Question

Calculate the amount, in moles, of Ag+ present at equilibrium when excess AgI(s) is added to 15.0 mL 0.88 M NaI(aq). Assume no change in volume. For AgI, Ksp = 8.51×10-17.
Answer

9.2×10-9 mol
1.5×10-18 mol
9.7×10-17 mol
1.4×10-10 mol
1.1×10-18 mol

Answer 1

AgI has low solubility, therefore:
Concentration of Ag+: 8.51x10-17 / 0.88 = 9.67x10-17 mol/l
Amount of Ag+ in 15 ml: 9.67x10-17 ∙ 0.015 = 1.5x10-18 mol
Answer:
1.5x10-18