NaCl}}\)
-
Step 2: Account for common ion from \(\mathrm{NaCl}\)
The solution contains \(10^{-2} \text{M} \mathrm{NaCl}\), which dissociates completely:
\[ \mathrm{NaCl} \rightarrow \mathrm{Na}^+ + \mathrm{Cl}^-, \quad [\mathrm{Cl}^-]_{\text{from NaCl}} = 10^{-2} \text{M} = 0.01 \text{M} \] Thus, the total \([\mathrm{Cl}^-]\) is: \[ [\mathrm{Cl}^-] = S + 0.01 \] -
Step 3: Apply \( K_{sp} \)
Substitute into the \( K_{sp} \) expression: \[ K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] = S \cdot (S + 0.01) = 1.8 \times 10^{-10} \] -
Step 4: Simplify using common ion effect
Since \(\mathrm{AgCl}\) is sparingly soluble, its solubility \( S \) is very small (\( S \ll 0.01 \)). Therefore, we can approximate: \[ S + 0.01 \approx 0.01 \] So: \[ S \cdot 0.01 \approx 1.8 \times 10^{-10} \] Solve for \( S \): \[ S = \frac{1.8 \times 10^{-10}}{0.01} = \frac{1.8 \times 10^{-10}}{10^{-2}} = 1.8 \times 10^{-10+2} = 1.8 \times 10^{-8} \text{M} \] -
Step 5: Verify approximation
If \( S = 1.8 \times 10^{-8} \), then: \[ S + 0.01 = 1.8 \times 10^{-8} + 0.01 \approx 0.01 \] The approximation holds since \( 1.8 \times 10^{-8} \) is much smaller than \( 0.01 \). Let’s confirm exactly: \[ K_{sp} = S (S + 0.01) = (1.8 \times 10^{-8})(1.8 \times 10^{-8} + 0.01) \approx (1.8 \times 10^{-8}) \cdot 0.01 = 1.8 \times 10^{-10} \] This matches the given \( K_{sp} \), confirming the solution.
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Step 6: Check options
The calculated solubility \( S = 1.8 \times 10^{-8} \text{M} \) matches option (C).