Question

If 800 mg of lead(11) iodide (MM 461 g/mol) are placed in 1000 mL of pure water, how many milligrams of lead (II) iodide dissolve? Ksp= 4.4x10-9 All 800 mg of lead(II) iodide dissolve Less than 450 mg do not dissolve, but some lead(11) iodide does O All 800 mg of lead(tl) iodide do not dissolve O 450 mg amount do not dissolve Less than 800 mg but more than 450 mg do not dissolve

Answer 1

The value of solubility product constant (K_sp) =   4.4 × 10⁻⁹

For lead(II) iodide, the solubility product is written following the dissociation :

PbI₂ (s) = Pb²⁺ (aq) + 2 I^- (aq)

So, K_sp = [Pb²⁺][I^-]²

(K_sp is the product of the concentration of the ions raised to their stoichiometric coefficients)

[…] denotes concentration in molarity.

Molar solubility = number of moles of solute ( PbI₂) which will dissolve in 1000 mL of solution.                                                             

Let molar solubility of lead(II) iodide = s M.

So, the concentration of Pb²⁺ = s M , and I ^- = 2s M

(since 1 mole of the salt gives 1 Pb²⁺ ions and 2 I ^- ions).

Putting the value of concentration of Pb²⁺ and I ^- in the equation for K_sp and solving for s :

K_sp = 4.4 × 10⁻⁹= (s)(2s)²

Or, 4.4 × 10⁻⁹= 4s³

Or, s = 1.032 x 10^-3 M (this is the molar solubility)

So, in 1000 mL of solution, 1.032 x 10^-3 moles of lead(II) iodide will dissolve.

Converting moles to mass :

Mass ( in g) = Number of moles x molar mass

Or, mass ( in g) = 1.032 x 10^-3 moles x 461 g/mol = 0.476 g = 476 mg ( 1g = 1000 mg)

Thus, in the 1000 mL solution, 476 mg of lead (II) iodide will dissolve .

The remaining mg of lead iodide ( 800 mg – 476 mg = 324 mg) will precipitate.

So, the last option is correct.