Use Silver nitrate.
A yellow precipitate of silver iodide will form /
AgNO3 + I^- = AgI(s) + NO3^-
This is one of the classic tests for halogens.
No, potassium iodide is a compound composed of the monatomic ion K+ (potassium cation) and the monatomic ion I- (iodide anion). It is not a polyatomic ion.
iodine and iron
One common method to detect the presence of Iodide ions (I-) is to use the Silver Nitrate test. When Silver Nitrate solution reacts with Iodide ions, a yellow precipitate of Silver Iodide is formed. This precipitate is a visual indicator that Iodide ions are present in the solution.
Ammonium Iodide
The formula for the compound formed by the combination of barium ion (Ba^2+) and iodide ion (I^-) is BaI2, which is barium iodide. This compound is formed when one barium ion combines with two iodide ions due to their respective charges.
It is not the anions (e.g. iodide) that are responsible for the flame test color, rather the cations such as sodium ion, potassium ion and calcium ion give you different colors.
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Iodide ion : I-
No, potassium iodide is a compound composed of the monatomic ion K+ (potassium cation) and the monatomic ion I- (iodide anion). It is not a polyatomic ion.
iodine and iron
One common method to detect the presence of Iodide ions (I-) is to use the Silver Nitrate test. When Silver Nitrate solution reacts with Iodide ions, a yellow precipitate of Silver Iodide is formed. This precipitate is a visual indicator that Iodide ions are present in the solution.
No, iodide is an ion formed by the nonmetal iodine.
Sodium peroxoborate is tested with sulfuric acid and potassium iodide to determine the presence of peroxides. When sodium peroxoborate reacts with sulfuric acid and potassium iodide, oxygen gas is released, causing iodine to be liberated from the potassium iodide. The presence of iodine can be observed by a color change from clear to brown or blue-black due to the formation of iodine.
The formula for ammonium iodide is NH4I. It is composed of one ammonium ion (NH4+) and one iodide ion (I-).
Ammonium Iodide
Testing for Iodide, I-Sodium iodide, NaI, is the source of iodide anion for this experiment.Reaction with bleach involves three steps. The brown color shows the presence of I3- ions.Hypochlorite ion yields chlorine:OCl- (aq) + Cl- (aq) + H2OCl2 (aq) + 2 OH-Chlorine reacts with iodide anion:Cl2 (aq) + 2 I- (aq)I2 (aq) + 2 Cl- (aq)Triiodide ion is formed:I2 (aq) + I- (aq) I3- (aq)Starch reacts with iodine and iodide to form a characteristic blue/black complex. A corn starch packing peanut is shown here.Silver ion reacts with iodide to form silver iodide, AgI.Ag+ (aq) + I- (aq) AgI (s)Reaction of iodide with sulfuric acid produces hydrogen sulfide gas and brown triiodide solution in a series of reactions:I- (aq) + H2SO4 (aq) HI (aq) + HSO4- (aq)8 HI + H2SO4 (aq) H2S (g) + 4 I2 (aq) + 4 H2OI2 (aq) + I- (aq) I3- (aq)Again, starch is used to confirm the presence of iodine and iodide.
The formula for the compound formed by the combination of barium ion (Ba^2+) and iodide ion (I^-) is BaI2, which is barium iodide. This compound is formed when one barium ion combines with two iodide ions due to their respective charges.