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).

Question

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

Answer 1

nvm, do i just multiply 1/kf * ksp? i think so.

Answer 2

It looks like Kf*Ksp to me.

Answer 3

But don't i need to reverse the complex ion equation in order for the lead 2 chloride ions to cancel out?

Answer 4

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.}

Answer 5

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

Answer 6

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.

Answer 7

well you were right, i just wished i had ur knowlege i suppose. lol

Answer 8

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.