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 1

Mass of lead iodide = 800 mg = 0.8 g

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide = Mass/Molar mass = 0.8 g/461 g/mol = 0.0017 moles

Volume of solvent = 5 L

Molarity of lead iodide = Moles/Volume = 0.0017/5 = 0.00035 M

We have, Ksp = 9.8 x 10^-9

Also, Ksp = [Pb²⁺][I^-]²

9.8 x 10^-9 = 0.00035[I^-]²

On solving,

[I^-] = 0.0053 M

Therefore, moles of I^- = Molarity x Volume = 0.0053 x 5 = 0.026 moles

Moles of I^- already present in the solution = 2 x 0.0017 = 0.0034 moles

Added moles of HI = 0.026-0.0034 = 0.023 moles

Mass of HI added = Moles of HI x Molar mass of HI = 128 g/mol x 0.023 mol = 2.95 g