This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Let's solve this step by step. The problem asks for the solubility of AgI in a 0.274 M solution of NaI. This is a common ion effect problem. Given: Ksp for AgI = 8.5 × 10⁻¹⁷ (from the image's calculation) Concentration of NaI = 0.274 M Step 1: Write the dissociation equations for AgI and NaI. AgI is a sparingly soluble ionic compound, so it establishes an equilibrium with its ions in solution: AgI(s) ⇌ Ag⁺(aq) + I⁻(aq) NaI is a soluble ionic compound, so it dissociates completely in solution: NaI(aq) → Na⁺(aq) + I⁻(aq) Step 2: Determine the initial concentrations of ions. From the dissociation of NaI, the concentration of iodide ions (I⁻) is: [I⁻] from NaI = 0.274 M Let 's' be the molar solubility of AgI in the NaI solution. This means that at equilibrium, 's' moles/L of AgI will dissolve, producing 's' moles/L of Ag⁺ and 's' moles/L of I⁻ from AgI. [Ag⁺] = s [I⁻] from AgI = s Step 3: Calculate the total concentration of the common ion. The common ion in this case is I⁻. The total concentration of I⁻ in the solution will be the sum of I⁻ from NaI and I⁻ from AgI: Total [I⁻] = [I⁻] from NaI + [I⁻] from AgI Total [I⁻] = 0.274 M + s Since AgI is very sparingly soluble (indicated by its small Ksp value), the amount of I⁻ contributed by AgI (s) will be much smaller than the initial concentration of I⁻ from NaI (0.274 M). Therefore, we can make the approximation: Total [I⁻] ≈ 0.274 M Step 4: Write the Ksp expression for AgI and substitute the equilibrium concentrations. The Ksp expression for AgI is: Ksp = [Ag⁺][I⁻] Substitute the equilibrium concentrations into the Ksp expression: 8.5 × 10⁻¹⁷ = (s)(0.274) Step 5: Solve for 's', the molar solubility of AgI. s = (8.5 × 10⁻¹⁷) / (0.274) s = 3.102189... × 10⁻¹⁶ M Rounding to two significant figures (consistent with 8.5 × 10⁻¹⁷ and 0.274 M): s = 3.1 × 10⁻¹⁶ M The molar solubility of AgI in 0.274 M NaI solution is 3.1 × 10⁻¹⁶ M.