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March 30, 2017

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Equal volumes of 0.240 M AgNO3 and 0.140 M ZnCl2 solution are mixed. Calculate the quilibrium concentrations of Ag+, and Zn2+.

  • Chemistry - ,

    ZnCl2(aq) + 2AgNO3(aq) ==> Zn(NO3)2(aq) + 2AgCl(s)
    I would assume a convenient volume of each, say 100 mL, then
    (ZnCl2) = (Zn^+2) = 100mL x 0.140M = 14 mmoles/200 mL = 0.070M Zn ion.
    (Cl^-) = 2 x 14 = 28 mmoles Cl^-

    The Ag^+ is trickier because AgCl is insoluble AND you have some of the Cl^-, which is in excess, acting as a common ion which decreases the solubility of AgCl.
    AgCl(s) ==> Ag^+ + Cl^-
    Ksp = (Ag^+)(Cl^-) = 1.82E-10
    (Ag^+) = Ksp/(Cl^-).
    You know Ksp. (Cl^-) is done this way. AgCl formed is 24 mmoles which uses up all of the Ag in the AgCl and 24 of the 28 Cl^- in the ZnCl2. So there are 4 mmoles Cl^- in excess and (Cl^-) = mmoles/mL = 4 mmoles/200 mL = ??
    Then substitute that into the (Ag^+) = Ksp/(Cl^-) and solve for Ag^+. I don't know if this was a trick question your prof gave you or if your prof wanted you to be on your toes and recognize that a ppt was formed and there was a common ion present.

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