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A) Find [Zn^2+] and [Fe(CN)6 ^4-] in saturated solution of Zn2Fe(CN)6.
B) Repeat the above question in 0.1 mM ZnSO4 saturated with Zn2Fe(CN)6.

  • Chemistry -

    A. The dissociation of Zn2Fe(CN)6 is:
    Zn2Fe(CN)6s --> 2Zn2+aq + Fe(CN)64-
    For Zn2Fe(CN)6,
    Solubility = s = 3.74x10-6 M and
    [Fe(CN)64-] = solubility
    [Zn2+] = (2)(solubility) <--- see equation above

    B. Ksp = [Zn2+]2[Fe(CN)64-]
    2.1x10-16 = (0.1)2x
    Because of the low solubility we assume that the added concentration of Zn2+ is roughly equal to the total Zn2+ concentration. "x" represents the ferrocyanide ion concentration which changes because of the equilibrium shift caused by the addition of Zn2+ cations. The approximate concentration of Zn2+ is 0.1. To get the concentration of Fe(CN)64- solve for x

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