Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag+ ion in the solution after equilibrium has been reestablished. For AgCl, Ksp = 1.8 × 10¯10

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  1. ........AgCl(s) ==> Ag^+(aq) + Cl^-(aq)

    How many moles KCl were added. That is 0.200g/molar mass KCl = about 0.003 but you need to be more accurate. In 1L that will be 0.003 M.
    ..........KCl ==> K^+ + Cl^-

    Now substitute into the Ksp expression for AgCl.
    1.8E-10 = (Ag^+)(Cl^-)
    (Ag^+) = x from AgCl.
    (Cl^-) = x from AgCl + 0.003 from KCl. You have one unknown, solve for that which is the solubility of AgCl under these conditions.

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