Determine the molar solubility of AgI in pure water.K_sp (AgI)= 8.51 × 10^-17. A)9.22 × 10^-9 mol L^-1 B)4.26 × 10^-17 mol L^-1 C)8.51 × 10^-17 mol L^-1 D)2.77 × 10^-6 mol L^-1 E)4.40 × 10^-6 mol L^-1

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Quizplus · Accepted Answer

The molar solubility of AgI in pure water can be calculated using the Ksp expression:Ksp = [Ag+][I-] = 8.51 × 10^-17Since AgI dissociates in a 1:1 ratio, the molar solubility (s) can be expressed as:s = [Ag+] = [I-]Substituting this into the Ksp expression, we get:Ksp = s^2 = 8.51 × 10^-17Solving for s, we get:s = √(8.51 × 10^-17) = 9.22 × 10^-9 mol L^-1Therefore, the molar solubility of AgI in pure water is 9.22 × 10^-9 mol L^-1.

Determine the Molar Solubility of AgI in Pure Water