Experiment 1. Detection of sulfate ions
Pour 1-2 ml of sodium sulfate solution into one test tube, and 1-2 ml of potassium sulfate solution into the other. Add barium chloride solution drop by drop to both test tubes. Explain what you observed.
Make up equations electrolytic dissociation taken salts and the equation of the exchange reaction. Write down the full and abbreviated ionic equations for the reaction.
What compounds can serve as a reagent for barium ions Ba 2+?
What is the essence of detecting ions using a reagent?
Experiment 2. Detection of Cl chloride ions -
Using the solubility table, find out which salts containing the chloride ion Cl - are insoluble (slightly soluble). Using the reagents you have, prove that chloride ions are present in the sodium chloride solution.
Make up equations for the dissociation of salts, exchange reactions, and complete and abbreviated ionic equations for the reactions carried out.
Experiment 3. Detection of sulfate ions and chloride ions Cl -
Two test tubes contain solutions of potassium chloride and magnesium sulfate. What reactions can be used to prove that one test tube contains a solution of potassium chloride, and the other contains a solution of magnesium sulfate?
Divide the solution from the first test tube in half and pour into two test tubes. Pour a solution of lead (II) nitrate into one test tube and a solution of barium chloride into the other. In which test tube did the precipitate form? Which salt - KCl or MgSO 4 - is contained in the first test tube?
Test the solution from the second test tube for the presence of an anion not found in the first test tube. To do this, add a solution of lead (II) nitrate to the test solution. Explain what you observed.
Create exchange equations for the reactions you have carried out and complete and abbreviated ionic equations for ion detection reactions.
Experience 4
Carry out reactions to confirm high-quality composition the following substances: a) barium chloride; b) magnesium sulfate; c) ammonium carbonate. Use Table 12 to perform this experiment.
Table 12
Ion determination
Let's imagine this situation:
You are working in a laboratory and have decided to conduct an experiment. To do this, you opened the cabinet with reagents and suddenly saw the following picture on one of the shelves. Two jars of reagents had their labels peeled off and safely remained lying nearby. At the same time, it is no longer possible to determine exactly which jar corresponds to which label, and the external signs of the substances by which they could be distinguished are the same.
In this case, the problem can be solved using the so-called qualitative reactions.
Qualitative reactions These are reactions that make it possible to distinguish one substance from another, as well as to find out the qualitative composition of unknown substances.
For example, it is known that cations of some metals, when their salts are added to the burner flame, color it a certain color:
This method can only work if the substances being distinguished change the color of the flame differently, or one of them does not change color at all.
But, let’s say, as luck would have it, the substances being determined do not color the flame, or color it the same color.
In these cases, it will be necessary to distinguish substances using other reagents.
In what case can we distinguish one substance from another using any reagent?
There are two options:
- One substance reacts with the added reagent, but the second does not. In this case, it must be clearly visible that the reaction of one of the starting substances with the added reagent actually took place, that is, some external sign of it is observed - a precipitate formed, a gas was released, a color change occurred, etc.
For example, it is impossible to distinguish water from a solution of sodium hydroxide using hydrochloric acid, despite the fact that alkalis react well with acids:
NaOH + HCl = NaCl + H2O
This is due to the lack of any external signs reactions. A clear, colorless solution of hydrochloric acid when mixed with a colorless hydroxide solution forms the same clear solution:
But then, you can have water from aqueous solution alkalis can be distinguished, for example, using a solution of magnesium chloride - in this reaction a white precipitate forms:
2NaOH + MgCl 2 = Mg(OH) 2 ↓+ 2NaCl
2) substances can also be distinguished from each other if they both react with the added reagent, but do so in different ways.
For example, you can distinguish a sodium carbonate solution from a silver nitrate solution using a hydrochloric acid solution.
with sodium carbonate hydrochloric acid reacts with the release of a colorless, odorless gas - carbon dioxide (CO 2):
2HCl + Na 2 CO 3 = 2NaCl + H 2 O + CO 2
and with silver nitrate to form a white cheesy precipitate AgCl
HCl + AgNO 3 = HNO 3 + AgCl↓
The tables below present various options for detecting specific ions:
Qualitative reactions to cations
| Cation | Reagent | Sign of reaction |
| Ba 2+ | SO 4 2- |
Ba 2+ + SO 4 2- = BaSO 4 ↓ |
| Cu 2+ | 1) Precipitation blue color:
Cu 2+ + 2OH − = Cu(OH) 2 ↓ 2) Black sediment: Cu 2+ + S 2- = CuS↓ |
|
| Pb 2+ | S 2- | Black precipitate: Pb 2+ + S 2- = PbS↓ |
| Ag+ | Cl − |
Precipitation of a white precipitate, insoluble in HNO 3, but soluble in ammonia NH 3 ·H 2 O: Ag + + Cl − → AgCl↓ |
| Fe 2+ |
2) Potassium hexacyanoferrate (III) (red blood salt) K 3 |
1) Precipitation of a white precipitate that turns green in air: Fe 2+ + 2OH − = Fe(OH) 2 ↓ 2) Precipitation of a blue precipitate (Turnboole blue): K + + Fe 2+ + 3- = KFe↓ |
| Fe 3+ |
2) Potassium hexacyanoferrate (II) (yellow blood salt) K 4 3) Rodanide ion SCN − |
1) Brown precipitate: Fe 3+ + 3OH − = Fe(OH) 3 ↓ 2) Precipitation of blue precipitate (Prussian blue): K + + Fe 3+ + 4- = KFe↓ 3) The appearance of intense red (blood red) coloring: Fe 3+ + 3SCN − = Fe(SCN) 3 |
| Al 3+ | Alkali ( amphoteric properties hydroxide) |
Precipitation of a white precipitate of aluminum hydroxide when adding a small amount of alkali: OH − + Al 3+ = Al(OH) 3 and its dissolution upon further pouring: Al(OH) 3 + NaOH = Na |
| NH4+ | OH − , heating | Emission of gas with a pungent odor: NH 4 + + OH − = NH 3 + H 2 O Blue turning of wet litmus paper |
| H+ (acidic environment) |
Indicators: − litmus − methyl orange |
Red staining |
Qualitative reactions to anions
| Anion | Impact or reagent | Sign of reaction. Reaction equation |
| SO 4 2- | Ba 2+ |
Precipitation of a white precipitate, insoluble in acids: Ba 2+ + SO 4 2- = BaSO 4 ↓ |
| NO 3 − |
1) Add H 2 SO 4 (conc.) and Cu, heat 2) Mixture of H 2 SO 4 + FeSO 4 |
1) Formation of a blue solution containing Cu 2+ ions, release of brown gas (NO 2) 2) The appearance of color of nitroso-iron (II) sulfate 2+. Color ranges from violet to brown (brown ring reaction) |
| PO 4 3- | Ag+ |
Precipitation of a light yellow precipitate in a neutral environment: 3Ag + + PO 4 3- = Ag 3 PO 4 ↓ |
| CrO 4 2- | Ba 2+ |
Formation of a yellow precipitate, insoluble in acetic acid, but soluble in HCl: Ba 2+ + CrO 4 2- = BaCrO 4 ↓ |
| S 2- | Pb 2+ |
Black precipitate: Pb 2+ + S 2- = PbS↓ |
| CO 3 2- |
1) Precipitation of a white precipitate, soluble in acids: Ca 2+ + CO 3 2- = CaCO 3 ↓ 2) The release of colorless gas (“boiling”), causing cloudiness of lime water: CO 3 2- + 2H + = CO 2 + H 2 O |
|
| CO2 | Lime water Ca(OH) 2 |
Precipitation of a white precipitate and its dissolution with further passage of CO 2: Ca(OH) 2 + CO 2 = CaCO 3 ↓ + H 2 O CaCO 3 + CO 2 + H 2 O = Ca(HCO 3) 2 |
| SO 3 2- | H+ |
Emission of SO 2 gas with a characteristic pungent odor (SO 2): 2H + + SO 3 2- = H 2 O + SO 2 |
| F − | Ca2+ |
White precipitate: Ca 2+ + 2F − = CaF 2 ↓ |
| Cl − | Ag+ |
Precipitation of a white cheesy precipitate, insoluble in HNO 3, but soluble in NH 3 ·H 2 O (conc.): Ag + + Cl − = AgCl↓ AgCl + 2(NH 3 ·H 2 O) = ) 
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