0620 · 7.3Preparation of salts

Choosing salts and solubility rules

0:00Making a salt is only part of the challenge. A successful preparation must also separate the product from leftover reactants and produce a pure sample. Some salts remain dissolved in water, while others form solids immediately. By the end of this video, you will understand how these differences determine the method. Think of salt preparation as a problem with three decisions. First, choose reactants that produce the salt you want.

0:22Next, decide how to remove anything unwanted. Finally, recover the product without introducing new impurities. The best method depends on whether the salt and the starting materials dissolve in water. A salt is an ionic compound containing positive and negative ions. In the reactions we will use, an acid supplies the negative ion. Hydrochloric acid produces chlorides, sulfuric acid produces sulfates, and nitric acid produces nitrates. The other reactant supplies the positive ion, which completes the salt’s identity.

0:52For example, copper(II) sulfate needs a source of copper(II) ions and a source of sulfate ions. Copper(II) oxide provides the copper, while sulfuric acid provides the sulfate. Their reaction produces copper(II) sulfate and water. Choosing reactants becomes much easier when you work backwards from the name of the required product. A soluble substance dissolves in water to form a solution. Insoluble means that it does not dissolve appreciably under the conditions we are considering.

1:18Dissolved ions pass through ordinary filter paper along with the water. An insoluble solid is trapped, so filtration separates solids from solutions, not dissolved substances from each other. The solubility rules help us predict which substances remain in solution. Sodium, potassium and ammonium salts are soluble. Nitrates are also soluble. These rules are especially useful when choosing solutions for a reaction, because each dissolved reactant must supply the ions we need.

1:45Chlorides are soluble, except lead chloride and silver chloride. Sulfates are soluble too, with three exceptions: barium sulfate, calcium sulfate and lead sulfate. Always check the exceptions before choosing your preparation method. A salt ending in chloride or sulfate is not automatically soluble, even though most examples in those families are. Carbonates are insoluble, except sodium, potassium and ammonium carbonates. The hydroxide rule also begins with insoluble, and its exceptions are sodium, potassium, ammonium and calcium.

2:16Calcium hydroxide is only partially soluble. Notice that these two rules start the opposite way to the chloride and sulfate rules, so their exceptions are the soluble ones. We can apply these rules rather than treating them as an isolated list. Potassium carbonate is soluble because potassium salts are soluble. Copper(II) carbonate is insoluble because it is not among the carbonate exceptions. Barium sulfate is insoluble, whereas sodium sulfate is soluble.

2:41These conclusions tell us which separation methods will work. For a soluble salt, we usually prepare a solution first and then obtain crystals.

Soluble salts using excess solids

2:49An acid can react with a suitable metal, an insoluble base, an insoluble carbonate, or an alkali. The first three routes allow us to add an excess solid. The alkali route needs a different strategy because both reactants are dissolved. Excess means that more reactant is added than the acid can use up. Once all the acid has reacted, some solid remains. That leftover material gives us a practical way to ensure the acid has been consumed.

3:14Because the excess is insoluble, we can remove it by filtration before collecting the salt. Let us work through the preparation of pure copper(II) sulfate crystals from copper(II) oxide and dilute sulfuric acid. The product is soluble because copper(II) sulfate is not a sulfate exception. Copper(II) oxide is an insoluble base, making the excess solid method suitable. We therefore plan a reaction, followed by filtration and crystallisation. Gently warm the dilute sulfuric acid, then add copper(II) oxide in small portions while stirring.

3:45Warming helps the reaction proceed, while gradual addition makes it easier to control. Copper(II) oxide plus sulfuric acid gives copper(II) sulfate plus water. The balanced symbol equation is on the board. Continue adding the oxide until no more reacts and some solid remains despite stirring. This indicates that copper(II) oxide is now in excess. Adding only enough for the mixture to look different would not reliably remove all the acid.

4:10The stopping condition matters because leftover acid would contaminate the final preparation. Filter the mixture to remove the unreacted copper(II) oxide. The solid left on the filter paper is called the residue. Liquid collected below is the filtrate, which contains the dissolved copper(II) sulfate. For this preparation, keep the filtrate because that is where the wanted substance is. Filtration does not give us crystals immediately because the salt is still dissolved.

4:35We must remove some water to make the solution more concentrated. Transfer the filtrate to an evaporating basin and heat gently to the point of crystallisation. At this stage, enough water has evaporated for crystals to form when the solution cools. One practical check is to place a drop of the concentrated solution on a cool glass rod. If crystals form as that drop cools, stop heating the main solution.

4:55Leave it to cool so crystals develop in the basin. Then filter the crystals and dry them between sheets of filter paper. Crystallisation works because the concentrated solution can hold less dissolved salt as it cools. Some of the dissolved material therefore joins the growing crystals. Do not simply boil the preparation completely dry. Controlled crystallisation produces crystals and avoids the strong heating that can damage some hydrated salts. The same sequence works with a suitable metal and an acid.

5:23For example, magnesium plus hydrochloric acid gives magnesium chloride plus hydrogen. Add magnesium in small portions until no more reacts and some remains. Filter off the excess metal, then heat the filtrate to the point of crystallisation, cool, filter and dry the crystals. Not every metal is suitable for this method. Copper does not react with dilute hydrochloric acid to produce a salt and hydrogen. Very reactive metals such as sodium are unsafe choices for this preparation.

5:52When hydrogen is produced, keep the reaction away from flames because the gas is flammable. An insoluble carbonate offers another route to a soluble salt. Copper(II) carbonate plus sulfuric acid gives copper(II) sulfate, carbon dioxide and water. Add the carbonate gradually because the gas causes bubbling. Continue until no more reacts and some carbonate remains, rather than relying on a temporary pause between additions. Filter off the excess carbonate and keep the salt solution.

6:20Heat the filtrate to the point of crystallisation, allow it to cool, then filter and dry the crystals. The gas product changes what you observe during the reaction, but not the overall separation method. The key is still complete acid consumption followed by removal of the excess solid.

Titration for soluble alkali salts

6:35An alkali is a soluble base, so the excess solid strategy cannot work here. Sodium hydroxide and hydrochloric acid are both present in solution. Adding too much alkali leaves dissolved sodium hydroxide mixed with the sodium chloride. Filter paper cannot separate them, so we need to find the reacting volumes accurately. A titration solves this problem by identifying the volumes needed for neutralisation. Measure a fixed volume of sodium hydroxide solution into a conical flask using a pipette.

7:03Add a few drops of a suitable indicator. Place hydrochloric acid in a burette so its delivered volume can be measured and controlled. Add acid while swirling the flask, using drops near the endpoint. Stop at the indicator’s appropriate colour change and record the volume delivered. Sodium hydroxide plus hydrochloric acid gives sodium chloride plus water. The indicator helps us find the reacting amounts, but it also introduces an unwanted substance into the solution.

7:28Repeat using the same measured volumes of acid and alkali, but without the indicator. This second preparation produces a salt solution without indicator contamination. Heat it to the point of crystallisation, cool, then filter and dry the crystals. The detailed technique is covered in the video on acid base titrations. The reason for repeating without indicator is purity, not an extra chemical reaction. We first use a visible signal to establish the correct quantities.

7:54Then we make a clean preparation using those quantities. Evaporating the original indicator mixture would leave unwanted indicator material with the salt, so it is not the preferred route to pure crystals. Titration is the usual route for sodium, potassium and ammonium salts. Their hydroxides and carbonates are all soluble, so none of them can be added as an excess solid. Sodium and potassium metals are far too reactive to add to an acid safely.

8:18Potassium sulfate, for example, is made by titrating potassium hydroxide solution with dilute sulfuric acid.

Precipitating insoluble salts

8:24This next part is for the Extended paper. An insoluble salt can be prepared by precipitation, which means forming an insoluble solid when suitable solutions are mixed. That solid is called a precipitate. Instead of removing water to obtain our product, we use its insolubility to separate it directly by filtration. Consider making barium sulfate from barium chloride solution and sodium sulfate solution. Barium chloride is soluble because barium is not one of the chloride exceptions.

8:51Sodium sulfate is soluble because sodium salts are soluble. The desired product is insoluble because barium sulfate is one of the sulfate exceptions. Mix the two solutions and stir. Barium ions meet sulfate ions and form solid barium sulfate. Sodium ions and chloride ions remain dissolved in the surrounding liquid. The word equation is barium chloride plus sodium sulfate gives barium sulfate plus sodium chloride. The balanced equations on the board show which ions form the solid.

9:20A barium ion has a 2+ charge, while a sulfate ion has a 2− charge. They combine in equal numbers to make electrically neutral barium sulfate. The other ions remain in solution because they do not form an insoluble product here. Filter the mixture and keep the residue, which is the barium sulfate precipitate. Wash it with distilled water to remove soluble substances remaining between the solid particles. The product stays on the paper because it is insoluble.

9:45Finally, dry the washed precipitate between filter papers or in a suitably warm place. Washing is essential because filtration alone leaves some solution clinging to the solid. That liquid contains dissolved substances which would remain as impurities after drying. Distilled water removes them without introducing additional dissolved salts. Notice that we keep the residue here, whereas the excess solid method first requires us to keep the filtrate. To choose the two solutions, pick one soluble salt that contains each ion of the product.

10:15Sodium, potassium and nitrate are useful partner ions because their salts are always soluble. Those partners stay dissolved in the filtrate and are washed away from the precipitate. A lead sulfate preparation could therefore mix lead nitrate solution with sodium sulfate solution. Silver chloride is prepared the same way. Silver nitrate dissolves because every nitrate is soluble, and sodium chloride dissolves because it is a sodium salt. Mixing their solutions forms silver chloride, one of the two chloride exceptions, as a solid.

10:43Silver nitrate plus sodium chloride gives silver chloride plus sodium nitrate. Filter, wash the residue with distilled water and dry it as before.

Water of crystallisation

10:51This next part is for the Extended paper. Water of crystallisation means the water molecules present in hydrated crystals. These molecules form part of the crystal structure rather than simply coating its surface. Removing surface moisture with filter paper therefore does not automatically remove the water of crystallisation. In hydrated copper(II) sulfate, each formula unit holds five water molecules. Cobalt(II) chloride crystals carry six water molecules per formula unit in their hydrated form.

11:17Their formulae are shown on screen. The dot indicates water of crystallisation in a fixed ratio, not a droplet of liquid attached to the outside. For all students, a hydrated substance is a substance that is chemically combined with water. An anhydrous substance is a substance containing no water. A hydrated crystal can feel dry because its water is incorporated into the structure. Drying a surface and producing an anhydrous substance are therefore different processes.

Avoiding errors and exam answers

11:44Several common mistakes come from losing track of where the product is. Filtering a salt solution does not collect the dissolved salt on the paper. In the excess solid method, the first residue is unwanted reactant. During precipitation, however, the residue is the product, so discarding it would throw away the prepared salt. Other mistakes involve incomplete purification. Stopping solid addition too early can leave excess acid in the solution.

12:06Keeping the indicator mixture introduces contamination, while failing to wash a precipitate leaves soluble impurities behind. For crystallisation, concentrate the solution and then cool it, rather than describing complete evaporation to dryness. An examination might ask you to describe the preparation of pure, dry copper(II) sulfate crystals from sulfuric acid and copper(II) oxide. Begin by identifying the soluble product and insoluble reactant. That immediately suggests adding an excess, filtering, and crystallising.

12:37Your answer should give the actual sequence rather than only naming the method. Warm the dilute sulfuric acid gently. Add copper(II) oxide in small portions while stirring until no more reacts and some solid remains. Filter off the excess oxide. Heat the filtrate to the point of crystallisation. Leave the concentrated solution to cool so crystals form. Filter to collect the crystals. Dry them between sheets of filter paper. Each sentence provides a separate practical step, and the sequence makes clear which material is kept.

13:07If asked to explain why excess copper(II) oxide is added, state that it ensures all the acid reacts. To explain filtration, identify removal of the excess insoluble solid. An instruction to state a solubility rule requires the rule and its exceptions, not a preparation method. These techniques can also be assessed in the practical paper or Alternative to Practical paper. The central idea is to choose a method that makes unwanted material easy to remove.

13:33For soluble salts made with a suitable solid, add excess reactant, filter, concentrate, cool, collect and dry. With an alkali, establish the reacting volumes by titration and repeat without indicator. An insoluble product is obtained by precipitation, followed by filtration, washing and drying. Use the solubility rules to decide which route is possible before starting. Keep track of whether the wanted substance is dissolved or present as a solid. Finally, remember that hydrated substances are chemically combined with water, while anhydrous substances contain none.

14:06Understanding these relationships lets you design a preparation rather than simply memorise a sequence.