Making salts

In a precipitation reaction, two aqueous solutions are combined and an insoluble salt forms as a solid. For instance, when silver nitrate is mixed with sodium chloride:

$AgN{O}_3(aq)+NaCl(aq)\to AgCl(s)+NaN{O}_3(aq)$

The solid silver chloride ($AgCl$) is the precipitate we are trying to collect.

Once the reaction is complete, filtration is used to separate the solid from the liquid. The solid left behind on the filter paper is called the residue.

Even after most liquid drains through, the residue stays wet. That liquid isn’t pure water—it’s leftover reaction solution containing dissolved spectator ions (for example $N{a}^{+}$ and $N{O}_3^{-}$ in the reaction above) that never became part of the solid.

The point of this method is to obtain a solid salt as the final product. If we tried to "dissolve the precipitate," we would undo the separation and lose the product we just made. So washing cannot be to dissolve the solid.

Consider the other incorrect ideas:

• Option C: Large, well-shaped crystals usually form when a solution cools slowly or evaporates slowly. A quick rinse on filter paper is not the stage where crystal size and shape are controlled.
• Option D: Some salt preparations involve acids, but many precipitation reactions do not. Distilled water is used to rinse, not as a specific acid-neutralising step.

Rinsing the residue with distilled water flushes out the trapped reaction solution between the solid particles. This carries away soluble spectator ions, so when the precipitate dries it is cleaner and not contaminated by dissolved ions left on its surface.

In precipitation, two aqueous ionic solutions react to produce an insoluble solid called a precipitate. We then use filtration to separate it, leaving the solid on the filter paper as the residue.

This links to Pure and impure substances (Filtration) and Types of chemical reaction (Formation of salts).

The whole purpose is to keep the solid precipitate. If it were dissolved, it would go through the filter paper with the liquid into the filtrate, meaning the product would be lost. That directly conflicts with the aim of the preparation.

Growing larger crystals needs time and the right conditions, such as slow cooling of a saturated solution or slow evaporation. Washing a residue with distilled water is a quick cleaning action and does not provide the environment needed for big, cubic crystals to develop.

Acids are not required for many precipitation reactions. For example, combining silver nitrate ($AgN{O}_3$) and sodium chloride ($NaCl$) forms a precipitate without any acid involved. Distilled water is used as a general rinse to clean the solid, not as a targeted neutralisation step.

After filtration, the residue is still damp with filtrate, which contains dissolved spectator ions (ions that stayed in solution instead of forming the solid). Rinsing with distilled water washes these soluble ions away, helping ensure the final dried salt is as pure as possible.

We need the reaction type that makes a salt as the only product. Any option that also creates water ($H_{2}O$), hydrogen gas ($H_2$), or carbon dioxide ($C{O}_2$) must be rejected. This can be answered by remembering the common word equations for acid reactions from Topic: Types of chemical reaction.

For an acid reacting with a metal, the general pattern is:

$\text{acid}+\text{metal}\to \text{salt}+\text{hydrogen gas }(H_2)$

Because hydrogen gas is always produced, this cannot be the reaction we want.

An acid plus a metal oxide is a neutralization reaction. The usual equation is:

$\text{acid}+\text{metal oxide}\to \text{salt}+\text{water }(H_{2}O)$

Since water forms along with the salt, this option does not fit the condition.

Acids reacting with metal carbonates follow this general equation:

$\text{acid}+\text{metal carbonate}\to \text{salt}+\text{water }(H_{2}O)+\text{carbon dioxide }(C{O}_2)$

This produces both water and carbon dioxide, so it fails the requirement.

Ammonia ($N{H}_3$) behaves as a base. In an acid-base reaction, it takes in a proton ($H^{+}$) to make ammonium ($N{H}_4^{+}$). The general equation is:

$\text{acid}+\text{ammonia}\to \text{ammonium salt}$

For example: $N{H}_3(aq)+HCl(aq)\to N{H}_{4}Cl(aq)$

Here, the ammonium salt is formed as the only product, with no water, hydrogen, or carbon dioxide made.

We are looking for a salt-forming reaction where no $H_{2}O$, $H_2$, or $C{O}_2$ is produced. That means the salt must be the sole product. We can eliminate options by matching each reaction type to its usual products.

If an acid reacts with a metal, the products include a salt and hydrogen gas ($H_2$).

General equation: $\text{acid}+\text{metal}\to \text{salt}+\text{hydrogen}$

Example: $2HCl(aq)+Mg(s)\to MgC{l}_2(aq)+H_2(g)$

Because $H_2$ is formed, Option A cannot be correct.

When an acid reacts with a metal oxide (a base), it is a neutralization that produces water.

General equation: $\text{acid}+\text{metal oxide}\to \text{salt}+\text{water}$

Example: $H_{2}S{O}_4(aq)+CuO(s)\to CuS{O}_4(aq)+H_{2}O(l)$

Since water is made, Option C is not suitable.

An acid reacting with a metal carbonate gives a salt plus water and carbon dioxide.

General equation: $\text{acid}+\text{metal carbonate}\to \text{salt}+\text{water}+\text{carbon dioxide}$

Example: $2HCl(aq)+CaC{O}_3(s)\to CaC{l}_2(aq)+H_{2}O(l)+C{O}_2(g)$

Because both $H_{2}O$ and $C{O}_2$ are produced, Option D is incorrect.

Ammonia ($N{H}_3$) acts as a base. When it reacts with an acid, it forms an ammonium salt and nothing else.

General equation: $\text{acid}+\text{ammonia}\to \text{ammonium salt}$

Example: $N{H}_3(aq)+HCl(aq)\to N{H}_{4}Cl(aq)$

Since no $H_{2}O$, $H_2$, or $C{O}_2$ is produced, Option B matches the requirement.

The situation is about making a soluble salt by reacting an acid with an insoluble reactant (the metal carbonate). This is the usual lab approach covered in Types of chemical reaction: Formation of salts.

The general equation for this reaction is: $\text{acid}+\text{metal carbonate}\to \text{salt}+\text{water}+\text{carbon dioxide}$

In this method, the insoluble metal carbonate is deliberately added in excess compared with the acid.

This is done so that every bit of the acid reacts (complete neutralization). Any leftover acid would stay dissolved and would be hard to remove later, so using excess carbonate helps ensure the final salt solution is not contaminated by unreacted acid.

Because extra carbonate was added, once bubbling stops there will still be some solid carbonate remaining in the beaker.

To get a clean solution of the soluble salt, we must separate the unreacted solid carbonate from the salt solution. Filtration is the standard way to separate an insoluble solid from a liquid.

Check the other answers:

• Option A: Salt crystals are not collected at this stage; crystals appear later during evaporation/crystallization.
• Option B: Filtration does not change evaporation speed; evaporation depends on temperature and surface area.
• Option C: Filtration cannot remove water; both the water and dissolved salt pass through the filter together.

Filtering is mainly done to remove the remaining solid carbonate that was added in excess to make sure the acid was completely used up.

When making a soluble salt, an acid is reacted with an insoluble metal carbonate. To avoid leaving any acid in the final mixture, the carbonate is added in excess. After the reaction ends, you have a salt solution plus some unused solid carbonate sitting at the bottom.

Option A says the filter is used to collect salt crystals. But at this point the salt is still dissolved (it’s soluble), so it passes through with the water. Crystals are only produced later in the crystallization stage after heating and then cooling.

Option B suggests filtration makes evaporation quicker. Filtration is simply a separation technique based on whether particles are insoluble and large enough to be trapped by the filter paper. It does not provide heat and does not control how fast water will evaporate.

Option C claims filtration removes water from the salt. In fact, the dissolved salt and water flow through the filter paper together as the filtrate. To remove water you would use evaporation (or distillation), not filtration.

So the correct purpose is to separate the liquid salt solution from the solid left behind. The excess, insoluble carbonate stays on the filter paper as the residue, leaving a pure salt solution.

To make a pure sample of potassium nitrate (${\text{KNO}}_3$), start by spotting the ions in the formula:

• Potassium provides the positive ion: ${\text{K}}^{+}$
• Nitrate provides the negative ion: ${\text{NO}}_3^{-}$

A titration is used to make soluble salts by carefully neutralizing an aqueous acid with an aqueous alkali (a soluble base) so neither reactant is left over.

$\text{Acid}+\text{Alkali}\to \text{Salt}+\text{Water}$

So, to form ${\text{KNO}}_3$, the acid must supply ${\text{NO}}_3^{-}$ and the alkali must supply ${\text{K}}^{+}$.

Option A combines potassium chloride and sodium nitrate, which are both already salts. Dissolving and mixing salts usually just gives a solution containing a mix of ions rather than a controlled neutralization that produces a pure new salt.

Option B uses potassium oxide and nitrous acid.

1. Nitrous acid (${\text{HNO}}_2$) would lead to potassium nitrite (${\text{KNO}}_2$), not potassium nitrate.
2. Potassium oxide is a solid and reacts very strongly with water, so it is not a typical aqueous reactant used for a standard titration.

Option C uses potassium metal with nitric acid. That is a metal + acid reaction, not a gentle neutralization. Potassium is extremely reactive, and combining it with a strong acid is highly unsafe and not something done as a titration.

Option D matches what titration needs: nitric acid (${\text{HNO}}_3$) supplies nitrate ions, and potassium hydroxide ($\text{KOH}$) supplies potassium ions. Both are soluble, so the endpoint can be found accurately.

${\text{HNO}}_3(aq)+\text{KOH}(aq)\to {\text{KNO}}_3(aq)+{\text{H}}_2\text{O}(l)$

This controlled neutralization lets the student reach exact neutrality and obtain a pure potassium nitrate solution.

Because the goal is a pure sample of potassium nitrate (${\text{KNO}}_3$) made by titration, we need an acid–base neutralization:

$\text{Acid}+\text{Base}\to \text{Salt}+\text{Water}$

For potassium nitrate, the acid must be the source of ${\text{NO}}_3^{-}$ and the base must be the source of ${\text{K}}^{+}$.

Option A is potassium chloride and sodium nitrate. When two soluble salts are mixed, the solution just contains ${\text{K}}^{+}$, ${\text{Cl}}^{-}$, ${\text{Na}}^{+}$, and ${\text{NO}}_3^{-}$ together. There is no neutralization step and no way to end up with a pure new product from a titration.

Option B includes nitrous acid (${\text{HNO}}_2$), but the ion needed in potassium nitrate is ${\text{NO}}_3^{-}$. Nitric acid provides nitrate; nitrous acid would instead form potassium nitrite (${\text{KNO}}_2$). Also, potassium oxide (${\text{K}}_2\text{O}$) is a reactive solid and is not normally used as the aqueous base in a titration.

Option C uses potassium metal and nitric acid. Potassium (Group 1) reacts extremely vigorously, and reacting a metal with a strong acid is not the careful, measurable neutralization that titration requires.

That leaves Option D: potassium hydroxide ($\text{KOH}$) and nitric acid (${\text{HNO}}_3$). This is a standard strong base + strong acid titration, producing potassium nitrate solution and water:

${\text{KOH}}_{(aq)}+{\text{HNO}}_{3(aq)}\to {\text{KNO}}_{3(aq)}+{\text{H}}_2{\text{O}}_{(l)}$

Using an indicator allows the exact endpoint to be found, helping ensure the potassium nitrate formed is pure.

Before choosing reactants, identify what potassium nitrate (${\text{KNO}}_3$) is made from.

It contains:

• Potassium ions: ${\text{K}}^{+}$
• Nitrate ions: ${\text{NO}}_3^{-}$

A titration for making a soluble salt uses an aqueous acid and an aqueous base (alkali) so the neutralization can be measured precisely.

$\text{Acid}+\text{Base}\to \text{Salt}+\text{Water}$

Now connect each ion in ${\text{KNO}}_3$ to where it should come from:

• The nitrate ion (${\text{NO}}_3^{-}$) should come from the acid, so the acid must be nitric acid (${\text{HNO}}_3$).
• The potassium ion (${\text{K}}^{+}$) should come from the base, so a suitable soluble base is potassium hydroxide ($\text{KOH}$).

So the needed pair is: ${\text{KOH}}_{\text{(aq)}}$ and ${\text{HNO}}_{3\text{(aq)}}$.

Checking each remaining option against the ion requirements and the titration method:

• Option A: Two salts mixed together won’t carry out an acid–base neutralization, so it won’t produce a pure new salt.
• Option B: ${\text{HNO}}_2$ contains ${\text{NO}}_2^{-}$ (nitrite), not ${\text{NO}}_3^{-}$ (nitrate), so it makes the wrong potassium salt.
• Option C: Potassium metal is not an aqueous base and its reaction with acid is dangerously violent, so it is not used for titration.

Option D fits perfectly: both reactants are soluble and allow a controlled titration to form potassium nitrate and water.

${\text{KOH}}_{\text{(aq)}}+{\text{HNO}}_{3\text{(aq)}}\to {\text{KNO}}_{3\text{(aq)}}+{\text{H}}_2{\text{O}}_{\text{(l)}}$

To decide how to make a salt, you start by looking up (or being told) whether it is soluble. Here, the question tells us that lead(II) sulfate (${\text{PbSO}}_4$) is insoluble in water.

In IB Chemistry, insoluble salts are normally prepared by a precipitation reaction, not by crystallising from solution after neutralisation or metal reactions.

A precipitation method works by mixing two solutions that each contain ions needed to form the target salt, and both starting compounds must be soluble so their ions are free in solution.

For instance, to produce ${\text{PbSO}}_4(s)$ we could mix:

1. A soluble lead(II) salt (like lead(II) nitrate, ${\text{Pb(NO}}_3{)}_2$)
2. A soluble sulfate (like sodium sulfate, ${\text{Na}}_2{\text{SO}}_4$)

The reaction would be: ${\text{Pb(NO}}_3{)}_2(aq)+{\text{Na}}_2{\text{SO}}_4(aq)\to {\text{PbSO}}_4(s)+2{\text{NaNO}}_3(aq)$

Using a metal oxide with an acid is mainly used to make soluble salts, because the salt stays dissolved and any extra solid oxide can be removed by filtration. If the desired salt is insoluble, you would end up with solid ${\text{PbSO}}_4$ mixed together with leftover oxide, giving two solids that are difficult to separate cleanly.

Titration is chosen when both reactants are solutions (acid and alkali) and the product salt remains soluble, so it can later be crystallised. Because ${\text{PbSO}}_4$ is insoluble, titration is not the correct approach for making it.

Reacting a metal with an acid is typically used to form soluble salts of reasonably reactive metals. If an insoluble salt forms (such as ${\text{PbSO}}_4$), it can coat the metal surface and block further reaction, so the process stops quickly and is not a reliable preparation method.

Since ${\text{PbSO}}_4$ is insoluble, the correct IB lab method is to mix two soluble salt solutions so the insoluble product forms as a precipitate. The solid can then be separated by filtration.

We are asked to prepare lead(II) sulfate (${\text{PbSO}}_4$), and we are told it is insoluble in water. That detail is crucial because many common salt-making procedures are designed for salts that stay dissolved.

In the syllabus section on Types of chemical reaction: Formation of salts, insoluble salts are made by precipitation rather than by methods intended to produce a soluble salt.

Heating a metal oxide with an acid (e.g., $\text{CuO}+{\text{H}}_2{\text{SO}}_4\to {\text{CuSO}}_4+{\text{H}}_2\text{O}$) works well when the salt produced is soluble, because the salt stays in solution and extra oxide can be filtered away. If the target is insoluble like ${\text{PbSO}}_4$, the product would be trapped as a solid together with any unreacted oxide, so you could not easily isolate a pure sample.

Titration is used when both reactants are soluble (acid + alkali), for example $\text{NaOH}+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}$. This is specifically a method to obtain soluble salts (often Group 1 or ammonium salts). It is not the correct method if the desired salt is insoluble.

The reaction of a metal with an acid (e.g., $\text{Mg}+{\text{H}}_2{\text{SO}}_4\to {\text{MgSO}}_4+{\text{H}}_2$) is a standard route to soluble salts. With lead and sulfuric acid, insoluble ${\text{PbSO}}_4$ would form on the metal surface, creating a barrier that prevents the acid from continuing to react, so the reaction would quickly slow or stop.

Because A, B, and D are all typical methods for preparing soluble salts, they are not appropriate here. The correct method is precipitation: mix two soluble salt solutions (such as lead(II) nitrate and sodium sulfate) so the insoluble salt forms:

${\text{Pb(NO}}_3{)}_2(aq)+{\text{Na}}_2{\text{SO}}_4(aq)\to {\text{PbSO}}_4(s)+2{\text{NaNO}}_3(aq)$

Then filter off the solid ${\text{PbSO}}_4$, rinse it with distilled water, and dry it.

In a precipitation reaction, two soluble salts are combined in water and an insoluble salt forms as a solid called a precipitate. To see why we use solutions at the start, we need to think about how particles behave in different states of matter, linking to Topic: Matter.

In a solid ionic compound, positive and negative ions are arranged in a giant ionic lattice and held by strong electrostatic attraction. Because the ions are fixed in place, they cannot travel to new positions easily. If you only stirred two solid salts together, very few ions would be able to reach each other, so the reaction would not proceed effectively.

When a soluble salt dissolves to make an aqueous solution ($aq$), water molecules pull ions away from the lattice. Once separated, the ions can move around the liquid, meaning they are mobile. This is why we begin with solutions of soluble salts.

Collision Theory (from Topic: Types of chemical reaction) states that particles must collide, with enough energy and the right orientation, for a reaction to happen.

In solution, the free-moving ions can collide repeatedly. When the particular ions that form the insoluble salt meet, they join into a new solid lattice and the solid separates from the mixture as a precipitate.

Check the remaining options:

• Option B cannot be right because precipitation reactions are used specifically to make a solid precipitate, not to keep everything as a liquid.
• Option C is not correct because these are usually calm double displacement (ion-exchange) reactions; using solutions is about allowing reaction to occur, not avoiding explosions.
• Option D is not correct because changing pH is the focus of neutralization/titration, whereas precipitation is mainly about ions swapping partners to create an insoluble solid.

Because ions need to be free to move around so they can collide and build the insoluble ionic lattice, the reactants must be in solution. This matches the first option.

A precipitation reaction is when two solutions containing soluble salts are mixed to create an insoluble salt (the precipitate) plus another salt that stays dissolved. To see why solutions are required, we focus on particle-level reactions, as in Topic 3: Collision theory.

The target is an insoluble salt, which means a solid that does not dissolve in the solvent. Since the purpose is to form a solid precipitate, any option claiming the final product is a liquid must be wrong.

Most precipitation reactions are simple ion-exchange (double displacement) reactions. They usually happen smoothly at room temperature and are not like highly reactive situations (for example, alkali metals with water). Therefore, using solutions is not mainly to stop an explosion.

Adjusting pH to neutral is the key idea in acid-base neutralization, not the key reason for precipitation. In precipitation, we rely on solubility and ion exchange to make a solid form from two ionic solutions, so pH control is not the main requirement here.

In solids, ions are stuck in a lattice and cannot travel to collide easily. When the salts dissolve, they dissociate into ions:

$\text{AB}(s)\to {\text{A}}^{+}(aq)+{\text{B}}^{-}(aq)$

Now the ions can move through the solution and collide. When the correct oppositely charged ions meet and form an insoluble combination, they build a solid precipitate. This is exactly what collision theory predicts.

We are being asked why gentle evaporation is recommended instead of strong heating when forming crystals of ammonium salts such as ${\text{NH}}_4\text{Cl}$. Crystallisation is meant to separate a solid from its solution without changing what the solid is. Overheating can cause chemical change rather than just removing water.

Ammonium chloride is an ionic compound that does not normally react with the porcelain/ceramic used for an evaporating dish under standard lab conditions. Since the container is designed to be unreactive, this cannot explain why strong heating should be avoided.

From typical solubility patterns you learn in the Matter unit, ammonium salts are generally very soluble in water. Heating water usually increases the solubility of many solid salts, so ${\text{NH}}_4\text{Cl}$ is not “insoluble in hot water.”

${\text{NH}}_4\text{Cl}$ is not a fuel and does not burn. It does not ignite or help a fire keep burning; it is even used in some situations where fire resistance is useful. So flammability is not the concern here.

A key fact is that ammonium salts can be thermally unstable. If ammonium chloride is heated strongly, it undergoes thermal decomposition and produces gases—ammonia and hydrogen chloride:

${\text{NH}}_4\text{Cl(s)}\stackrel{\Delta }{\to }{\text{NH}}_3\text{(g)}+\text{HCl(g)}$

With strong heating during crystallisation, the salt can decompose and leave the dish as gases instead of remaining as a solid to form crystals. Gentle evaporation removes water while keeping the salt intact so crystals can grow.

Because ammonium salts like ammonium chloride can decompose into gases when overheated, the correct reason for gentle evaporation is that strong heating may cause the salt to break down and escape as gases.

The aim is to collect crystals of ammonium chloride (${\text{NH}}_4\text{Cl}$) from a solution. That means we should remove solvent (water) by evaporation but avoid conditions that would destroy or change the solute. So we check each option for why strong heating is not used.

Evaporating basins are usually porcelain/ceramic, chosen because they are chemically inert to many common lab chemicals. Common salts do not normally react with them, even when warmed, so the salt reacting with the dish is not the reason.

In the Pure and impure substances topic, you learn that many solid salts dissolve better at higher temperatures. ${\text{NH}}_4\text{Cl}$ is soluble, and heating tends to increase its solubility rather than make it insoluble. So this option does not fit.

A flammable substance burns in oxygen and acts as a fuel. ${\text{NH}}_4\text{Cl}$ is an inorganic ionic salt, not an organic fuel, and it will not catch fire like ethanol or paper. So flammability is not the issue.

Ammonium salts can break down when heated strongly. Ammonium chloride undergoes thermal decomposition to form ammonia and hydrogen chloride gases:

${\text{NH}}_4\text{Cl(s)}\to {\text{NH}}_3\text{(g)}+\text{HCl(g)}$

If the salt is overheated, it can turn into gases and disappear instead of crystallising. Gentle heating focuses on evaporating the water while leaving the salt to form crystals.

To answer this, start by thinking about what a salt is. Salts are ionic compounds built from cations (positive ions) and anions (negative ions) arranged together.

From Topic 3: Types of chemical reaction, especially the Formation of salts section, you may remember that salts are often formed when an acid reacts with a base.

Metallic bonding is the attraction between positive metal ions and a mobile “sea” of delocalized electrons. That bonding explains the structure of metals and alloys, but it does not describe how salts are held together.

Hydrogen bonding is an intermolecular attraction (a force between molecules), not the main type of bonding that builds a solid crystal lattice. It is much too weak to be the primary force holding a giant, rigid salt lattice together.

Covalent bonding involves atoms (usually non-metals) sharing electrons. Some substances can form giant covalent networks, but salts are not made from atoms sharing electrons—they are made from ions.

Because salts consist of ions, the bonding that produces their structure must be ionic bonding. Ionic bonding is the strong electrostatic attraction between oppositely charged ions ($M^{+}$ and $X^{-}$).

These attractions act in every direction, arranging ions into a repeating 3D pattern called a regular giant lattice structure.

Using elimination, begin by clarifying the key word: a salt is an overall neutral compound formed when a positive ion (cation) and a negative ion (anion) attract each other. The question asks for the bonding that produces the large, repeating 3D arrangement known as a giant lattice.

Metallic bonding forms when metal atoms contribute outer electrons to a shared, delocalized electron ‘sea’. Although this can create a giant lattice, it describes metals (such as copper or gold) and alloys—not salts. Salts contain ions (often from a metal and a non-metal, or polyatomic ions), so this option does not fit.

Hydrogen bonding is an intermolecular force, meaning it acts between molecules (for example, between water molecules, $H_{2}O$). It is not the main bond that constructs a giant lattice. It is also far too weak to explain the strong, rigid, high-melting-point structure typical of salts.

Covalent bonding is the sharing of electron pairs between non-metal atoms, commonly producing small molecules (like $C{O}_2$ or $C{H}_4$) or network solids (like diamond or graphite). Even though networks are giant structures, salts are defined by electron transfer and ion formation, not by electron sharing.

After eliminating the other choices, the remaining option is ionic bonding. This matches the definition: electrostatic attraction between oppositely charged ions ($M^{+}$ and $X^{-}$). Because these forces pull in all directions, they create a regular giant ionic lattice, which is characteristic of salts.