If your cuvette had been dirty, how would this have affected the calculated valu

Question

If your cuvette had been dirty, how would this have affected the calculated value of K_sp? The solubility of lead(II) iodide in water at 15 degree C is 0.46 g dm^-3. For a saturated solution of lead(II) iodide at 15 degree C, calculate the concentration in mol dm^-3 of lead(II) ions, the concentration in mol dm^-3 of iodide ions, and the K_sp of lead(II) iodide. The experiment procedure for this experiment has you add 5 mL of 0.0040 M AgNO_3 to 5 mL of 0.0024 M K_2CrO_4. Is either of these reagents in excess? If so, which one?

Explanation / Answer

Question 3.

Solubility of Lead Iodide, PbI2(s), is 0.46 g/dm3, or 0.46 g/L

mol = mass/MW = 0.46/461.01 = 0.0009978 mol per liter

Calculate

Lead(II) ions --> Pb+2

[Pb+2] = [PbI2] = 0.0009978 mol / liter = 0.0009978 mol dm^-3

for I-2 ions, we have

[I-] = 2*[PbI2] = 2*0.0009978 = 0.0019956 M or mol dm^-3

finally

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

so

Ksp = (0.0009978)(0.0019956 ^2) = 3.973658*10^-9

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