Calculate the solubility of AgI in a 0.274 M NaI solution, given Ksp(AgI) = 8.5 x 10⁻¹⁷.

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)

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

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

The Ksp expression for AgI is:
Ksp = [Ag⁺][I⁻]

Substitute the equilibrium concentrations into the Ksp expression:

8.5 × 10⁻¹⁷ = (s)(0.274)

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.