These two compounds, silver diamine fluoride and potassium iodide (actually a solution called Lugol’s solution), are being used together in dentistry.
When combined they form a yellow precipitate. I cannot find any articles that state specifically what is formed. I would like to know what the yellow precipitate is and the steps involved in the reaction of the two chemicals. The equation is: $$\ce{Ag(NH3)2F + ssKI -> ?}$$
The yellow precipitate that forms is silver(I) iodide, $\ce{AgI}$. It is very, very insoluble and hence will precipitate immediately if silver ions and iodide ions meet in solution.
In fact, all three common halides other than fluoride form insoluble salts with silver: silver(I) chloride is a white precipitate, silver(I) bromide is somewhere between white and yellow and silver(I) iodide is yellow. These precipitates are so characteristic that they were historically (and still in qualitative inorganic analysis lab courses) used to identify the different halide ions in solution. To further improve the distinction, silver(I) chloride can be dissolved in dilute solutions of ammonia while silver(I) bromide needs concentrated ones. Silver(I) iodide does not dissolve in ammonia.
$$\begin{align}\ce{Ag+ (aq) + Cl- (aq) -> AgCl v &->[NH3 (dil)] [Ag(NH3)2]+ (aq) + Cl- (aq)}\tag{1}\\ \ce{Ag+ (aq) + Br- (aq) -> AgBr v &->[NH3 (conc)] [Ag(NH3)2]+ (aq) + Br- (aq)}\tag{2}\\ \ce{Ag+ (aq) + I- (aq) -> AgI v &->[NH3] \text{no reaction}}\tag{3}\end{align}$$
I assume that the original silver solution was the diammine complex to prevent silver chloride from precipitating as soon as chloride ions are added (due to not perfectly deionised water being used).
|
asked |
|
|
viewed |
8,979 times |
|
active |
