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  • AgCl Solubility in NaCl Solutions: A Detailed Calculation
    Here's how to determine the solubility of AgCl in a 0.35 M NaCl solution:

    Understanding the Concepts

    * Solubility Product (Ksp): The solubility product constant (Ksp) is an equilibrium constant that describes the solubility of a sparingly soluble ionic compound. For AgCl, the Ksp is 1.8 x 10⁻¹⁰.

    * Common Ion Effect: The common ion effect describes the decrease in solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution.

    Calculations

    1. Write the dissolution equilibrium:

    AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

    2. Set up an ICE table (Initial, Change, Equilibrium):

    | | Ag⁺ | Cl⁻ |

    |-------------|-------|--------|

    | Initial | 0 | 0.35 | (From NaCl)

    | Change | +s | +s |

    | Equilibrium | s | 0.35+s |

    3. Write the Ksp expression:

    Ksp = [Ag⁺][Cl⁻] = 1.8 x 10⁻¹⁰

    4. Substitute the equilibrium concentrations into the Ksp expression:

    1.8 x 10⁻¹⁰ = (s)(0.35 + s)

    5. Since Ksp is very small, we can assume that 's' is negligible compared to 0.35:

    1.8 x 10⁻¹⁰ ≈ (s)(0.35)

    6. Solve for 's', which represents the solubility of AgCl:

    s ≈ (1.8 x 10⁻¹⁰) / 0.35 ≈ 5.1 x 10⁻¹⁰ M

    Conclusion

    The solubility of AgCl in a 0.35 M NaCl solution is approximately 5.1 x 10⁻¹⁰ M. This value is significantly lower than the solubility of AgCl in pure water, demonstrating the common ion effect.

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