For chemistry students · 4 minute read
What precipitate forms when silver nitrate reacts with potassium iodide?
Silver nitrate and potassium iodide form silver iodide, an insoluble yellow precipitate. Potassium and nitrate ions remain dissolved in the solution.
Try the yellow precipitate setupThe exact supported pair opens in a beaker. Nothing runs until you select Analyze.
Supported prediction
AgNO3(aq) + KI(aq) -> AgI(s) + KNO3(aq)
What the equation means
Silver nitrate and potassium iodide undergo a precipitation reaction. Insoluble silver iodide forms as a yellow solid while potassium and nitrate ions remain in solution. The molecular equation has a one-to-one reactant ratio.
Expected observation
A yellow silver iodide precipitate
OmniLab shows a bounded yellow precipitate cue inside the beaker to represent the expected silver iodide solid.
Read the equation
Three details that explain the result
Connect the coefficients and formulas to the gas, solution, and visible cue.
-
Find the insoluble product
AgI has the (s) state symbol. Silver iodide separates from the solution as a yellow solid.
-
Identify the dissolved ions
KNO3 remains aqueous. Potassium and nitrate ions stay in solution while silver and iodide form AgI.
-
Check the one-to-one ratio
One AgNO3 unit and one KI unit give one AgI unit and one KNO3 unit. The equation needs no larger coefficients.
Three safety notes
Read these before any physical experiment
- 01
Wear gloves and splash goggles when handling silver nitrate.
- 02
Protect the silver iodide precipitate from strong light.
- 03
Collect silver-containing waste for approved disposal.
Keep exploring
Connect this observation to another reaction
These guides use the same pattern: predict the equation, identify the visible cue, then check what the virtual model leaves out.
See how the full virtual lab works