What is the solubility, in moles per liter, of AgCl (K_sp = 1.8 × 10^-¹⁰), in 0.0100 molar aqueous potassium chloride solution?Hint: Be sure to account for the presence of a common ion. You should also be able to simplify the math because the solubility is very low.

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The solubility of AgCl in the presence of a common ion (Cl⁻ from KCl) is calculated using the Ksp expression: Ksp = [Ag⁺][Cl⁻]. With [Cl⁻] = 0.0100 M from KCl, the solubility of Ag⁺ is x, so Ksp = x(0.0100) = 1.8 × 10⁻¹⁰. Solving for x gives x = 1.8 × 10⁻⁸ mol L⁻¹.

What Is the Solubility, in Moles Per Liter, of AgCl