What is the molar solubility of silver(I)iodide at 25°C? The solubility product constant for silver(I)iodide is 8.3 × 10^-17 at 25°C. A) 9.1 × 10^-9 M B) 4.2 × 10^-17 M C) 4.8 × 10^-5 M D) 2.7 × 10^-6 M E) 8.3 × 10^-17 M

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The solubility product (Ksp) for silver(I) iodide (AgI) is given as 8.3 × 10^-17. The dissolution of AgI in water can be represented as AgI(s) ⇌ Ag+(aq) + I-(aq). Since each formula unit of AgI produces one Ag+ ion and one I- ion, if s is the molar solubility of AgI, then the concentrations of Ag+ and I- at equilibrium are both s. The Ksp expression is Ksp = [Ag+][I-] = s^2. Solving for s gives s = sqrt(Ksp) = sqrt(8.3 × 10^-17) = 9.1 × 10^-9 M.

What Is the Molar Solubility of Silver(I)iodide at 25°C? the Solubility