A solution is prepared by adding some solid PbF2 to water and allowing the solid to come to equilibrium with the solution. The equilibrium constant for the reaction below is Kc = 3.6x10−8.
PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)
What is the equilibrium concentration of Pb2+ in the solution? Assume that some of the solid remains undissolved.
......................PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)
I.....................solid.............0......................0
C....................solid.............x.....................2x
E.....................solid.............x.....................2x
K = (Pb^2+)(F^-)^2
Plug the E line into the Kc expression and solve for x = (Pb^2+) = ?
what do you mean plug it into the kc expression? do i just use x and 2x to plug into K = (Pb^2+)(F^-)^2?
3.6x10−8 = x(2x)^2
x^3=9.0x10^-9
x=0.002
Well, I always like to take a crack at solving chemistry problems, but I must admit, equilibrium constants make me feel a little unstable.
But fear not, I can still try my best to help you out! Now, let's see. The equilibrium constant, Kc, is given as 3.6 x 10^-8. That's a small number, like trying to fit into a clown car with way too many clowns!
To find the equilibrium concentration of Pb2+, we'll need some more information. Specifically, we'll need to know the initial concentration of PbF2 that was added to the water.
So, while I can't directly solve for the equilibrium concentration of Pb2+ without that initial concentration, I hope this explanation tickles your funny bone a little and helps you along your chemistry journey. Keep clowning around with those calculations, and you'll be a chemistry pro in no time, my friend!
To determine the equilibrium concentration of Pb2+ in the solution, we need to use the equilibrium constant and the stoichiometry of the given reaction.
The equilibrium constant (Kc) is a measure of the ratio of products to reactants at equilibrium. In this case, the equilibrium constant (Kc) is given as 3.6x10^-8.
The stoichiometric equation shows that for every one mole of PbF2 that dissolves, one mole of Pb2+ and two moles of F- ions are produced. So, the concentration of Pb2+ at equilibrium will be directly related to the concentration of PbF2 that dissolves.
Let's assign a variable (x) to the equilibrium concentration of Pb2+. Since the stoichiometric coefficient of Pb2+ in the balanced equation is 1, the concentration of PbF2 that dissolves will also be x.
So, the concentration of Pb2+ at equilibrium is x moles/liter.
From the stoichiometry of the reaction, the concentration of F- ions will be 2x moles/liter.
The concentration of undissolved PbF2 at equilibrium will be the initial concentration of PbF2 minus x moles/liter.
Now we can write the expression for the equilibrium constant (Kc) using these concentrations:
Kc = [Pb2+][F-]^2 / [PbF2]
Substituting the concentration expressions, we have:
3.6x10^-8 = x * (2x)^2 / (initial concentration of PbF2 - x)
Simplifying and rearranging the equation:
3.6x10^-8 = 4x^3 / (initial concentration of PbF2 - x)
At this point, you will need to know the initial concentration of PbF2 to solve for x. Without that information, we can't give you a specific numerical value for the equilibrium concentration of Pb2+.
Once you have the initial concentration of PbF2, you can solve the equation for x using numerical methods or an appropriate calculator to find the equilibrium concentration of Pb2+ in the solution.