The solubility of insoluble substances changes depending on the nature of the solution. Below are two solutions
in which Cu(OH)2 is dissolved; in each case, the solubility of Cu(OH)2 in those solutions is not the same as it is
in pure water.
(a) For each Cu(OH)2 solution below, state whether the solubility is greater than or less than that of pure water
and briefly account for the change in solubility.
i. 1-molar HCl(aq)
ii. 1-molar Cu(NO3)2(aq)
A saturated solution of lead(II) iodide in pure water has a lead ion concentration of 1.3 × 10-3 M at 25°C.
(b) Calculate the value for the solubility-product constant of PbI2 at 25°C.
(c) Calculate the molar solubility of PbI2. in a 0.10 M Pb(NO3)2 solution at 25°C.
(d) To 333 milliliters of a 0.12 M Pb(NO3)2 solution, 667 milliliters of 0.015 M KI is added. Does a precipitate form?
Support your answer with calculations and a concluding statement from those calculations.

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asked by Henry
  1. #1.
    i. more soluble in HCl because
    Cu(OH)2 + 2HCl ==> CuCl2 + 2H2O
    ii. less soluble in Cu(NO3)2 because of the common ion effect of Cu^2+ from the Cu(NO3)2. In pure water,
    .......Cu(OH)2 ==> Cu^2+ + 2OH^-
    If we add extra Cu^2+ from Cu(NO3)2, that disturbs the equilibrium and Le Chatelier's principle says the reaction will shift to the left which means less solubility.

    .........PbI2 ==> Pb^2+ + 2I^-

    Substitute the E line into Ksp expression. The problem tells you the value of x, calculate value for 2x, substitute into Ksp expression and solve for Ksp.

    c. Same process as b bu Pb is not x + 0.1. Substitute into Ksp and solve for x which gives you the molar solubility of PbI2.

    d. Calculate M Pb(NO3)2 and M KI in the mixed solution. Then calculate Qsp and compare with Ksp.
    Post your work if you get stuck.

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