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Oduvanchick [21]
3 years ago
13

The molar solubility of pbi2 is 1.5 103 m.

Chemistry
1 answer:
Vsevolod [243]3 years ago
8 0

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

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