Question

Determine the equilibrium concentrations for Pb2+ and Cl- if you initially have 0.750 grams of PbCl2 (s) and 0.10 M Cl- in a volume of 0.500 L

Answer 1

@aaronq

Answer 2

So you need to write the Ksp equation from the dissociation reaction.

Can you start with the dissociation of \(PbCl_2\)?

Answer 3

do you mean products/reactants

Answer 4

i forgot to mention that it gives me my k
K=1.7x10^-5

Answer 5

that would be the Ksp, the solubility product constant - it explains how soluble a compound is (at a given temperature, pressure, etc.)

The dissociation in water in this case, would be:

\(PbCl_2\rightarrow Pb^{2+}+2Cl^-\)

Answer 6

would i then do the ice chart
and would i=I ignore my reactant because it is a solid

Answer 7

yep! that's right.

Answer 8

but if i did the Ksp equation what would i input for my solid

Answer 9

the Ksp works out to the same thing as the equilibrium constant Keq because the concentration of any solid is 1. it's just a specific name given to it.

\(\sf K_{eq}=\dfrac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}=\dfrac{[Pb^{2+}][Cl^-]^2}{1}=[Pb^{2+}][Cl^-]^2=K_{sp}\)

Answer 10

oh so my solid would just be one

Answer 11

yep!