Chemistry

In which aqueous system is PbI2 least soluble and why?
[A] 0.5 M HI [B] 0.2 M HI [C] 0.8 M KI

I understand that I- is the common ion. I don't understand why [C] 0.8 M KI is the correct answer.

  1. 👍
  2. 👎
  3. 👁
  1. Thanks for telling me what the problem is. That helps a lot.
    This is the common ion effect.
    HI is a strong acid. On ionization it give a (I^-) = 0.5 M for A and 0.2 M for B and 0.8 M for C. So here's the way it works, and I'll use the 0.8 M KI for the reasoning.
    .......................PbI2(s) ==> Pb^2+ + 2I^-
    I........................solid.............0...........0
    C.......................solid.............x...........2x
    E........................solid.............x...........2x
    So, if I want to know the solubility of PbI2 in a solution, you see it is
    Ksp = (Pb^2+)(I^-)^2 or (x)(2x) = 4x^3 = Ksp.
    But, when you add a common ion, in this case I'm using KI, the KI (its true the other HI too) ionizes completely and it dissolves completely to give
    ..................KI(s) ==> K^+ + I^-
    I..................0.8 M.......0........0
    C.................-0.8 M.....0.8M.....0.8 M
    E....................0.........0.8..........0.8 M
    So now that Ksp expression looks this way:
    Ksp = (Pb^2+)(I^-)^2
    For Pb you substitute x.
    For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
    Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above. The bottom line is this.
    .......................PbI2(s) ==> Pb^2+ + 2I^-
    I........................solid.............0...........0
    C.......................solid.............x...........2x
    E........................solid.............x...........2x
    Adding the I^- as a common ion, you INCREASE the 2I^- on the right which forces the equilibrium to the left to reduce the I^- anyway it can and that makes the PbI2 solid much less soluble. C is the answer because the I^- in KI is the largest of the three.
    Explanation is long because of the typing but it's all there. Follow up if necessary. I'm here for another few hours.

    1. 👍
    2. 👎
    👤
    DrBob222
  2. oops.
    I didn't proof when I should have proofed and this
    "For Pb you substitute x.
    For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
    Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above." should have been this.
    "For Pb you substitute x.
    For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
    Ksp = (x)(x+0.8)^2 so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/(0.8)^2 = much larger number than above when solving for x = solubility PbI2." Sorry about that.

    1. 👍
    2. 👎
    👤
    DrBob222

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    So I'm not exactly sure how to answer this problem. NaC2H3O2 is soluble Express your answers as ions separated by a comma. If the compound is not soluble, enter noreaction

  2. Chemistry

    For each substance, write the formula(s) of all species you expect to be present in aqueous solution in the first column. If there are significant minor species, write their formulas in the last column. If the substance is only

  3. chemistry

    Write the balanced chemical equation for each of these reactions. Include phases. When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. However, when additional aqueous

  4. Chemistry

    Is Glucose soluble in HCl, NaOH and NaHCO3 because it's soluble in water? - I put soluble for all because it dissolves in water. Is p-toluidine soluble in HCl, NaOH, and NaHCO3? I put soluble in HCl, insoluble in NaOH and

  1. Chemistry 110

    3. Suppose that part (II) of the experiment were carried out with aqueous NaOH instead of Ba(OH)2 to form soluble Na2SO4 along with H2O as products. (a) Write the overall balanced equation. (b) Write the net ionic equation. (c)

  2. chemistry

    The reaction of aqueous cobalt(II) iodide and aqueous lead(II) nitrate is represented by the balanced formula equation. CoI2(aq) + Pb(NO3)2(aq) → PbI2(s) + Co(NO3)2(aq) Give the balanced ionic equation for the reaction. Include

  3. Chemistry

    How do these react? Some may have no reaction. a. aqueous chromium(III) nitrate plus magnesium metal b. aqueous lithium sulfate plus aqueous barium chloride c. solid lead metal plus aqueous potassium chloride d. aqueous lithium

  4. Chemistry

    Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12× 10–12. What is the solubility (in mol/L) of silver chromate in 1.4 M potassium chromate aqueous solution? In 1.4 M silver nitrate aqueous

  1. Chemistry

    The net ionic equation for ALL aqueous reaction of strong acids and strong soluble bases that form soluble salts is 1. the formation of H2O from H+ and OH−. 2. the formation of water from the ele-ments. 3. the formation of water

  2. chemistry

    Name two metals which form both soluble trioxocarbonate(iv) acid and soluble tetraoxosulphate(vi).

  3. Chemistry

    Is PbI2 soluble in water or not? Following the solubility rules for ionic compounds, salts containing Pb are insoluble and iodide salts are soluble, so can it be both?

  4. Chemistry

    Which one of the following pairs of 0.100 mol L-1 solutions, when mixed, will produce a buffer solution? A. 50. mL of aqueous CH3COOH and 25. mL of aqueous CH3COONa B. 50. mL of aqueous CH3COOH and 25. mL of aqueous HCl C. 50. mL

You can view more similar questions or ask a new question.