Question

A solution contains 2.0 x 10⁻³ M Pb(II). What concentration of I- is necessary for the precipitation of PbI2? The Ksp for PbI₂ at 25 degrees C is 1.4 x 10⁻⁸.

a) 1.2 x 10⁻⁵ M

b) 2.8 x 10⁻⁴ M

c) 5.6 x 10⁻⁶ M

d) 7.1 x 10⁻⁹ M

Answer 1

The concentration of I- necessary for the precipitation of PbI2 is approximately 1.2 x 10⁻⁵ M.

Step-by-step explanation:

The concentration of the iodide ions necessary for the precipitation of PbI2 can be determined using the solubility product constant (Ksp) equation. The balanced chemical equation for the precipitation reaction is:

Pb²+ (aq) + 2I¯ (aq) → PbI2 (s)

From the equation, we can see that 2 moles of iodide ions are required for every 1 mole of lead(II) ion. Given that the concentration of Pb²+ is 2.0 x 10⁻³ M, the concentration of I¯ needed can be calculated as follows:

Ksp = [Pb²+][I¯]²

1.4 x 10⁻⁸ = (2.0 x 10⁻³) [I¯]²

[I¯]² = (1.4 x 10⁻⁸) / (2.0 x 10⁻³)

[I¯] = √((1.4 x 10⁻⁸) / (2.0 x 10⁻³))

[I¯] ≈ 1.2 x 10⁻⁵ M