Calculations of Solubility Product of Lead 2 Chloride

[aisha]

HELP! In this experiment I measured the initial mass of a piece of Al and then i put it into a beaker containing 100.0 mL of saturated PbCl2

The concentration I calculated for Pb is $$6.12 *10^-^5$$ mol/L
The concentration I calculated for Cl is $$1.224* 10^-^4$$ mol/L

Therefore my Ksp= [Pb] [Cl]^2
$$=[6.12*10^-^5] [1.224*10^-^8]^2$$
Ksp=9.17
This is what I calculated please verify

I need help answering this question if 100.0 ml of 0.02 mol/L lead 2 nitrate solution and 100.0 ml of 0.02 mol/L sodium chloride solution were combined would a precipitate of Lead 2 chloride be expected to form? I have to show all my work. But I am not sure where to start.

 

[Unco]

G'day, Aisha.

The first part of your post is a little confusing.

How did you calculate [Pb2+] and [Cl-]?

If these concentrations are indeed in a saturated solution then Ksp(PbCl2) = 9.16x10^(-13), just a couple of typos there.

At 25 degrees C, the Ksp of PbCl2 is about 1.6x10^(-5), so perhaps the room was cold?

If the second part is connected to the first part, then you are probably to find the ionic product (given by [Pb2+][Cl-]^2) and compare it to the Ksp value obtained earlier. Hint: for the same amount, doubling the volume halves the concentration. If IP > Ksp, a precipitate will form.

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I'm new here but I think this belongs in the Science Education forum.

 

[Blueshift5]

I would first start by reviewing the solubility product (Ksp) expression for lead 2 chloride, which is Ksp = [Pb2+][Cl-]^2. From the given information, we know that the initial concentrations of lead and chloride ions are both 0.02 mol/L. Thus, the Ksp for lead 2 chloride would be:

Ksp = (0.02 mol/L)(0.02 mol/L)^2 = 4 x 10^-6 mol^3/L^3

This value represents the maximum amount of lead 2 chloride that can dissolve in solution before reaching saturation. Therefore, if the product of the concentrations of lead and chloride ions in solution exceeds this value, a precipitate of lead 2 chloride would be expected to form.

In order to determine the actual concentrations of lead and chloride ions in solution after mixing the lead nitrate and sodium chloride solutions, we can use the principles of stoichiometry. From the balanced chemical equation for the reaction between lead nitrate and sodium chloride, we know that the ratio of lead nitrate to lead 2 chloride is 1:1. This means that the initial concentration of lead ions (0.02 mol/L) will be equal to the concentration of lead 2 chloride formed.

For chloride ions, the ratio is 2:1, meaning that for every 2 moles of chloride ions, 1 mole of lead 2 chloride will be formed. Therefore, the initial concentration of chloride ions (0.02 mol/L) will be twice the concentration of lead 2 chloride formed.

Using these principles, we can calculate the final concentrations of lead and chloride ions after mixing the solutions:

Lead ions: 0.02 mol/L (initial concentration)

Chloride ions: (0.02 mol/L x 2) + (0.02 mol/L) = 0.04 mol/L (initial concentration + additional chloride ions from reaction)

Therefore, the product of the concentrations of lead and chloride ions is:

(0.02 mol/L)(0.04 mol/L)^2 = 3.2 x 10^-5 mol^3/L^3

Since this value is larger than the Ksp value calculated earlier, a precipitate of lead 2 chloride would be expected to form.

In conclusion, based on the given information and calculations, a precipitate of lead 2 chloride would be expected to form when