posted by jimmy .
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.
A. The dissociation of Zn2Fe(CN)6 is:
Zn2Fe(CN)6s --> 2Zn2+aq + Fe(CN)64-
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