Explain what is the concentration of cn- at equilibrium

Reference no: EM13233541

A 50.0-mL sample of 2.0 x 10-4 M CuNO3 is added to 50.0 mL of 4.0 M NaCN. Cu+ reacts with CN- to form the complex ion [Cu(CN)3]^2:
[Cu(CN)3]^2- <--> Cu+ (+) 3CN- K = 1.0 x 10^-9 What is the concentration of CN- at equilibrium?

Reference no: EM13233541

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