Question

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.

Answer 1

`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).

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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))`