Calc Solubility of PbCl2: 0.12g/0.015L

[henry3369]

Homework Statement

Suppose 200 mg of PbCl2 was added to 15.0 mL of water in a flask, and the solution was allowed to reach equilibrium at 20.0 C. Some solute remained at the bottom of the flask after equilibrium, and the solution was filtered to collect the remaining PbCl2, which had a mass of 80 mg . What is the solubility of PbCl2 (in g/L)?

Homework Equations

Not really sure:
g solute/L of solution
or
g solute/L of solvent

The Attempt at a Solution

I know how to solve this, but I was wondering if I'm supposed to include the the solute to the volume when finding solubility
So amount dissolved = 200-80 = 120 mg
120 mg = 0.12 g
15 ml = 0.015 ml
So is the answer:
0.12g/0.015 ml
OR:
0.012g/(0.015 ml + 0.012 ml)
Where 0.012 ml came from converting 0.012 g to ml.

 

[Chestermiller]

No. Solubility is defined based only on the volume of solvent.

Chet

 

[James Pelezo]

Chet, that's true for a constant mass of solute, but it can be shown that for salts like PbCl₂, the solubility in water at 25^oC is proportional to its K_sp value. For a 1:2 ionization ratio at 25^oC, the Solubility = (Ksp/4)^1/3. The published Ksp-value for PbCl₂ at saturation from standard K_sp tabels = 1.6 x 10^-5. This gives Solubility PbCl₂ = (1.6 x 10^-5/4)^1/3 = 0.016M = 2.00 gms/L (max solubility at saturation at 25^oC in DI water ).

Now, if I understand the above problem data of adding 200-mg PbCl₂ into 15-ml water and recovering 80-mg by filtration gives a mass of 120-mg of the salt that remains dissolved in solution. Based on this and assuming 120-mg did dissolve, then the concentration of PbCl₂ would be [(0.120-gm/15-ml) = [(0.120/277)mole/(0.015L)] = 0.029M in PbCl₂ x (277-gm PbCl₂/mole) would equal 8.00 g/L. This suggests that Lead(II) Chloride in the problem is 4 times more soluble than what would be calculated from published standard Ksp tables. This might be achievable by increasing the pH with OH^- to generate Pb-Hydroxide complex that would shift the PbCl₂ <=> Pb⁺² + 2Cl^- to the right, but unless something else is going on this problem, the most PbCl₂ will dissolve is 2.00-g/L = 2.00-mg/ml = 30.0-mg/15-ml max.; not 120-mg/15-ml.

Also, for salts having published Ksp values, S = solubility in moles/L:
Salts with 1:1 ionization ratios (AgCl) => S = (Ksp)^1/2
Salts with 1:2 or 2:1 ionization ratios (PbCl₂) => S = (Ksp/4)^1/3
Salts with 1:3 or 3:1 ionization ratios (Al(OH)₃) => S = (Ksp/27)^1/4
Salts with 2:3 or 3:2 ionization ratios (Pb(AsO₄)₂) => S = (Ksp/105)^1/5
________________________________________________
*Conditions are deionized water, 25^oC, 1.00 Atm. No common ion effect or complex-ion effect.
Caution: Do not use Ksp values to compare solubility unless all being considered have the same ionization ratio. Otherwise, use the appropriate solubility equation associated with the ionization ratio to compare solubility.

 

[Chestermiller]

Thanks James. I never imagined that they would give them problem data that would not be consistent with the solubility constant. Good catch.

Chet

 

[James Pelezo]

Thanks James. I never imagined that they would give them problem data that would not be consistent with the solubility constant. Good catch.

Chet

It surprised me too, I was going to play with the numbers and noticed it didn't match the theoretical calculated value. I though I'd overlooked some element 'different' condition I missed in the problem, but is was a plane jane solubility calculation.

 

[Borek]

Question is not about using Ksp (of which OP most likely have no idea), but about using data given.

Sadly, it is nothing unusual for the questions of this type to be not based on the real life data, but on whataeverIpickedoutfrommynose data.

15 ml = 0.015 ml

Beware: you mean liters on the right.

 

[James Pelezo]

Ha! Good point, Borek.