Calculate the value of the equilibrium constant for the following reaction: PbI2 (s) + 2I- (aq) <-> PbI4 ^ 2- (aq) The solubility product constant Ksp for PbI2 is 9.8e-9 and the formation constant Kf for PbI4^2- is 3.0e4
I have Ksp=[PbI4^2-]
Kf= [PbI4^2-] / [PbI2][I-]^2
Kc= [PbI4^2-] / [I-]^2
I don't know where to go from here
You need to re-write the equations as such.
PbI2(s) ==> Pb^2+ + 2I^-
Pb^2+ + 4I^- --> [PbI4]^2-
--------------------------
add the two equation to get this.
PbI2(s) + 2I^- ==? [PbI4]^2- for which
Keq = Ksp*Kf
You can plug in Ksp and Kf and see that Keq is what it should be.
Ksp = (Pb^2+)(I^-)^2
Kf = [PbI4^2-]/(Pb^2+)(I^-)^4
Keq = [PbI4^2-]/(I^-)^2
thank you!!!!
Well, let's have some fun with this! It looks like you're a bit stuck, so let me give you a clownish hand.
First, let's rewrite the equilibrium expression using the given information:
Kc = [PbI4^2-] / [I-]^2
Now, let's substitute the values we know:
Kc = (3.0e4) / ([I-]^2)
But wait a second, where's PbI2? We need to take into account its solubility product constant, Ksp. Since we don't know the concentration of PbI4^2-, we'll have to express it in terms of the concentration of PbI2.
Knowing that the reaction is in equilibrium, we can say that:
[PbI4^2-] = [PbI2] * (1/x)
where "x" represents the concentration of PbI4^2- in relation to PbI2.
Now, let's substitute this expression back into Kc:
Kc = (3.0e4) / ([I-]^2) = ([PbI2] * (1/x)) / ([I-]^2)
Still with me? Great!
Now, we need to express [I-] in terms of "x." Looking at the stoichiometry of the reaction, we can see that the concentration of I- decreases by a factor of 2 for every factor of x that [PbI4^2-] increases:
[I-] = 2x
Substituting this back into Kc:
Kc = ([PbI2] * (1/x)) / ((2x)^2)
Simplifying:
Kc = ([PbI2] / 4x^2)
Now, let's bring Ksp into the mix. Ksp is the product of the concentrations of the products over the reactant:
Ksp = [PbI2] * [I-]^2
Substituting [I-] = 2x:
Ksp = [PbI2] * (2x)^2
Simplifying:
Ksp = 4[PbI2] * x^2
Now, let's rearrange this equation to solve for [PbI2]:
[PbI2] = Ksp / (4x^2)
We can substitute this back into the expression for Kc:
Kc = (Ksp / (4x^2)) / 4x^2
Simplifying:
Kc = Ksp / (16x^4)
Wow, we made it through! So, the equilibrium constant, Kc, for this reaction is Kc = Ksp / (16x^4), where "x" represents the concentration of PbI4^2- relative to PbI2.
And there you have it! Now go out there and show off your clownish chemistry skills!
To calculate the value of the equilibrium constant (Kc) for the given reaction, you can start by using the information already provided:
1. Write the expression for Ksp: Ksp = [PbI4^2-] = 9.8e-9
2. Write the expression for Kf: Kf = [PbI4^2-] / ([PbI2] * [I-]^2) = 3.0e4
Now, you want to find the expression for Kc in terms of the concentration of PbI4^2- and I- ions. To do this, we can use the concept of the stoichiometry of the balanced equation:
3. Because there is no coefficient in front of PbI4^2- in the balanced equation, we can conclude that the stoichiometric coefficient for PbI4^2- is 1.
4. Because there is a coefficient of 2 in front of I- in the balanced equation, we can conclude that the stoichiometric coefficient for I- is 2.
Based on this information, we can write the expression for Kc:
Kc = [PbI4^2-] / [I-]^2
Since you already have the expression for Ksp ([PbI4^2-] = 9.8e-9) and Kf (Kf = [PbI4^2-] / ([PbI2] * [I-]^2) = 3.0e4), you can substitute these values into the expression for Kc:
Kc = (9.8e-9) / [I-]^2
Therefore, the value of the equilibrium constant (Kc) for the given reaction is (9.8e-9) divided by the square of the concentration of I- ions.