Use the appropriate values of Ksp and Kf to find the equilibrium constant for the following reaction: PbCl2(s)+ 3OH(aq) >> Pb(OH)3 (aq) + 2Cl(aq).
ksp for PbCl2 is 1.17*10^-5
kf for complex ion is 8*10^13.
anyways...I know that the dissociation of the first constant is PbCl2 > Pb + 2Cl. the thing is that if im going to find the equlibrium constant what do i do about the PbCl being a solid? Does that affect it at all? Thx
Write Ksp (on a sheet of paper).
Write Kf (same thing).
Multiply Ksp*Kf on the paper.
Now write Krxn from the problem and compare with Ksp*Kf.[The (Pb+2 cancels. There is no PbCl2(s) and you need (Cl-)2 and it doesn't cancel]. My eyes aren't all that great but Ksp*Kf looks ok to me.
nvm, do i just multiply 1/kf * ksp? i think so.
It looks like Kf*Ksp to me.
But don't i need to reverse the complex ion equation in order for the lead 2 chloride ions to cancel out?
ok i tried it and you were right. i guess the PbCl2 did catch me off guard. I'll have to review this over. thanks
Well, excuse me, I just proved my point about my eyes I guess. I forgot and didn't turn off the superscripts. Such is life.
well you were right, i just wished i had ur knowlege i suppose. lol
Four years of college, six more in graduate school, and 60 years experience will get it. And there is a lot of chemistry I don't know.
To find the equilibrium constant for the given reaction, you need to write the balanced chemical equation:
PbCl2(s) + 3OH-(aq) >> Pb(OH)3(aq) + 2Cl-(aq)
First, you need to consider the dissociation of PbCl2. Although PbCl2 is a solid in its pure form, it still dissociates into its constituent ions in a saturated solution. The dissociation equation for PbCl2 is:
PbCl2(s) >> Pb2+(aq) + 2Cl-(aq)
Now you have two equations: one for the dissociation of PbCl2 and one for the overall reaction. To find the equilibrium constant for the overall reaction, you can multiply the equilibrium constants for each step of the reaction.
The equilibrium constant for the dissociation of PbCl2 (K1) can be calculated as follows:
K1 = [Pb2+(aq)][Cl-(aq)]^2 / [PbCl2(s)]
Here, [ ] represents the concentration of the species. Since PbCl2 is a solid, its concentration does not change and can be considered constant, so you can omit it from the equation. Therefore, the expression becomes:
K1 = [Pb2+(aq)][Cl-(aq)]^2
Similarly, the equilibrium constant for the overall reaction (Koverall) can be calculated using the expression:
Koverall = [Pb(OH)3(aq)][Cl-(aq)]^2 / [OH-(aq)]^3
Now, you may notice that in the reaction given, OH-(aq) is consumed and Cl-(aq) is produced in equal amounts. Therefore, the concentration of OH-(aq) in terms of [Cl-(aq)] can be written as:
[OH-(aq)] = [Cl-(aq)] / 2
Substituting this expression into the equation for Koverall, we get:
Koverall = [Pb(OH)3(aq)][Cl-(aq)]^2 / ([Cl-(aq)] / 2)^3
= 8 * [Pb(OH)3(aq)][Cl-(aq)]^2
Now, to find the equilibrium constant (Koverall), you need to multiply K1 and Koverall:
K = K1 * Koverall
= [Pb2+(aq)][Cl-(aq)]^2 * 8 * [Pb(OH)3(aq)][Cl-(aq)]^2
Next, you need to substitute the given values of Ksp and Kf:
Ksp for PbCl2 = 1.17 * 10^-5
Kf for complex ion = 8 * 10^13
Since Ksp represents the solubility product of PbCl2 and Kf represents the stability constant for the complex ion, you can write the following relationship:
Ksp = [Pb2+(aq)][Cl-(aq)]^2
Kf = [Pb(OH)3(aq)][OH-(aq)]^3
Substituting these values into the equation for K:
K = K1 * Koverall
= Ksp * (Kf / [OH-(aq)]^3) * 8 * Kf
= Ksp * (8 * Kf / ([Cl-(aq)] / 2)^3)
Now, you can substitute the given values of Ksp and Kf to calculate K. The concentration of OH-(aq) can be determined based on the starting concentrations of PbCl2 and OH-.