Question

1. Calculate the solubility of PbI₂ in units of both moles x liters_-1and g x(100mL)^-1 using your average K_sp value

Average K_sp= 7.40x10^-7

                                    

2.      Explain why the solid PbI2 does not appear in the equilibrium expression for solubility.

3.      What effect would the use of KNO₃ in Part III Step 4, instead of KNO₂ have on the experimental absorbance values? Explain your answer.

Answer 1

1) PbI2 <----> Pb2+ + 2I-
Let x = moles/L of PbI2 that dissolve. This will give us x mol/L Pb2+ and 2x mol/L I-
7.4 x 10^-7 = (x)(2x)^2 = 4x^3
x = molar solubility = 5.7 x10^-3 M => 5.7 x 10^-3 moles per Liter

in  g x(100mL)-1 so pbi2 molar mass is 461

molar solubility = 46.1g/100ml

2)

lead(II) iodide, salts with limited solubility in water. The chemical equations for their equilibria in distilled water are as follows:

PbI₂(s)    <---------->    Pb²⁺(aq)    +    2I^-(aq)       K_sp = 9.8 x 10^-9

PbI₂ has a substantially smaller Ksp Therefore, PbI₂has a lower molar solubility, and I^- ions have the same starting concentration, PbI₂will be the first to precipitate when Pb²⁺ ions are added to the solution.

3)

Discussion: In this experiment, you will determine the solubility product, Ksp value, for a relatively insoluble compound, PbI2. The saturated solution contains solid PbI2 in equilibrium with its ions. The equilibrium system can be established by simply dissolving solid PbI2 in distilled water until no more dissolves or by mixing solutions of Pb(NO3)2 and KI to form the precipitate. The solid PbI2 will be removed from the saturated mixture using a centrifuge, then the [I-] remaining in solution at equilibrium can be determined spectrophotometrically by reacting it with NO2- ions in an acidic environment. This will form an orange-brown solution of I2, which absorbs a sufficient amount of light with a wavelength of ~525 nm to make an accurate analysis of its concentration. You will be using a Spectrophotometer, the Spec-20.