You are tasked with determining the equilibrium constant of the Ni^2+ and CN^-=

Question

You are tasked with determining the equilibrium constant of the Ni^2+ and CN^-= complexation reaction. This experiment contains several steps. Ni^2+ (aq) + 4 Cn^- (aq) rightarrow Ni(CN)_4^2- (aq) What is the mathematical expression for K_eq for the above reaction? 10 mL of a 0.25 M Ni^2+ solution and 5 mL of a 0.60 M CN solution are combined and diluted to a final volume of 25 mL. What are [NI^2+] and (CN^-] in this mixture? After making the above mixture, the absorbance of this solution was measured. Based on this absorbance. [Ni(CN)_4^2-]_eq was determined to be 0.097 M. Based on this information and the above reaction equation, what are [Ni^2+]_eq and [CN^-]_eq? Based on your expression for K_eq and the equilibrium concentrations [Ni^2+]_eq [CN^-]_eq and [Ni(CN)_4^2-]_eq compute K_eq for this reaction.

Explanation / Answer

Ni^2+ (aq) + 4CN^- <==> Ni(CN)4 ^2-

Keq = [Ni(CN)4]^2-/[Ni^2+][CN^-]^4

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concentration of Ni^2+ in the diluted solution = 10mL * 0.25 M/25 mL = 0.1 M

concentration of CN^- in the diluted solution = 5mL * 0.60 M /25mL = 0.12 M

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[Ni^2+] eqm = 0.1M - 0.097 = 0.003 M

[CN^-] eqm = 0.12-0.097 = 0.023 M

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Keq = 0.097 / 0.003 * (0.023) ^4 = 61121.6

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