PbCl2(s) ==> Pb^2+ + 2Cl-
Ksp =(Pb^+2)(Cl-)^2
All of the problems are an example of the common ion effect as well as Le Chatelier's principle.
NaCl ==> Na+ + Cl-
Set up an ICE chart for the PbCl2.
(Pb2+) = x
(Cl-) from the PbCl2 = x
Cl- from the NaCl is 0.25 M (since 0.250 mole was placed in 1L).
Therefore, (x)(x+0.25) = Ksp. Substitute Ksp and solve for x. You will have a quadratic equation unless you make the simplifying assumption that x + 0.25 = 0.25.
Calculate the equilibrium vale of Pb +2 (aq) in 1.00 L of saturated PbCl2 solution to which 0.250 mole of NaCl has been added. 60.0 ml of 0.0300 M NaCl 60.0mL of 0.0500 M Pb(NO3)2
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