Question

precipitate Q=S ox 10-14 3. Determine the concentration of fluoride ion in a saturated solution of PbF2 (in moles/liter), if the Ksp = 3.6 x 10-8 for PbF2

Answer 1

Let the solubility of PbF₂ be S moles/Litre in the saturated solution of PbF₂

First of all write the equilibrium equation between PbF₂ and its ions as given below:

PbF₂ (s)  $\Leftrightarrow$  PbF₂ (aq) $\rightarrow$ Pb²⁺ +    2F^-

at equilibrium    S moles/Litre    S moles/Litre     2S moles/Litre

Therefore, the concentration of [Pb²⁺] = S moles/Litre

and the concentration of [F^-] = 2S moles/Litre

Also given K_sp = 3.6 * 10^-8

Solubilty Product K_sp = [Pb²⁺] * [F^-]²,

on substituting the above values we get,  3.6 * 10^-8 = S * (2S)²

$\therefore$ 4S³ = 3.6 * 10^-8

$\therefore$ S³= {(3.6 * 10^-8)/4}

Or S = 2.08 * 10^-3 moles/Litre

Since, [F^-] = 2S moles/Litre

Therefore [F^-] = 2 * 2.08 * 10^-3 moles/Litre

[F^-] = 4.16 * 10^-3 moles/Litre

The concentration of Fluoride ion in saturated solution of PbF₂ = 4.16 * 10^-3 moles/Litre