Question 12.EP.3: In Example Problem 12.2, we saw that cyanide ions precipitat...
In Example Problem 12.2, we saw that cyanide ions precipitate with silver. Bromide ions show similar behavior. The two pertinent equilibria are
Ag^{+}(aq)+CN^{-}(aq)\rightleftarrows AgCN(s) K = 1.7 × 10^{16}
Ag^{+}(aq)+Br^{-}(aq)\rightleftarrows AgBr(s) K = 1.9 × 10^{12}
Which ion is more likely to precipitate silver from a solution, assuming roughly equal concentrations of cyanide and bromide ions?
Strategy
The likely extent of each reaction can be predicted by the size of the equilibrium constant. By comparing the two numbers, we can see which anion is more likely to precipitate the silver.
Both equilibrium constants are much greater than one, but the value for the reaction with cyanide ions is significantly larger than that for bromide. Thus silver cyanide has a much greater tendency to form and will be the more likely product.