Unlabeled tubes contain solutions of sodium chloride, bromide, iodide and sulfide.

Unlabeled tubes contain solutions of sodium chloride, bromide, iodide and sulfide. What reactions can be used to distinguish between these solutions?

We will add silver nitrate or argentum en about three – AgNO3 to each salt residue.

Silver and Argentum the name of the same substance – Ag

The first salt is sodium chloride – NaCl, we write down the reaction:

NaCl + AgNO3 = NaNO3 + AgCl ↓

(white, cheesy)

Down arrows indicate that precipitation is occurring. In our case, silver chloride precipitated into a white precipitate, curdled in consistency. Under AgCl in the equation, we must sign what color the sediment is and do not forget to put the down arrow. So we formulate other equations as well. We also get NaNO3 – sodium nitrate.

Our next substance is sodium bromide – NaBr, we write down the reaction:

NaBr + AgNO3 = NaNO3 + AgBr ↓

( light yellow)

AgBr -silver bromide precipitates in a light yellow precipitate.

Our next substance is sodium iodide – NaI.

NaI + AgNO3 = NaNO3 + AgI ↓

(yellow color)

AgI ↓ – silver iodide precipitates yellow.

Our next substance is sodium sulfide – NaS

2AgNO3 + Na2S → 2NaNO3 + Ag2S ↓

(black)

Formed argentum sulfide or silver sulfide – a black precipitate.