6-17 From the solubility product of Zinc ferrocyanide Zn2 Fe (CN)6,calculate the

Question

6-17 From the solubility product of Zinc ferrocyanide Zn2 Fe (CN)6,calculate the concentration of   Fe(CN)6 in .1mM ZnSO4saturated with Zn2 Fe (CN)6 .Assume that Zn2 Fe(CN)6 is a neglible source of Zn2+.
I found the Ksp as 2.1x10-16
I have the equation 2ZnSO4 +Fe (CN)6 yields 2SO4 +Zn2 Fe (CN)6 I am not sure if this is right equation but any helpwould be appreciated Also the answer is 2.1 x10-8 I need help getting the right concentrations
Part B : What concentration of K4Fe(CN)6 should be insuspension of solid Zn2(CN)6 in water to give [Zn2+]=5 x 10-7M. Answer:8.4 x 10-4
I found the Ksp as 2.1x10-16
I have the equation 2ZnSO4 +Fe (CN)6 yields 2SO4 +Zn2 Fe (CN)6 I am not sure if this is right equation but any helpwould be appreciated Also the answer is 2.1 x10-8 I need help getting the right concentrations
Part B : What concentration of K4Fe(CN)6 should be insuspension of solid Zn2(CN)6 in water to give [Zn2+]=5 x 10-7M. Answer:8.4 x 10-4

Explanation / Answer

Part A
The solubility product equation for precipitation reactiondescribed by chemicalequation 2(Zn^2+) + (Fe(CN)6)^4- =>Zn2(FeCN)6) is: 2.1x10^-16=[Zn2+]^2 x [Fe(CN)6] Thus, 2.1x10^-16=(0.1x10^-3)^2 x [Fe(CN)6] Next, 2.1x10^-16=10^-8 x [Fe(CN)6] 2.1x10^-8=[Fe(CN)6] which is the answer you are looking for. Part B: Equilibrium [Fe(CN)6] concentrations is given by itsanalytical concentration (as K4[Fe(CN)6] is very soluble in water)and solubility product equation becomes: 2.1x10^-16=(5x10^-7)^2 x [Fe(CN)6] Thus, [Fe(CN)6]=2.1x10^-16 / 25x10^-14=8.4x10^-4 mol/L

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