What is the solubility, in moles per liter, of AgCl (K_sp = 1.8 × 10^-10), in 0.0300 M CaCl₂ solution?Hint: Be sure to account for the presence of a common ion and carefully organize. You should also be able to simplify the math because the solubility is very low.

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The solubility of AgCl in a solution with a common ion (Cl⁻ from CaCl₂) is calculated using the Ksp expression: Ksp = [Ag⁺][Cl⁻]. In 0.0300 M CaCl₂, [Cl⁻] is approximately 0.0600 M. Solving 1.8 × 10⁻¹⁰ = s × 0.0600 gives s ≈ 3.0 × 10⁻⁹ M.

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