[jmg12]

1. A solution is initially containing 0.050 M Pb(NO3)2 and 0.20 M NaF. What will be the concentration of all ions after precipitation of PbF2 occurs? (Ksp PbF2 = 3.3 x 10-6)

2. A solution is made of 0.10 M Zn 2+ and 0.1 M in NH3. This causes the [Zn(NH3)4]2+ complex ion to form, as shown in the equation below. Determine the concentration of free Zn2+ ion in this solution. (Kformation = 2.9 x 109)

3. A 0.010 M solution of AgNO3 is made 0.50 M in NH3, thus forming the [Ag(NH3)2]+ complex ion. Will AgCl precipitate if enough NaCl is added to make the solution 0.010 M in Cl-. (Kformation for complex ion = 1.1 x 107)

[rabolisk]

No. You have to put in some effort for us to help you.

[jmg12]

If I asked how to go about them, I simply do not know where to start. My paper is full of eraser marks everywhere. I need a starting point then I can do the rest on my own.

[rabolisk]

OK. What is Ksp? Mathematically, what is it? What do you know about lead nitrate and sodium fluoride's solubility?

[jmg12]

I know they will react to make PbF2 which is insoluble. The Ksp is a measure of the solubility.
The Ksp expression would be
3.3x10 -8 (correction it isn't 6) = [Pb +2][F-]²
That's as far as I pretty much go for all of them lol

[rabolisk]

That's good. So, all of the Pb²⁺ will have to come from Pb(NO₃)₂ and all of the F^- will have to come from NaF. Use that fact, along with an ICE table to figure out [Pb²⁺] and [F^-] at equilibrium. Write out a reaction to help you.

[jmg12]

so there would be 0.05 M Pb ²⁺ and 0.20 M F -
PbF2 ---> Pb + 2F
             0.05 0.20
             +x    +2x
             0.05 + x (0.20+2x)
(0.05 +x)(0.20 +2x)² = 3.3x10-8
I solved for X and I am not getting the write answers when I add them.
My book says Pb is 3.3 x 10-6 M
and F is 0.10 M
I'm so confused lol

[rabolisk]

That is the right way of going about it. Some people would write (0.05 - x)(0.20 - 2x)² = 3.3 x 10^-8, but it doesn't really matter. What would help, though, is some intuition to make the math easier. What you can do is assume that [F^-] is 0.10. This is because x will be approximately -0.05, and 0.20 - 0.10 = 0.10. You can assume this because Ksp is very small. Using this yields (0.05 + x)(0.10)² = 3.3 x 10^-8. You will get the right answer.

[jmg12]

When I solve for that I get -0.05 = x
So then when I try to get the Pb concentration I do 0.05-0.05 = 0
Makes no sense to me lol

[rabolisk]

-.0499967 is close to -.05, but still different.

[jmg12]

oh!!!! I see it now. Thanks a lot. Any tips on the next ones? I've been trying to use Knet = Ksp x Kformation , but it isn't really helping. Sorry to keep bothering you like this.

[rabolisk]

First, perhaps a simpler way to do 1 would be to assume that the reaction goes to completion and then do the reverse.

Pb²⁺ + 2F^-  PbF₂

Assume that this goes to completion, but also note that Pb²⁺ is the limiting reactant, and thus, there is 0.10 M of F^- left. Now write the reverse reaction, and do Ksp, except that the "initial" concentration of F^- is 0.10, not 0.

PbF₂  Pb²⁺ + 2F^-
I                                  0                        0.10
C                                 +x                       +2x
E                                  x                        0.10 +2x

Set Ksp up, but you can assume that 2x <<<< 0.10, and drop that term.
Then the math becomes really easy.

2 is the same as 1, except we are talking about Kformation rather than Ksp. The math is slightly tougher because in Ksp, the solid doesn't appear in the equilibrium constant, but in 2, [Zn(NH₃)₄]²⁺ does. I would use the above method of assuming the reaction goes to completion, then doing the calculations for the reverse reaction.

For 3, you first need to figure out the equilibrium [Ag⁺]. Then I suppose you need Ksp for AgCl.

[jmg12]

Great. I understand it now. Thank you so much.