Solubility Product Constant Lab

[xavior6]

Homework Statement

Hello. I recently did a lab with my chemistry class in which we mixed Potassium Iodide and Lead(II) Nitrate of different dilutions. We started off with 10mL of each solution and mixed them, giving us a final volume of 20mL. We then gradually decreased each solution's volume, making the next trial 8mL of each, but we used water to raise the final level to 20mL again. The one after that was 6mL/6mL, and the one after that was 4mL/4mL respectively. It is important to note that a precipitate formed for the 10mL, the 8mL, and the 6mL. The 4mL was the first to not produce a precipitate.

The initial concentration of Potassium Iodide was 0.020M
The initial concentration of Lead (II) Nitrate was 0.010M.

The question is determine the solubility product constant of Lead (II) Iodide over a range of values.

Homework Equations

CV=CV

Q = [a]

The Attempt at a Solution

It is obvious that the Ksp must be greater than the Trial Ion Product of the first flask to not produce a precipitate, but less than the Trial Ion product of the first flask to product a precipitate. In this case, the former is the 4mL/4mL, and the latter is the 6mL/6mL. Therefore, if I find the final Concentration of Iodide and Lead for each case, I can then use the Ksp formula to determine their trial ion product.

The problem is that BOTH of the answers I obtained were GREATER than the accepted Ksp of Lead (II) Iodide. I cannot logically explain this to myself, and I can't understand if I made a flaw in my calculations. Can someone please give me a hand? Please keep in mind that the final volume is ALWAYS 20mL (we add water).

 

[Borek]

At first sight your calculations look OK.

What was temperature of the solution?

There is always a possibility that concentrations given were wrong.

Have you used distilled water?

Perhaps amount of the precipitate was so small, that it wasn't visible?

 

[GCT]

Homework Statement

Hello. I recently did a lab with my chemistry class in which we mixed Potassium Iodide and Lead(II) Nitrate of different dilutions. We started off with 10mL of each solution and mixed them, giving us a final volume of 20mL. We then gradually decreased each solution's volume, making the next trial 8mL of each, but we used water to raise the final level to 20mL again. The one after that was 6mL/6mL, and the one after that was 4mL/4mL respectively. It is important to note that a precipitate formed for the 10mL, the 8mL, and the 6mL. The 4mL was the first to not produce a precipitate.

The initial concentration of Potassium Iodide was 0.020M
The initial concentration of Lead (II) Nitrate was 0.010M.

The question is determine the solubility product constant of Lead (II) Iodide over a range of values.

Homework Equations

CV=CV

Q = [a]

The Attempt at a Solution

It is obvious that the Ksp must be greater than the Trial Ion Product of the first flask to not produce a precipitate, but less than the Trial Ion product of the first flask to product a precipitate. In this case, the former is the 4mL/4mL, and the latter is the 6mL/6mL. Therefore, if I find the final Concentration of Iodide and Lead for each case, I can then use the Ksp formula to determine their trial ion product.

The problem is that BOTH of the answers I obtained were GREATER than the accepted Ksp of Lead (II) Iodide. I cannot logically explain this to myself, and I can't understand if I made a flaw in my calculations. Can someone please give me a hand? Please keep in mind that the final volume is ALWAYS 20mL (we add water).

Write the actual Ksp equation with respect to the precipitate.