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8 May, 08:16

We kept adding pbi2 into water until no more will dissolve. what is the concentration of lead (ii) iodide in the solution

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  1. 8 May, 09:45
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    PbI (ii) ionization in the solution of PBI (ii) into water is:

    PbI ₂ (solution) Pb₂⁺ + 2I⁻

    If the conc. of PbI (ii) in the sol. is xM then the conc. of Lead (ii) will be x M and conc. of iodide will be 2 x M.

    Therefore,

    Ksp = [Pb ²⁺][I-]²

    Plugging the values:

    1.4*10⁻⁸ = x ⋅ (2x) ²

    1.4*10⁻⁸ = 4x³

    x³ = {1.4*10⁻⁸}:4

    x³ = 0.35 x 10⁻⁸

    or

    x³ = 3.5 x 10⁻⁹

    x = 1.51 x 10⁻³

    Hence,

    Concentration of iodide ions in the solution:

    2x = 3.02 x 10⁻³
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