Which one of the following equilibrium reactions corresponds to the β2 constant for the complexation reaction described above? Part a) Ag+(aq) + 2Cn^-1(aq)

Groundwater near mining operations must be monitored to ensure that the cyanide ion, CN^- used to extract gold from ore does not enter the environment. Cyanide rapidly forms a stable complex with silver ion, Ag+, to produce Ag(CN)2^-1 with a β2 constant of 1.3*10^21. The large formation constant makes Ag+ useful as a titrant for cyanide ion.

1 Answer
May 20, 2016

#beta_2# is the cumulative complex formation constant for #K_1K_2#, i.e. the reaction up and including to the second step (not JUST the second step).


For the overall reaction

#\mathbf("Ag"^(+)(aq) + 2"CN"^(-)(aq) stackrel(beta_2" ")(rightleftharpoons) ["Ag"("CN")_2]^(-)(aq)),#

we have that:

#\mathbf(beta_2 = 1.3xx10^(21) = \frac([["Ag"("CN")_2]^(-)])(["Ag"^(+)]["CN"^(-)]^2))#

If you were confused and were actually asking for #K_2#, then we have a simple two-step complexation, wherein two equivalents of cyanide, as the stronger-field ligand than water, complex with silver (implicitly displacing water).

(However, there is no need to include water in the equation.)

For each step, we have

#"Ag"^(+)(aq) + "CN"^(-)(aq) stackrel(K_1" ")(rightleftharpoons) "AgCN"(aq),#

#K_1 = \frac(["AgCN"])(["Ag"^(+)]["CN"^(-)]),#

and

#"AgCN"(aq) + "CN"^(-)(aq) stackrel(K_2" ")(rightleftharpoons) ["Ag"("CN")_2]^(-)(aq),#

#K_2 = \frac([["Ag"("CN")_2]^(-)])(["AgCN"]["CN"^(-)]).#

Thus, multiplying them gives:

#color(blue)(beta_2)#

#= color(blue)(K_1K_2)#

#= \frac(cancel(["AgCN"]))(["Ag"^(+)]["CN"^(-)])\frac([["Ag"("CN")_2]^(-)])(cancel(["AgCN"])["CN"^(-)])#

#= color(blue)(\frac([["Ag"("CN")_2]^(-)])(["Ag"^(+)]["CN"^(-)]^2))#