This is a precipitation reaction. Halides of silver are insoluble in water (except silver fluoride) whereas all nitrates are soluble in water. Sodium salts are soluble. Thus, silver iodide is the precipitate.
Formula:
AgNO3(aq) + NaI(aq) -> AgI(s) + NaNO3(aq)
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
A white precipitate of silver iodide forms due to the reaction between silver ions and iodide ions, leaving potassium nitrate in solution. This reaction is a double displacement reaction and is used as a test for iodide ions.
Silver nitrate solution and potassium iodide solution can be mixed to form silver iodide due to a double displacement reaction where the silver ions from silver nitrate react with the iodide ions from potassium iodide to form insoluble silver iodide precipitate. The balanced chemical equation for this reaction is: AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq).
A yellow precipitate of silver iodide is formed due to the reaction between potassium iodide and silver nitrate, as silver iodide is insoluble. The reaction can be described by the equation: 2KI (aq) + AgNO3 (aq) → AgI (s) + 2KNO3 (aq)
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Hydrogen iodide can be tested using silver nitrate solution. When hydrogen iodide is bubbled through silver nitrate solution, a yellow precipitate of silver iodide is formed. This confirms the presence of iodide ions in the sample.
When silver nitrate and strontium iodide react, a double displacement reaction occurs. Silver iodide and strontium nitrate are formed as products. Silver iodide is a yellow precipitate while strontium nitrate remains in solution.
A white precipitate of silver iodide forms due to the reaction between silver ions and iodide ions, leaving potassium nitrate in solution. This reaction is a double displacement reaction and is used as a test for iodide ions.
You can test for the presence of iodide ions using silver nitrate. When silver nitrate is added to a solution containing iodide ions, a yellow precipitate of silver iodide forms. This precipitate confirms the presence of iodide ions in the solution.
Silver nitrate solution and potassium iodide solution can be mixed to form silver iodide due to a double displacement reaction where the silver ions from silver nitrate react with the iodide ions from potassium iodide to form insoluble silver iodide precipitate. The balanced chemical equation for this reaction is: AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq).
Dissolve each of the silver nitrate and potassium iodide separately in water, then mix the two solutions slowly with stirring. Silver iodide will precipitate and can be separated by filtering it from the liquid.
A yellow precipitate of silver iodide is formed due to the reaction between potassium iodide and silver nitrate, as silver iodide is insoluble. The reaction can be described by the equation: 2KI (aq) + AgNO3 (aq) → AgI (s) + 2KNO3 (aq)
Potassium iodide is added in excess to ensure that all available lead nitrate has reacted to form lead iodide. This helps to maximize the yield of lead iodide and ensures that there is no excess lead nitrate remaining in the solution.
They form Silver Iodide and Poassium nitrate. Silver Iodide is a yellow coloured compound which forms precipitate in the solution. Potassium Nitrate remains dissolved. These reactions only take place in solution. When both reactants are in solid state then reaction may not occur, or it is too slow to be observed.
Silver nitrate is commonly used to test for the presence of halide ions, such as chloride, bromide, and iodide ions. When silver nitrate is added to a solution containing these ions, characteristic precipitates are formed: white for chloride ions, cream for bromide ions, and yellow for iodide ions.
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.