Please use the values in the resources listed below instead of the textbook values. Calculate the equilibrium concentration (in M) of Ni2+ in a 0.71 M solution of Ni(CN),² (K = 1.7x 1030) %D
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Please use the values in the resources listed below instead of the textbook values.
Calculate the equilibrium concentration (in M) of Ni2+ in a 0.71 M solution of Ni(CN),2-. (K, = 1.7x 1030)
%D
4
4.0
t.
Supporting Materials
Periodic Table
Constants and
Supplemental Data
Factors
Additional Materials
Section 15.3"
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Given : Concentration of Ni(CN)42- = 0.71 M
And Kf of Ni(CN)42- = 1.7 X 1030
The formation reaction of Ni(CN)42- complex can be written as,
=> Ni2+ (aq) + 4 CN- (aq) -------> Ni(CN)42- (aq) Kf = 1.7 X 1030
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