Question

The concentration of iodide ions in a saturated solution of lead (ii) iodide is ________ m. The solubility product constant of pbi2 is?

Answer 1

The concentration of iodide ion in a saturated solution of lead (ii) iodide is 3.04 × 10⁻³ M.

What is a Chemical Reaction ?

A chemical reaction is a process in which chemical bonds between atoms to break and reorganize to form other new substances.

Now write the equation:

PbI₂ (s) ↔ Pb⁺² (aq) + 2I⁻ (aq)

_                x                 2x

How to find the concentration when solubility product is given ?

It is expressed as:

$K_{sp} = [A^{+}]^{a} [B^{-}]^{b}$

where,

$K_{sp}$ = Solubility product constant

A⁺ = Cation in an aqueous solution

B⁻ = Anion in an aqueous solution

a, b = Relative concentration of A and B.

Now put the values in above formula we get

$K_{sp} = [A^{+}]^{a} [B^{-}]^{b}$

1.4 × 10⁻⁸ = [Pb⁺²]¹ [I⁻]²

1.4 × 10⁻⁸ = [x]¹ [2x]²

1.4 × 10⁻⁸ = 4x³

x³ = 3.5 × 10⁻⁹

x = 1.52 × 10⁻³

So,

Pb⁺² = x = 1.52 × 10⁻³

I⁻ = 2x = 2 × 1.52 × 10⁻³

          = 3.04 × 10⁻³ M

Thus from the above conclusion we can say that The concentration of iodide ion in a saturated solution of lead (ii) iodide is 3.04 × 10⁻³ M.

Disclaimer: Th question was given incomplete on the portal. Here is the complete question.

Question: The concentration of iodide ions in a saturated solution of lead (ii) iodide is ________ m. The solubility product constant of PbI₂ is 1.4 × 10⁻⁸.

Learn more about the Solubility Product here: brainly.com/question/1419865

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