concentration of PbI ions in SLN saturated with PbI solid containing dissolved I- with a concentration of 0.085M

To determine the concentration of PbI ions in a saturated solution (SLN) that is saturated with PbI2 solid and containing dissolved I- ions, you will need to use the solubility product constant (Ksp) of PbI2.

The balanced equation for the dissolution of PbI2 in water is:

PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)

The solubility product constant, Ksp, for this reaction is the product of the concentrations of the Pb2+ and I- ions raised to their stoichiometric coefficients:

Ksp = [Pb2+][I-]^2

In this case, since the initial concentration of I- ions is known to be 0.085 M, and assuming the solubility of PbI2 is reached, the concentration of Pb2+ ions can be calculated. Let's denote the concentration of Pb2+ ions as x M.

Therefore, the Ksp expression becomes:

Ksp = x * (0.085)^2

Now, you need to look up the value of Ksp for PbI2. The Ksp value for PbI2 is 1.4 × 10^-8 at 25°C.

Once you have the Ksp value, you can solve for x by rearranging the equation:

1.4 × 10^-8 = x * (0.085)^2

Solving this equation will give you the concentration of Pb2+ ions (x) in the SLN saturated with PbI2 solid and containing dissolved I- ions.

To summarize:
1. Write the balanced equation for the dissolution of PbI2.
2. Use the Ksp expression to set up an equation.
3. Look up the Ksp value for PbI2.
4. Solve the equation for x to find the concentration of Pb2+ ions in the SLN.