A student attempts to separate Cl⁻ and I⁻ from a solution that is 0.05 M Cl⁻ and 0.10 M I⁻ with Pb(NO₃)₂.As Pb(NO₃)₂ is added,which anion will precipitate first and what will the concentration of the anion be when the second anion begins to precipitate? K_sp = 1.7 x 10⁻⁵ for PbCl₂ and K_sp = 9.8 x 10⁻⁹ for PbI₂.

Question

Quizplus · Accepted Answer

First, we need to compare the solubility product (Ksp) values for PbCl2 and PbI2. PbI2 has a much lower Ksp value than PbCl2, which means it is less soluble and will precipitate first.

Next, we need to determine the concentration of iodide ion when the second anion begins to precipitate. This occurs when the concentration of Pb2+ ions added is greater than or equal to the molar solubility of PbI2 (which is equal to the concentration of iodide ions). At this point, PbI2 will begin to precipitate and the concentration of iodide ions will remain constant.

Using the Ksp value for PbI2, we can set up an ICE table to determine the concentration of iodide ion when PbI2 just begins to precipitate:

PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)

Initial:     0        0.10      0.05 
Change:  -x        +x        +2x
Equilibrium: -x     0.10+x   0.05+2x

Ksp = [Pb2+][I-]^2 
9.8 x 10^-19 = (0.10 + x)(0.05 + 2x)^2

We can assume that x is very small compared to 0.05 and 0.10:

9.8 x 10^-19 = (0.10)(0.05)^2 
---------------------------- 
            (0.05)^2

Solving for x:

x = 0.0012 M

Therefore, the first anion to precipitate is iodide, with a concentration of 0.0012 M when the second anion begins to precipitate.

A Student Attempts to Separate Cl− and I− from a Solution

A Student Attempts to Separate Cl⁻ and I⁻ from a Solution