Question

Solubility of lead iodide PbI2 is 1.20 × 10–3 mol dm–3 at 298K. Calculate its solubility product constant

Answer 1

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Explanation:

Answer 2

Answer:

The solubility product constant of the reaction is 6.912×10⁻⁹M.

Explanation:

First, let's see how the ionization of lead iodide takes place,

$\mathrm{PbI_2\rightleftharpoons Pb^{2+}+2I^{-}}$             (1)

Where,

PbI₂= lead iodide

Pb⁺²=lead ions

I⁻=iodide ions

• The equilibrium constant for the dissolution of a solid substance into an aqueous solution is the solubility product constant. Ksp is used to represent it.

The solubility product constant of the equation (1) is given as,

$\mathrm{K_{sp}=[Pb^{+2}][2I^{-}]^2}$

$\mathrm{K_{sp}=[s][2s]^2}$

$\mathrm{K_{sp}=4s^3}$       (3)

From the question, we have that the solubility of lead iodide is(s)=1.20×10⁻³ mol dm⁻³

By inserting the value of "s" in equation (3) we get;

$\mathrm{K_{sp}=4(1.20\times 10^{-3})^3}$

$\mathrm{K_{sp}=4\times 1.728\times 10^{-9}}$

$\mathrm{K_{sp}=6.912\times 10^{-9}M}$

Hence, the solubility product constant of the reaction is 6.912×10⁻⁹M.

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