Question

Calculate the solubility in g/l of lead (ii) fluoride in water at 25 oc if ksp of pbf2 is 4. 1 x 10-8.

Answer 1

The solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L in water at 25°C.

At a specific solution temperature, a solid salt compound can entirely dissolve in pure water up to a predetermined molar solubility limit. The dissociation stoichiometry ensures that the molarities of the constituent ions are proportionate to one another. The saturable nature of the solution causes them to also coexist in a solubility equilibrium with the solid component. At this temperature, a solubility product constant Ksp is calculated using the solubility product of their molarity values.

Lead (II) fluoride has the following solubility equilibrium for its saturated solution:

                   $PbF_2(s)$ ⇄ $Pb^2^+ (aq) + 2F^- (aq)$

                   $K_s_p = [Pb^2^+][F^-]^2$

This compound dissociates in a 1:2 ratio of ions. For the compound dissolved in pure water, the Ksp is expressed in terms of the molar solubility "x" as:

$K_s_p = (x) (2x)^2$

$K_s_p = 4x^3$

Here,

$K_s_p= 4.1$ × $10^-^8$

4.1 × 10⁻⁸ = 4 x³

x³ = 1.025 × 10⁻⁸

x³ = 10.25 × 10⁻⁹

x = 2.17 × 10⁻³ g/L

Therefore, the solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L.

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