Problem about solubility/precipitation

[mx41]

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An aqueous .500 Liter solution contains Pb(NO3)2 (Ksp PbF2 : 2.7*10^-8) and Sr(NO3)2 (Ksp SrF2 : 2.5*10^-9), both with concentration of .000035 M. NaF is slowly added to this solution with 2.00g of NaF. Will a solid precipitate form after the NaF has been completely stirred into solution? If so, what salt(s) comprise this precipitate? Justify with calculations.

Thanks

 

[UCB05]

gotta get a real calculator, paper, and pencil for this... screwing stuff up left and right. Ill try this again later.

2.00g NaF = 2/42 = .0476mol F- produced.
[Pb+2]=[Sr+2]=.000035M

Ksp PbF2 = [Pb+2]F-]^2 = 2.7x10^-8, [F-] = .0278, .0139mol F- would be consumed
Ksp SrF2 = [Sr+2]F-]^2 = 2.5x10^-9, [F-] = .0085, .0042mol F- would be consumed

This is the part I'm hazy on. Since enough NaF is added to saturate both reactions but SrF2 is less soluble, it seems like SrF2 will precipitate out. I'm not sure if the F- that remains dissociated is the sum of both reactions or just based on SrF2

*edit again* but now that i think about it, it doesnt seem right for so much F- to be consumed in each reaction, considering there's only .000035M of each ion... ugh

 

[mx41]

Thanks for your help I really appreciate it.

When I did the problem, I got the same Molarities as you did for SrF2 and PbF2

Now, I said that both salts precipitate not just SrF2. I said that because there was .0953M NaF added to the solution
2.00g NaF/ 41.98g/mol = .04764mol NaF
.04764mol NaF/0.50L = .0953 M NaF

So both salts have [F-] of less than .0953 M, so doesnt that mean that they both precipitate out? If not, how do you determine if one will precipitate and the other does not?
Thanks again

 

[UCB05]

Thanks for your help I really appreciate it.

When I did the problem, I got the same Molarities as you did for SrF2 and PbF2

Now, I said that both salts precipitate not just SrF2. I said that because there was .0953M NaF added to the solution
2.00g NaF/ 41.98g/mol = .04764mol NaF
.04764mol NaF/0.50L = .0953 M NaF

So both salts have [F-] of less than .0953 M, so doesnt that mean that they both precipitate out? If not, how do you determine if one will precipitate and the other does not?
Thanks again

Well, the question says that the NaF is added in slowly, so at the point of [F-] between the saturation point of SrF2 and PbF2, I'm thinking that SrF2 will precipitate out and thus remove the extra [F-] before PbF2 can precipitate. Buuuuut... there's only .000035M of Sr+2. With a large excess of F-, PbF2 would precipitate out once nearly all the Sr+2 has precipitated.

I think my answer to this problem is \😵/

 

[mx41]

Yeah I see what you are saying..


I think we're over thinking the problem.
I think that enough NaF was added so that both salts precipitated