Calculate the concentration of free Ag+ ions at equilibrium?

Commercial silver plating operations frequently use a solution containing the complex [Ag(CN)2]- ion. Because the formation constant Kf is quite large, this procedure ensures that the free Ag+ concentration in solution is low to promote uniform electrodeposition. In one process, a chemist added 9.0 L of 5.5 M NaCN to 90.0 L of .12 M AgNO3. Calculate the concentration of free Ag+ ions at equilibrium. Kf for this reaction is 1.0 x 10^21.

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  • Fern
    Lv 7
    4 years ago
    Favorite Answer

    Ag+ + 2CN- ===⇒ [Ag(CN)2]-

    5.5 moles NaCN/ 1liter x 9.0 liters x 1mole CN-/1mole NaCN = 49.5 moles CN-

    0.12 moles AgNO3/ 1liter x 90.0 liters x 1 mole Ag+/1mole AgNO3 = 10.8 moles Ag+

    10.8 moles Ag+ x 2 moles CN-. 1mole Ag+ = 21.6 moles CN- are consumed

    10.8 moles Ag+ x 1 mole [Ag(CN)2]- / 1mole Ag+ = 10.8 moles [Ag(CN)2]- are formed.

    Moles of CN- that remain: 49.5 moles – 21.6 moles = 27.9 moles CN-

    Final volume = 9.0 L + 90.0 L = 99 Liters

    [CN-] = 27.9 moles CN-/ 99 liters = 0.282

    [Ag(CN)2]- = 10.8 moles / 99 liters = 0.109

    Assume that [Ag(CN)2]- is 0.109 since the Kf is so large.

    Kf = [Ag(CN)2]- / [Ag+][CN-]^2 = 1.0 x 10^21

    0.109 / [Ag+](0.282)^2 = 1.0 x 10^21

    [Ag+] = 0.109/(0.282)^2(1.0 x 10^21)

    [Ag+] = 1.37 x 10^-21

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