Poppy - 23-10-2012 at 17:57
Hello. I have performed recently a reaction between lead nitrate and table salt, so that I could measure the reaction overral.
54g lead crystals were dissolved in 100mL. Another solution was comprised of 19g NaCl plus 60mL water.
Then both solutions were mixed and the precipitate collected and dried.
From the theorical yield of 45g I got only 33g PbCl2!!!!
The supernatant liquid was kept. It didn't precipitate anything else, and it has been a week since the reaction has proceded.
Those 12g of PbCl2 are missing! Could sodium nitrate ionic strenght be so astonishing as to increase lead chloride solubility this much?
Or maybe the source of lead nitrate was actually composed of hydrated crystals?
platedish29 - 23-10-2012 at 18:49
You should always roast powdered salts before stablish a measure for its mass.
Salts, specially fine and freshly prepared samples, tend to contain a lot of moisture aside water of hydration.
Moreover lead nitrate has not been assigned as a hydrate as per wiki.
[Edited on 24-10-2012 by platedish29]
AJKOER - 24-10-2012 at 04:42
If the acidic solution has dissolved some of the PbCl2, add a known amount of Na2CO3 and see if a precipitate is formed.
If yes, collect the PbCO3, dry and weigh.
Poppy - 24-10-2012 at 18:55
Yes something has precipitated.
Thanks
-
Before pushing with the weigh, perhaps some of the precipitate was not pure PbCl2 but contained Pb(OH)2 to some extent?
Treating this with Debye-Huckel equations for aqueous ionic solutions, sorry for the horribly long post
In the supernatant solution there should contain:
20.33g sodium nitrate along 160g water.
20.33g - 160mL
Xg - 1000mL
molarity in terms of sodium nitrate: 127,11g/L
1,495mol/L NaNO3
ionic strenght calculation:
u = 1,4954
sqrt( u) = 1,222879 (here named Uroot)
hydrated lead 2+ ion diameter: 0,45nm
hydrated chloride -1 ion diameter: 0,3nm
PbCl2 solubility: 0,03596
Lead chloride Kps: 1,86x10^(-4)
Kps(PbCl2) = a[Pb2+].(a[Clo])^2
Kps(PbCl2) = yPb.[Pb].(yCl-)^2.[Cl-]^2
[Cl-] = 0,07194
[Pb2+] = 0,03596
[Cl-]^2 = 0,005176
Kps(PbCl2) = yPb.(yCl-)^2.0,005176.0,03596
LogYPb = -(0,51.2^2.Uroot)/(1+3,3.0,45.Uroot)
LogYPb = -(0,51.4.1,222879)/(1+3,3.0,45.1,222879
LogYPb = -2,49467/19,15975
LogYPb = -0,13020
YPb = 10^(-0,13020) = 0,74097
LogYCl- = -(0,51.1^2.Uroot)/(1+3,3*0,3.Uroot)
LogYCl- = -(0,51.1,222879)/(1+3,3.0,3.1,222879)
LogYCl- = -0,623668/2,21065
LogYCl- = -0,2821197
Ycl- = 10^(-0,2821197) = 0,522252
(YCl-)^2 = 0,27275
Kps(PbCl2) = 0,74097.0,27275.4.S^3
pois [Pb2+].[Cl-].[Cl-] = [Pb2+].[2.Pb2+].[2.Pb2+] = 4[Pb2+]^3
1,86x10^(-4) = 0,80839.S^3
2,300868127.10^(-4) = S^3
S = 0,061277mol/L (PbCl2 solubility doubled)
That was an stimative considering the ionic strenght of the nitrate alone.
As long as lead dissolved lead itself contributes to strenghten the solution. This would be a shitty limit to pounder.
Considering the missing 12g of lead chloride are in the supernatant, let us use this formula once again to reconcile rather than foresee the results.
For this u changes to:
1,495mol/L NaNO3 + 0,2697mol/L PbCl2
calculating U: 1/2 (1,495.1^2 + 1,495.(-1)^2 + 0,2697.2^2 + 0,53937.1^2)
U = 2,304087
rootU = 1,51792
LogYPb = -(0,51.2^2.1,51792)/(1+3,3.0,45.1,51792)
LogYPb = -3,096557/3,25411
LogYPb = -0,951583
YPb = 10^(-0,951583) = 0,1117936
LogYCl- = -(0,51.1^2.1,51792)/(1+3,3.0,3.1,51792)
LogYCl- = -0,7741392/2,50274
LogYCl- = -0,3093166
YCl- = 10^(-0,3093166) = 0,49055
(YCl-)^2 = 0,24064
Kps(PbCl2) = 0,24064.0,1117936.4.S^3
1,86x10^(-4) = 0,10760805.S^3
1,72849.10^(-3) = S^3
S = 0,12001 so this is 3,33 times greater, so be it, the solubility of lead chloride would increase to 3,3/100L or 5,3g contained in
that ammount of water. Its worth noticing this formula gives the solubility ion terms of actual solution, not in terms of water content, so the
formula probably deals with a little more than 5,3g.
If there is still some NaCl and Pb(NO3)2 this value should rise further.
I can tell that once I put my hands on the equilibrium constants for all the reactions concerned, its a rather more straight way to workout than
simply assuming coefficients in terms of individual solubilities.
Poppy - 26-10-2012 at 20:01
The precipitated comprised 2g of lead salts, so there is still missing products .
Another experiment will be carried with "roasted" lead nitrate. I just don't want overdone salt as it easily decomposes to lead oxide and nitrogen
dioxides.
The ammounts will be improved also. Right.