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Calculating Solubility

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data
    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)?

    2. Relevant equations
    Not really sure:
    g solute/L of solution
    g solute/L of solvent

    3. 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
    0.012g/(0.015 ml + 0.012 ml)
    Where 0.012 ml came from converting 0.012 g to ml.
  2. jcsd
  3. May 13, 2015 #2
    No. Solubility is defined based only on the volume of solvent.

  4. May 13, 2015 #3

    James Pelezo

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    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 25oC is proportional to its Ksp value. For a 1:2 ionization ratio at 25oC, the Solubility = (Ksp/4)1/3. The published Ksp-value for PbCl2 at saturation from standard Ksp 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 25oC 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, 25oC, 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.
    Last edited: May 13, 2015
  5. May 13, 2015 #4
    Thanks James. I never imagined that they would give them problem data that would not be consistent with the solubility constant. Good catch.

  6. May 14, 2015 #5

    James Pelezo

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    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.
  7. May 14, 2015 #6


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    Staff: Mentor

    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.

    Beware: you mean liters on the right.
  8. May 14, 2015 #7

    James Pelezo

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    Gold Member

    Ha! Good point, Borek.
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