......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.
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.
1 answer