required practical 4 - inorganic test for ions Flashcards
method for testing cations - used on group 2 metal ions and transtition metal ions
-add diltue NaOH
-10 drops of 0.1moldm-3 ion solution in tt- barium chloride
-add 10 drops of NaOH mix well + record obvs
-continue ot add NaOH dropwise whilst shaking until inexcess and record observations
-repeat with others - calcium bromide, magnesium chloride, strontium chloride
results for cations test - group 2
-gorup 2 hydroxide get more soluble down the group
-magnesium hydroxide - insoluble and appear as white ppt - Mg2+ + 2OH- = Mg(OH)2 - may appear slightly alkaline - OH produced
-calcium hyrdoxide - partially insoluble and will appear as white ppt - may need more NaOH to compare ewith Mg - same eq as mG
- ^ more alkaline than Mg as its more soluble so more OH-
- strontium- may have very light ppt and barium - no ppt due to high solubility
- highly alkaline
results for test for cations - group 3 and transition metals
group 3
-aluminium salts will form a white ppt of aluminium hydroxide
[Al(H2O)6]3+(aq) + 3OH-(aq) → Al(H2O)3(OH)3(s) + 3H2O(l)
-which will dissolve in excess NaOH to form colourless solution
Al(H2O)3(OH)3(s) + OH- (aq) → [Al(OH)4]-(aq) + 3H2O(l)
transtition metals
-copper solutions - colourless to blue ppt
[Cu(H2O)6]2+(aq) + 2OH-(aq) → Cu(H2O)4(OH)2(s) + 2H2O(l)
-iron(II) solution - colourless to green ppt
[Fe(H2O)6]2+ (aq) + 2OH-(aq) → Fe(H2O)4(OH)2(s) + 2H2O(l)
-iron (III) - colourless to brown/orange ppt
[Fe(H2O)6]3+(aq) + 3OH-(aq) → Fe(H2O)3(OH)3(s) + 3H2O (l)
test for ammonium ions
. place about 10 drops of 0.1 mol dm-3 ammonium chloride/ unknown solution in a test tube
2. add about 10 drops of 0.4 mol dm-3 sodium hydroxide solution
3. shake the mixture
4. warm the mixture in the test tube gently using a water bath (more gentle than flame)
5. test the fumes released from the mixture by holding a piece of damp red litmus paper in the mouth of the test tube (use forceps)
results: alkaline ammonia gas is released which turns the red litmus paper blue
method for testing for cations - using sulfate ions
method: adding sulfate ions
1. place about 10 drops of 0.1 mol dm-3 metal ion solution (e.g. barium chloride) in a test tube
2. add about 10 drops of 1.0 mol dm-3 sulfuric acid (or other soluble sulfate solution), mix + record observations
3. continue to add sulfuric acid solution, dropwise with gentle shaking, until in excess
4. repeat test with the calcium bromide, magnesium chloride and strontium chloride
trend: Group II sulfates become less soluble down the group
magnesium and calcium salts:
- magnesium will remain colourless due to high solubility even upon adding excess
- calcium - intial will be colourless but if add excess - may get slight white ppt
strontium and barium solutions:
- will form white precipitates with addition of sulfate ions:
SrCl2(aq) + Na2SO4(aq) → 2NaCl(aq) + SrSO4(s) ionic equation: Sr2+(aq) + SO4 2-(aq) → SrSO4(s) ionic equation: Ba2+(aq) + SO4 2-(aq) → BaSO4(s)
test for sulphate ions
reagent to test for sulphate ions: BaCl2 solution acidified with dilute hydrochloric acid (equal volumes)
- if acidified barium chloride is added to a solution that contains sulfate ions a white precipitate of barium sulfate forms
simplest ionic equation:
Ba2+(aq) + SO42-(aq) → BaSO4(s)
hydrochloric acid is needed to react with carbonate impurities often found in salts → would form a white barium carbonate precipitate = false result
2HCl + Na2CO3 → 2NaCl + H2O + CO2
- fizzing due to CO2 would be observed if a carbonate was present
- cannot use sulfuric acid → contains sulfate ions and so would give a false positive result
test for halide ions
- test solution is made acidic with nitric acid, and
then silver nitrate solution is added dropwise - role of nitric acid: to react with any carbonates present to prevent the formation of the precipitate Ag2CO3 → would mask the desired observations
2HNO3 + Na2CO3 → 2 NaNO3 + H2O + CO2 - fluorides produce no precipitate
- chlorides produce a white precipitate:
Ag+(aq) + Cl-(aq) → AgCl(s) - bromides produce a cream precipitate:
Ag+(aq) + Br-(aq) → AgBr(s) - iodides produce a pale yellow precipitate:
Ag+(aq) + I-(aq) → AgI(s)
test to distinguish between halide ions
silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar:
- silver chloride dissolves in dilute ammonia to form a complex ion (→ colourless solution):
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)
- silver bromide dissolves in concentrated ammonia to form a complex ion (→ colourless solution): AgBr(s) + 2NH3 (aq) → [Ag(NH3)2]+(aq) + Br-(aq)
- silver iodide does not react with ammonia - it is too insoluble
test for the prescence of carbonate ions
- add equal small volume of any dilute acid (e.g. HCl) to (sodium carbonate) solution in test tube
- observe effervescence (fizzing due to CO2 would be observed if a carbonate was present)
bubble gas through limewater to test for CO2 → will turn limewater cloudy:
- use a delivery tube (with bung to prevent product loss) to transfer gas produced into a second test tube containing a small volume of calcium hydroxide solution (limewater)
- put a stopper into the test tube + shake tube
- limewater goes cloudy if carbonate ions present (CO2)
2HCl + Na2CO3 → 2NaCl + H2O + CO2
test for hydroxide ions
test a 1 cm depth of solution in a test tube with red litmus paper or universal indicator paper
- result: alkaline hydroxide ions will turn red litmus paper blue
explanation of the trend of the reactions of group 7 with conc H2SO4
the halides show increasing power as reducing agents as go down the group:
- 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
fluoride and chloride ions with conc H2SO4
- F- and Cl- ions 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 evolved
NaCl(s) + H2SO4 (l) → NaHSO4 (s) + HCl(g) observations: white steamy fumes of HCl evolved
bromide and conc H2SO4
Br- ions = stronger reducing agents than Cl- and F- - after the initial acid-base reaction Br- ions reduce the sulfur in H2SO4 from +6 to +4 in SO2
acid-base step (H2SO4 plays the role of acid):
NaBr(s) + H2SO4(l) → NaHSO4(s) + HBr(g)
redox step (H2SO4 acts as an oxidising agent):
2HBr + H2SO4 → Br2 (g) + SO2 (g) + 2H2O(l) observations:
- white steamy fumes of HBr evolved
- orange fumes of bromine evolved
- colourless, acidic gas SO2 (reduction product)
Ox ½ equation: 2Br- → Br2 + 2e-
Re ½ equation: H2SO4 + 2H+ + 2e- → SO2 + 2H2O
iodide and conc H2SO4
I- ions = strongest halide reducing agents
- can reduce the sulfur from +6 in H2SO4 to +4 in SO2, to 0 in S and -2 in H2S
acid-base step (H2SO4 plays the role of acid):
NaI(s) + H2SO4(l) → NaHSO4(s) + HI(g)
3 redox steps (H2SO4 acts as an oxidising agent):
2HI + H2SO4 → I2(s) + SO2(g) + 2H2O(l)
6HI + H2SO4 → 3I2 + S(s) + 4H2O (l)
8HI + H2SO4 → 4I2(s) + H2S(g) + 4H2O(l)
observations:
- white steamy fumes of HI evolved
- black/grey solid and purple fumes of iodine evolved
observations of reduction products:
- colourless, acidic gas SO2 (sulfur dioxide)-
- yellow solid of sulfur
- H2S (hydrogen sulfide) = gas with a bad egg smell
Ox ½ equation: 2I- → I2 + 2e-
Re ½ equation: H2SO4 + 2H+ + 2e- → SO2 + 2H2O Re ½ equation: H2SO4 + 6H+ + 6e- → S + 4H2O
Re ½ equation: H2SO4 + 8H+ + 8e- → H2S + 4H2O
types of filtration that can be used
gravitational filtration - use if small amounts of solid are formed (include: filter paper, filter funnel)
vacuum filtration - apparatus connected to a water pump which will provide a vacuum, use if larger amounts of solid are formed (include: buchner funnel, buchner flask (has thicker glass walls than a normal flask to cope with the vacuum), filter paper, air outlet to water pump)
