test for anions

Question 1

to detect the presence of chloride ions in aqueous solution.

Answer 1

chloride salt dissolved in deionized water.
add few drops of silver nitrate to the solution (AgNO3)
test tube turns cloudy, white precipitate silver chloride.

Question 2

to detect the presence of chloride ions in aqueous solution formula.

Answer 2

Ag+ + Cl- = AgCl, with precipitate

silver nitrate and cloride salt = silver chloride

Question 3

to detect the presence of chloride ions in aqueous solution, confirmatory test.

Answer 3

to confirm the precipitate is silver chloride, add dilute ammonia solution, the solution should be restored to its clear state if the precipitate is silver chloride bc its soluble in dilute ammonia solution.

Question 4

detect the presence of sulfate ions SO4-2 and sulfite ions SO3-2 in aqueous solution.

Answer 4

sulfate ion dissolved in deionized water; sulfite ion dissolved in water.
barium chloride [BaCl2] added to both solutions.
a white precipitate is formed in both test tubes.

Question 5

detect the presence of sulfate ions SO4-2 white precipitate

Answer 5

the white precipitate formed in the sulfate ion test tube is barium sulfate.
formula: Ba+2 + SO4-2= BaSO4 with precipitate
or barium chloride + sulfate ion = barium sulfate.

Question 6

detect the presence of sulfite ions SO3-2 white precipitate

Answer 6

the white precipitate formed in the sulfite ion test tube is barium sulfite.
formula: Ba+2 + SO3-2= BaSO3 with precipitate
or barium chloride + sulfite ion = barium sulfite.

Question 7

distinguish between sulfate and sulfite ions

Answer 7

dilute HCL is used to distinguish between barium sulfate and barium sulfite precipitates.
barium sulfite is soluble in dilute HCL so the cloudiness will disappear, think because it has a more simple chemical structure so it’s easier to dissolve.
the barium sulfate remains though because it’s not soluble in HCL

Question 8

to detect the presence of carbonate ions

[CO3-2] and hydro carbonate ions [HCO3-] in aqueous solution

Answer 8

place carbonate ions in a boiling tube and cork with 1 hole rubber stopper. have the glass one hole stopper feeding into another test tube with line water. remove the one-hole stopper and add dilute HCL acid, quickly replacing it and setting it back up as it was before.
an effervescence is observed as the acid comes in contact with the carbonate to liberate CO2.
formula- CO3-2 + 2H+ = CO2 + H2O