# 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.

May 20, 2016

${\beta}_{2}$ is the cumulative complex formation constant for ${K}_{1} {K}_{2}$, 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

$\text{Ag"^(+)(aq) + "CN"^(-)(aq) stackrel(K_1" ")(rightleftharpoons) "AgCN} \left(a q\right) ,$

${K}_{1} = \setminus \frac{\left[{\text{AgCN"])(["Ag"^(+)]["CN}}^{-}\right]}{,}$

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:

$\textcolor{b l u e}{{\beta}_{2}}$

$= \textcolor{b l u e}{{K}_{1} {K}_{2}}$

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

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