Asked 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.
Answered by GK
A. The dissociation of Zn2Fe(CN)6 is:
Zn<sub>2</sub>Fe(CN)<sub>6</sub><sub>s</sub> --> 2Zn<sup>2+</sup><sub>aq</sub> + Fe(CN)<sub>6</sub><sup>4-</sup>
For Zn2Fe(CN)6,
Solubility = s = 3.74x10<sup>-6</sup> M and
K<sub>sp</sub>=2.1x10<sup>-16</sup>
[Fe(CN)<sub>6</sub><sup>4-</sup>] = solubility
[Zn<sup>2+</sup>] = (2)(solubility) <--- see equation above

B. K<sub>sp</sub> = [Zn<sup>2+</sup>]<sup>2</sup>[Fe(CN)<sub>6</sub><sup>4-</sup>]
2.1x10<sup>-16</sup> = (0.1)<sup>2</sup>x
Because of the low solubility we assume that the added concentration of Zn<sup>2+</sup> is roughly equal to the total Zn<sup>2+</sup> concentration. "x" represents the ferrocyanide ion concentration which changes because of the equilibrium shift caused by the addition of Zn<sup>2+</sup> cations. The approximate concentration of Zn<sup>2+</sup> is 0.1. To get the concentration of Fe(CN)<sub>6</sub><sup>4-</sup> solve for x
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