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: $$\ce{Ag(NH3)2F + ssKI -> ?}$$

  • 2
    $\begingroup$ Note that Lugol’s solution is not $\ce{KI}$ but $\ce{KI/I2}$ or $\ce{KI3}$. $\endgroup$
    – Jan
    Feb 9, 2017 at 12:28
  • $\begingroup$ Silver iodide is formed which is yellow in color. $\endgroup$ Feb 9, 2017 at 13:00

2 Answers 2


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).


To add further to your question, for a silver diamine dental product (Riva Star by SDI), the white precipitate disappears after applying more potassium iodide to the mixture. The excess solution in the tooth cavity is then rinsed off with water. There is some cool chemistry going on here but I am unsure what it is. Riva Star comes in 2 solutions that are mixed together in the tooth cavity. Solution (1): Silver diamine fluoride has a pH of 10. Solution (2): Potassium iodide has a pH of 7. Tooth: Consists of collagen, proteins, hydroxyapatite, bacteria.

When solution (1) is first applied to the tooth, the high pH has been shown to aid in the formation of covalent bonds of phosphate groups onto proteins and crystallites to grow. It has been postulated that the fluoride usually reacts with hydroxyapatite, by replacing the calcium to form fluoroapatite, while the silver diamine exerts an antibacterial effect. The silver ions also react with the tooth structure to form the dark black silver phosphates. So, the possible by-products could be Calcium ions, fluoride ions, ammonium ions, silver ions, silver phosphate.

The addition of solution (2) is to remove the excess silver ions that may form the dark silver phosphates. This is to avoid black stains on the tooth. Solution (2) is added into the cavity to form white precipitates and more solution is added until the white precipitate disappears. What chemical reaction(s) could be going to explain the formation and dissolution of the white precipitate?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.