Question

A 800.0 mg sample of solid lead(II) iodide Ksp =9.8x10^-9 is dissolved in 5.0L of pure water at 25 degree C. How much hydroiodic acid, a strong acid, must be added to the solution (in g to two decimal places) to begin precipitating lead(II) iodide out of solution? ( ignore volume changes by the addition solutes)
answer 2.95g but I need to know the work.

Answer 1

Mass of solid lead iodide = 800.0 mg = 0.800 g

Molar mass of PbI2 = 461 g/mol

Moles of PbI2 = mass/molar mass = 0.800/461 = 0.0017 mol

Volume of water = 5.0 L

Concentration of [PbI2] = moles/volume = 0.0017/5.0 = 0.00035 M

Ksp = 9.8*10^-9

Ksp = [Pb2+][I-]²

9.8*10^-9 = 0.00035[I-]²

[I-]² = 2.8*10^-5

[I-] = 0.0053 M

Moles of I- = concentration*volume = 0.0053*5.0 = 0.026 mol

Moles of I- already present = 2[PbI2] = 2*0.0017 = 0.0034 mol

Moles of hydroiodic acid added = 0.026-0.0034 = 0.023 mol

Molar mass of HI = 128 g/mol

Mass of HI added = 128*0.023 = 2.95 g