A lab technician adds 0.035 mol of KCl to 1.00 L of 0.0050 M Pb(NO3)2. Ksp = 1.7 × 10–5 for PbCl2. Which of the following statements is correct?
a) Lead chloride precipitates until the solution is saturated.
b) The solution is unsaturated and no precipitate forms.
c) The concentration of lead ions is reduced by the addition of the chloride ions.
d)One must know Ksp for lead nitrate to make meaningful predictions on this system.
e) The presence of KCl will raise the solubility of Pb(NO3)2.
......pbCl2 ==> Pb^2+ + 2Cl^-
I.....solid.....0........0
C.....solid.....x........2x
E.....solid.....x........2x
Qsp = (Pb^2+)(Cl^-)^2
(Pb^2+) = 0.005 mols/L
(Cl^-) = 2x from the PbCl2 if it ppts and 0.035 from the KCl so
We can ignore the 2x.
Qsp = (0.0050)(0.035)^2 = about 6E-6 which is smaller than Ksp of 1.7E-5 so no ppt occurs. I've worked the problem; make your choices.
To determine the correct statement, we need to compare the calculated ion product (Qsp) with the solubility product constant (Ksp) for lead chloride (PbCl2).
First, let's calculate the concentration of chloride ions (Cl-) in the solution. Since 0.035 mol of KCl is added to a 1.00 L solution, the concentration of chloride ions is 0.035 M.
Next, let's calculate the concentration of lead ions (Pb2+) in the solution. The initial concentration of Pb(NO3)2 is given as 0.0050 M. As Pb(NO3)2 dissociates, it forms 1 mole of Pb2+ ions for every mole of Pb(NO3)2. Therefore, the concentration of Pb2+ ions is also 0.0050 M.
Now, let's calculate the ion product, Qsp, which is the product of the concentrations of the ions involved in the equilibrium expression. For PbCl2, the expression is Pb2+ + 2Cl-. Thus, Qsp = [Pb2+][Cl-]^2 = (0.0050 M)(0.035 M)^2 = 6.125 × 10^-6.
Comparing Qsp to the given Ksp value of 1.7 × 10^-5, we find that Qsp < Ksp. This means that the ion product is less than the solubility product, indicating that the solution is unsaturated and no precipitate of lead chloride will form.
Therefore, the correct statement is:
b) The solution is unsaturated and no precipitate forms.