Required practical 4: testing for anions- Group 7 (halide ions), OH-, CO3^2-, SO4^2-

Question 1

test for sulphate ions

Answer 1

BaCl2 solution acidified with HCl

if acidified BaCl2 is added to a solution that contains sulfate ions a white ppt of barium sulfate forms

Question 2

why is HCl needed when testing for sulfate ions?

Answer 2

to react with carbonate impurities that are often found in salts which would form a white barium carbonate precipitate and so give a false result

• you cant use sulfuric acid because it contains sulfate ions and so would give a false positive result
• fizzing due to CO2 would be observed f a carbonate was present

Question 3

testing for the presence of halide ions with silver nitrate ions

• F
• Cl
• Br
• I

Answer 3

the test solutions is made acidic with nitric acid, and then silver nitrate solution is added dropwise
-Florides produce no precipate 
-Chlorides produce a white ppt
Ag+ (aq) + Cl- (aq) -> AgCl (s)
-Bromides produce a cream precipitate
Ag+ (aq) + Br- (aq) -> AgBr (s)
-Iodides produce a pale yellow precipiate
Ag+ (aq) + I- (aq) -> AgI (s)

Question 4

how do we differentiate between the different silver halide precipiates?

Answer 4

-sliver chloride dissolves in dilute ammonia to form a complex ion- colourless solution
AgCl (s) + 2NH3 (aq) -> [Ag(NH3)2]+ (aq) + Cl-(aq) colourless solution
-silver bromide dissolves in conc ammonia to form a complex ion
AgBr (s) + 2NH3 (aq) -> [Ag(NH3)2]+ (aq) + Br- (aq) colourless solution
-Silver iodide does not react with ammonia- it is too insoluble

Question 5

whats the role of nitric acid when testing for halide ions in silver nitrate?

Answer 5

to react with any carbonates present to prevent formation of the precipitate AG2CO3. This would mask the desired observations
2HNO3 + NA2CO3 -> 2NANO3 + H2O + Co2

Question 6

how doe we test for the presence of carbonate ions?

Answer 6

• add any dilute acid and observe effervescence
• bubble gas through limewater to test for CO2 - will turn limewater cloudy

-fizzing due to CO2 would be observed if a carbonate was present

2HCl + NaC3 -> 2NaCl + H2O + CO2

Question 7

how do we test for the presence of a hydroxide ion?

Answer 7

alkaline hydroxide ions will turn red litmus paper blue

Question 8

explain the trend in differing reducing power of halides

Answer 8

• a reducing agent donates electrons
• the reducing power of the halides increases down group 7
• they have a greater tendency to donate electrons
• this is because as the ions get bigger it is easier for the outer electrons to be given away as the pull from the nucleus on them becomes smaller

Question 9

what happens when F- and Cl- ions react with H2SO4?

Answer 9

they are not strong enough reducing agents to reduce the S in H2SO4
no redox reactions occur
only acid- base reactions occur
H2SO4 plays the role of an acid (proton donor)

NaF(s) + H2SO4(l) -> NaHSO4(s) + HF(g)
Observations: White steamy fumes of HF are evolved.
NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g)
Observations: White steamy fumes of HCl are evolved.

Question 10

what happens when Br- ions react with H2SO4?

Answer 10

Br- ions are stronger reducing agents than Cl- and F- and after the initial acid-base reaction reduce the sulfur in H2SO4from +6 to + 4 in SO2
Acid- base step: NaBr(s) + H2SO4(l)  NaHSO4(s) +HBr(g)
Redox step: 2HBr + H2SO4  Br2(g) + SO2(g)+2H2O(l)
Ox ½ equation 2Br - ->Br2 + 2e-
Re ½ equation H2SO4 + 2 H+ + 2 e- —> SO2 + 2 H2O

H2SO4 plays the role of acid in the first step producing HBr and then acts as an oxidising agent in the second redox step.

Observations: White steamy fumes of HBr are evolved.
Orange fumes of bromine are also evolved and a colourless, acidic gas SO2

Reduction product = sulfur dioxide

Question 11

what happens when I- ions react with H2SO4?

Answer 11

I- ions are the strongest halide reducing agents. They can reduce the sulfur from +6 in H2SO4 to + 4 in SO2, to 0 in S and -2 in H2S.

NaI(s) + H2SO4(l) –> NaHSO4(s) + HI(g)
2HI + H2SO4 –> I2(s) + SO2(g) + 2H2O(l)
6HI + H2SO4 –> 3 I2 + S (s) + 4 H2O (l)
8HI + H2SO4 –>4I2(s) + H2S(g) + 4H2O(l)

Ox ½ equation 2I - –> I2 + 2e-
Re ½ equation H2SO4 + 2 H+ + 2 e- –> SO2 + 2 H2O
Re ½ equation H2SO4 + 6 H+ + 6 e- –> S + 4 H2O
Re ½ equation H2SO4 + 8 H+ + 8 e- —>H2S + 4H2O

the H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in the three redox steps.

Reduction products = sulfur dioxide, sulfur and hydrogen sulfide

-Observations:
White steamy fumes of HI are evolved.
Black solid and purple fumes of iodine are also evolved
A colourless, acidic gas SO2 A yellow solid of sulfur H2S (hydrogen sulfide), a gas with a bad egg smell

Question 12

how do you make insoluble salts

Answer 12

Insoluble salts can be made by mixing appropriate solutions of ions so that a precipitate is formed barium nitrate (aq) + sodium sulfate (aq) –> barium sulfate (s) + + sodium nitrate (aq)
These are called precipitation reactions. A precipitate is a solid

Question 13

how do we remove salts when making insoluble salts

Answer 13

normally the salt would be removed by filtration, washed with distilled water to remove soluble impurities and then dried on filter paper

Question 14

what are spectator ions?

Answer 14

ions that :

• not changing state
• not changing oxidation number