Ion exchange reactions are reactions in aqueous solutions between electrolytes that occur without changes in the oxidation states of their constituent elements

A necessary condition for the reaction between electrolytes (salts, acids and bases) is the formation of a slightly dissociating substance (water, weak acid, ammonium hydroxide), precipitate or gas.

Let's consider the reaction that results in the formation of water. Such reactions include all reactions between any acid and any base. For example, the reaction of nitric acid with potassium hydroxide:

HNO 3 + KOH = KNO 3 + H 2 O (1)

Starting materials, i.e. nitric acid and potassium hydroxide, as well as one of the products, namely potassium nitrate, are strong electrolytes, i.e. V aqueous solution they exist almost exclusively in the form of ions. The resulting water belongs to weak electrolytes, i.e. practically does not disintegrate into ions. Thus, the equation above can be rewritten more accurately by indicating the real state of substances in an aqueous solution, i.e. in the form of ions:

H + + NO 3 − + K + + OH ‑ = K + + NO 3 − + H 2 O (2)

As can be seen from equation (2), both before and after the reaction, NO 3 − and K + ions are present in the solution. In other words, essentially, nitrate ions and potassium ions did not participate in the reaction at all. The reaction occurred only due to the combination of H + and OH − particles into water molecules. Thus, by performing an algebraic reduction of identical ions in equation (2):

H + + NO 3 − + K + + OH ‑ = K + + NO 3 − + H 2 O

we will get:

H + + OH ‑ = H 2 O (3)

Equations of the form (3) are called abbreviated ionic equations, type (2) - complete ionic equations, and type (1) - molecular reaction equations.

Actually ionic equation reaction reflects its essence as much as possible, precisely what makes its occurrence possible. It should be noted that many different reactions can correspond to one abbreviated ionic equation. Indeed, if we take, for example, not nitric acid, but hydrochloric acid, and instead of potassium hydroxide we use, say, barium hydroxide, we have the following molecular equation of the reaction:

2HCl+ Ba(OH) 2 = BaCl 2 + 2H 2 O

Hydrochloric acid, barium hydroxide and barium chloride are strong electrolytes, that is, they exist in solution primarily in the form of ions. Water, as discussed above, is a weak electrolyte, that is, it exists in solution almost only in the form of molecules. Thus, complete ionic equation This reaction will look like this:

2H + + 2Cl − + Ba 2+ + 2OH − = Ba 2+ + 2Cl − + 2H 2 O

Let's cancel the same ions on the left and right and get:

2H + + 2OH − = 2H 2 O

Dividing both the left and right sides by 2, we get:

H + + OH − = H 2 O,

Received abbreviated ionic equation completely coincides with the abbreviated ionic equation for the interaction of nitric acid and potassium hydroxide.

When composing ionic equations in the form of ions, write only the formulas:

1) strong acids (HCl, HBr, HI, H 2 SO 4, HNO 3, HClO 4) (the list of strong acids must be learned!)

2) strong bases (hydroxides of alkali (ALM) and alkaline earth metals (ALM))

3) soluble salts

The formulas are written in molecular form:

1) Water H 2 O

2) Weak acids (H 2 S, H 2 CO 3, HF, HCN, CH 3 COOH (and others, almost all organic))

3) Weak bases (NH 4 OH and almost all metal hydroxides except alkali metal and alkali metal

4) Slightly soluble salts (↓) (“M” or “H” in the solubility table).

5) Oxides (and other substances that are not electrolytes)

Let's try to write down the equation between iron (III) hydroxide and sulfuric acid. In molecular form, the equation of their interaction is written as follows:

2Fe(OH) 3 + 3H 2 SO 4 = Fe 2 (SO 4) 3 + 6H 2 O

Iron (III) hydroxide corresponds to the designation “H” in the solubility table, which tells us about its insolubility, i.e. in the ionic equation it must be written in its entirety, i.e. as Fe(OH) 3 . Sulfuric acid is soluble and belongs to strong electrolytes, that is, it exists in solution mainly in a dissociated state. Iron(III) sulfate, like almost all other salts, is a strong electrolyte, and since it is soluble in water, it must be written as an ion in the ionic equation. Taking into account all of the above, we obtain a complete ionic equation of the following form:

2Fe(OH) 3 + 6H + + 3SO 4 2- = 2Fe 3+ + 3SO 4 2- + 6H 2 O

Reducing the sulfate ions on the left and right, we get:

2Fe(OH) 3 + 6H + = 2Fe 3+ + 6H 2 O

Dividing both sides of the equation by 2 we get the abbreviated ionic equation:

Fe(OH) 3 + 3H + = Fe 3+ + 3H 2 O

Now let's look at the ion exchange reaction that produces a precipitate. For example, the interaction of two soluble salts:

All three salts - sodium carbonate, calcium chloride, sodium chloride and calcium carbonate (yes, that too) - are strong electrolytes and all except calcium carbonate are soluble in water, i.e. are involved in this reaction in the form of ions:

2Na + + CO 3 2- + Ca 2+ + 2Cl − = CaCO 3 ↓+ 2Na + + 2Cl −

By canceling the same ions on the left and right in this equation, we get the abbreviated ionic equation:

CO 3 2- + Ca 2+ = CaCO 3 ↓

The last equation reflects the reason for the interaction of solutions of sodium carbonate and calcium chloride. Calcium ions and carbonate ions combine into neutral calcium carbonate molecules, which, when combined with each other, give rise to small crystals of CaCO 3 precipitate of ionic structure.

Important note for passing the Unified State Exam in Chemistry For the reaction of salt1 with salt2 to proceed, in addition to basic requirements to the occurrence of ionic reactions (gas, sediment or water in the reaction products), such reactions are subject to another requirement - the starting salts must be soluble. That is, for example, CuS + Fe(NO 3) 2 ≠ FeS + Cu(NO 3) 2 no reaction thoughFeS – could potentially form a precipitate, because insoluble. The reason that the reaction does not proceed is the insolubility of one of the starting salts (CuS). But, for example, Na 2 CO 3 + CaCl 2 = CaCO 3 ↓+ 2NaCl occurs because calcium carbonate is insoluble and the starting salts are soluble. The same applies to the interaction of salts with bases. In addition to the basic requirements for the occurrence of ion exchange reactions, in order for a salt to react with a base, the solubility of both of them is necessary. Thus: Cu(OH) 2 + Na 2 S – does not leak, becauseCu(OH) 2 is insoluble, although a potential productCuS would be a precipitate. Here's the reaction betweenNaOH andCu(NO 3) 2 proceeds, so both starting substances are soluble and give a precipitateCu(OH) 2: 2NaOH + Cu(NO 3) 2 = Cu(OH) 2 ↓+ 2NaNO 3 Attention! In no case should you extend the requirement of solubility of starting substances beyond the reactions salt1 + salt2 and salt + base. For example, with acids this requirement is not necessary. In particular, all soluble acids react well with all carbonates, including insoluble ones. In other words: 1) Salt1 + salt2 - the reaction occurs if the original salts are soluble, but there is a precipitate in the products 2) Salt + metal hydroxide - the reaction occurs if the starting substances are soluble and the products contain sediment or ammonium hydroxide.

Let's consider the third condition for the occurrence of ion exchange reactions - the formation of gas. Strictly speaking, only as a result of ion exchange, gas formation is possible only in rare cases, for example, during the formation of hydrogen sulfide gas:

K 2 S + 2HBr = 2KBr + H 2 S

In most other cases, gas is formed as a result of the decomposition of one of the products of the ion exchange reaction. For example, you need to know for sure as part of the Unified State Examination that with the formation of gas, due to instability, products such as H 2 CO 3, NH 4 OH and H 2 SO 3 decompose:

H 2 CO 3 = H 2 O + CO 2

NH 4 OH = H 2 O + NH 3

H 2 SO 3 = H 2 O + SO 2

In other words, if an ion exchange produces carbonic acid, ammonium hydroxide, or sulfurous acid, the ion exchange reaction proceeds due to the formation of a gaseous product:

Let us write down the ionic equations for all the above reactions leading to the formation of gases. 1) For reaction:

K 2 S + 2HBr = 2KBr + H 2 S

Potassium sulfide and potassium bromide will be written in ionic form, because are soluble salts, as well as hydrobromic acid, because refers to strong acids. Hydrogen sulfide, being a poorly soluble gas that dissociates poorly into ions, will be written in molecular form:

2K + + S 2- + 2H + + 2Br — = 2K + + 2Br — + H 2 S

Reducing identical ions we get:

S 2- + 2H + = H 2 S

2) For the equation:

Na 2 CO 3 + H 2 SO 4 = Na 2 SO 4 + H 2 O + CO 2

In ionic form, Na 2 CO 3, Na 2 SO 4 will be written as highly soluble salts and H 2 SO 4 as a strong acid. Water is a poorly dissociating substance, and CO 2 is not an electrolyte at all, so their formulas will be written in molecular form:

2Na + + CO 3 2- + 2H + + SO 4 2- = 2Na + + SO 4 2 + H 2 O + CO 2

CO 3 2- + 2H + = H 2 O + CO 2

3) for the equation:

NH 4 NO 3 + KOH = KNO 3 + H 2 O + NH 3

Molecules of water and ammonia will be written in their entirety, and NH 4 NO 3, KNO 3 and KOH will be written in ionic form, because all nitrates are highly soluble salts, and KOH is an alkali metal hydroxide, i.e. strong base:

NH 4 + + NO 3 − + K + + OH − = K + + NO 3 − + H 2 O + NH 3

NH 4 + + OH − = H 2 O + NH 3

For the equation:

Na 2 SO 3 + 2HCl = 2NaCl + H 2 O + SO 2

The full and abbreviated equation will look like:

2Na + + SO 3 2- + 2H + + 2Cl − = 2Na + + 2Cl − + H 2 O + SO 2

When composing ionic equations, one should be guided by the fact that the formulas of slightly dissociating, insoluble and gaseous substances are written in molecular form. If a substance precipitates, then, as you already know, an arrow pointing downward (↓) is placed next to its formula, and if a gaseous substance is released during the reaction, then an arrow pointing upward () is placed next to its formula.

For example, if a solution of barium chloride BaCl 2 is added to a solution of sodium sulfate Na 2 SO 4 (Fig. 132), then as a result of the reaction a white precipitate of barium sulfate BaSO 4 is formed. Let's write the molecular equation of the reaction:

Rice. 132.
Reaction between sodium sulfate and barium chloride

Let's rewrite this equation, depicting strong electrolytes in the form of ions, and reactions leaving the sphere as molecules:

We have thus written down the complete ionic equation of the reaction. If we exclude identical ions from both sides of the equation, i.e. ions that do not participate in the reaction (2Na + and 2Cl - on the left and right sides of the equation), we obtain the abbreviated ionic equation of the reaction:

This equation shows that the essence of the reaction is reduced to the interaction of barium ions Ba 2+ and sulfate ions, as a result of which a precipitate of BaSO 4 is formed. In this case, it does not matter at all which electrolytes contained these ions before the reaction. A similar interaction can be observed between K 2 SO 4 and Ba(NO 3) 2, H 2 SO 4 and BaCl 2.

Laboratory experiment No. 17
Interaction between solutions of sodium chloride and silver nitrate

To 1 ml of sodium chloride solution in a test tube, add a few drops of silver nitrate solution using a pipette. What are you observing? Write down the molecular and ionic equations for the reaction. Using the abbreviated ionic equation, suggest several options for carrying out such a reaction with other electrolytes. Write down the molecular equations for the reactions performed.

Thus, the abbreviated ionic equations are the equations in general view, which characterize the essence of a chemical reaction and show which ions react and what substance is formed as a result.

Rice. 133.
Reaction between nitric acid and sodium hydroxide

If an excess of nitric acid solution is added to a solution of sodium hydroxide colored crimson by phenolphthalein (Fig. 133), the solution will become discolored, which will serve as a signal for a chemical reaction to occur:

NaOH + HNO 3 = NaNO 3 + H 2 O.

The complete ionic equation for this reaction is:

Na + + OH - + H + + NO 3 = Na + + NO - 3 + H 2 O.

But since the Na + and NO - 3 ions in the solution remain unchanged, they can not be written, and ultimately the abbreviated ionic equation of the reaction is written as follows:

H + + OH - = H 2 O.

It shows that the interaction of a strong acid and alkali is reduced to the interaction of H + ions and OH - ions, as a result of which a low-dissociating substance is formed - water.

Such an exchange reaction can occur not only between acids and alkalis, but also between acids and insoluble bases. For example, if you obtain a blue precipitate of insoluble copper (II) hydroxide by reacting copper (II) sulfate with alkali (Fig. 134):

and then divide the resulting precipitate into three parts and add a solution of sulfuric acid to the precipitate in the first test tube, hydrochloric acid to the precipitate in the second test tube, and a solution of nitric acid to the precipitate in the third test tube, then the precipitate will dissolve in all three test tubes (Fig. 135) .

Rice. 135.
Reaction of copper (II) hydroxide with acids:
a - sulfur; b - salt; c - nitrogen

This will mean that in all cases a chemical reaction took place, the essence of which is reflected using the same ionic equation.

Cu(OH) 2 + 2H + = Cu 2+ + 2H 2 O.

To verify this, write down the molecular, complete and abbreviated ionic equations of the above reactions.

Laboratory experiment No. 18
Preparation of insoluble hydroxide and its interaction with acids

Pour 1 ml of iron (III) chloride or sulfate solution into three test tubes. Pour 1 ml of alkali solution into each test tube. What are you observing? Then add solutions of sulfuric, nitric and hydrochloric acids to the test tubes, respectively, until the precipitate disappears. Write down the molecular and ionic equations for the reaction.

Suggest several options for carrying out such a reaction with other electrolytes. Write down the molecular equations for the proposed reactions.

Let's consider ionic reactions that occur with the formation of gas.

Pour 2 ml of solutions of sodium carbonate and potassium carbonate into two test tubes. Then pour hydrochloric acid into the first, and a solution of nitric acid into the second (Fig. 136). In both cases, we will notice a characteristic “boiling” due to the released carbon dioxide.

Rice. 136.
Interaction of soluble carbonates:
a - c hydrochloric acid; b - with nitric acid

Let's write down the molecular and ionic reaction equations for the first case:

Reactions occurring in electrolyte solutions are written using ionic equations. These reactions are called ion exchange reactions, since electrolytes exchange their ions in solutions. Thus, two conclusions can be drawn.

Key words and phrases

1. Molecular and ionic reaction equations.
2. Ion exchange reactions.
3. Neutralization reactions.

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>> Chemistry: Ionic equations

Ionic equations

As you already know from previous chemistry lessons, most chemical reactions occurs in solutions. And since all electrolyte solutions include ions, we can say that reactions in electrolyte solutions are reduced to reactions between ions.

These reactions that occur between ions are called ionic reactions. And ionic equations are precisely the equations of these reactions.

As a rule, ionic reaction equations are obtained from molecular equations, but this occurs subject to the following rules:

Firstly, the formulas of weak electrolytes, as well as insoluble and slightly soluble substances, gases, oxides, etc. are not recorded in the form of ions; the exception to this rule is the HSO−4 ion, and then in diluted form.

Secondly, the formulas of strong acids, alkalis, and also water-soluble salts are usually presented in the form of ions. It should also be noted that a formula such as Ca(OH)2 is presented in the form of ions if lime water is used. If lime milk is used, which contains insoluble Ca(OH)2 particles, then the formula in the form of ions is also not written down.

When composing ionic equations, as a rule, the full ionic and abbreviated, that is, brief ionic reaction equations are used. If we consider the ionic equation, which has an abbreviated form, then we do not observe ions in it, that is, they are absent from both parts of the complete ionic equation.

Let's look at examples of how molecular, full and abbreviated ionic equations are written:

Therefore, it should be remembered that the formulas of substances that do not decompose, as well as insoluble and gaseous ones, when drawing up ionic equations are usually written in molecular form.

Also, it should be remembered that if a substance precipitates, a downward arrow (↓) is drawn next to such a formula. Well, in the case when a gaseous substance is released during the reaction, then next to the formula there should be an icon like an upward arrow ().

Let's take a closer look with an example. If we have a solution of sodium sulfate Na2SO4, and we add a solution of barium chloride BaCl2 to it (Fig. 132), we will see that we have formed a white precipitate of barium sulfate BaSO4.

Look closely at the image that shows the interaction between sodium sulfate and barium chloride:

Now let's write the molecular equation for the reaction:

Well, now let's rewrite this equation, where strong electrolytes will be depicted in the form of ions, and reactions that leave the sphere are presented in the form of molecules:

We have written down the complete ionic equation for the reaction.

Now let’s try to remove identical ions from one and the other part of the equality, that is, those ions that do not take part in the reaction 2Na+ and 2Cl, then we will get an abbreviated ionic equation of the reaction, which will look like this:

From this equation we see that the whole essence of this reaction comes down to the interaction of barium ions Ba2+ and sulfate ions

and that as a result, a BaSO4 precipitate is formed, even regardless of which electrolytes contained these ions before the reaction.

How to solve ionic equations

And finally, let's summarize our lesson and determine how to solve ionic equations. You and I already know that all reactions that occur in electrolyte solutions between ions are ionic reactions. These reactions are usually solved or described using ionic equations.

Also, it should be remembered that all those compounds that are volatile, difficult to dissolve or slightly dissociated find a solution in molecular form. Also, we should not forget that in the case when none of the above types of compounds are formed during the interaction of electrolyte solutions, this means that the reactions practically do not occur.

Rules for solving ionic equations

For clear example Let us take the formation of a sparingly soluble compound such as:

Na2SO4 + BaCl2 = BaSO4 + 2NaCl

In ionic form, this expression will look like:

2Na+ +SO42- + Ba2+ + 2Cl- = BaSO4 + 2Na+ + 2Cl-

Since you and I observe that only barium ions and sulfate ions reacted, and the remaining ions did not react and their state remained the same. It follows from this that we can simplify this equation and write it in abbreviated form:

Ba2+ + SO42- = BaSO4

Now let's remember what we should do when solving ionic equations:

First, it is necessary to eliminate the same ions from both sides of the equation;

Secondly, we should not forget that the sum of the electric charges of the equation must be the same, both on its right side and also on the left.

11. Electrolytic dissociation. Ionic reaction equations

11.5. Ionic reaction equations

Since electrolytes in aqueous solutions break down into ions, it can be argued that reactions in aqueous solutions of electrolytes are reactions between ions. Such reactions can occur with a change in the oxidation state of atoms:

Fe 0  + 2 H + 1 Cl = Fe + 2 Cl 2 + H 0 2

and without change:

NaOH + HCl = NaCl + H2O

In general, reactions between ions in solutions are called ionic, and if they are exchange reactions, then ion exchange reactions. Ion exchange reactions occur only when substances are formed that leave the reaction sphere in the form of: a) a weak electrolyte (for example, water, acetic acid); b) gas (CO 2, SO 2); c) sparingly soluble substance (precipitate). The formulas of sparingly soluble substances are determined from the solubility table (AgCl, BaSO 4, H 2 SiO 3, Mg(OH) 2, Cu(OH) 2, etc.). The formulas of gases and weak electrolytes need to be memorized. Note that weak electrolytes can be highly soluble in water: for example, CH 3 COOH, H 3 PO 4, HNO 2.

The essence of ion exchange reactions is reflected ionic reaction equations, which are obtained from molecular equations following the following rules:

1) the formulas of weak electrolytes, insoluble and slightly soluble substances, gases, oxides, hydroanions of weak acids (HS − , HSO 3 − , HCO 3 − , H 2 PO 4 − , HPO 4 2 − ; exception - HSO ion) are not written in the form of ions 4 – in a dilute solution); hydroxocations of weak bases (MgOH +, CuOH +); complex ions ( 3− , 2− , 2− );

2) the formulas of strong acids, alkalis, and water-soluble salts are represented in the form of ions. The formula Ca(OH) 2 is written as ions if lime water is used, but is not written as ions in the case of lime milk containing insoluble Ca(OH) 2 particles.

There are full ionic and abbreviated (short) ionic reaction equations. The abbreviated ionic equation is missing the ions present on both sides of the full ionic equation. Examples of writing molecular, full ionic and abbreviated ionic equations:

• NaHCO 3 + HCl = NaCl + H 2 O + CO 2 - molecular,

Na + + HCO 3 − + H + + Cl − = Na + + Cl − + H 2 O + CO 2   - complete ionic,

HCO 3 − + H + = H 2 O + CO 2   - abbreviated ionic;

• BaCl 2 + K 2 SO 4 = BaSO 4 ↓ + 2KCl - molecular,

Ba 2 + + 2 Cl − + 2 K + + SO 4 2 − = BaSO 4   ↓ + 2 K + + 2 Cl − - complete ionic,

Ba 2 + + SO 4 2 − = BaSO 4   ↓ - abbreviated ionic.

Sometimes the full ionic equation and the abbreviated ionic equation are the same:

Ba(OH) 2 + H 2 SO 4 = BaSO 4 ↓ + 2H 2 O

Ba 2+ + 2OH − + 2H + + SO 4 2 − = BaSO 4 ↓ + 2H 2 O,

and for some reactions the ionic equation cannot be compiled at all:

3Mg(OH) 2 + 3H 3 PO 4 = Mg 3 (PO 4) 2 ↓ + 6H 2 O

Example 11.5. Indicate a pair of ions that can be present in the full ion-molecular equation if it corresponds to the abbreviated ion-molecular equation

Ca 2 + + SO 4 2 − = CaSO 4 .

1) SO 3 2 − and H +; 3) CO 3 2 − and K + ; 2) HCO 3 − and K + ; 4) Cl− and Pb 2+.

Solution. The correct answer is 2):

Ca 2 + + 2 HCO 3 − + 2 K + + SO 4 2 − = CaSO 4   ↓ + 2 HCO 3 − + 2 K + (Ca(HCO 3) 2 salt is soluble) or Ca 2+ + SO 4 2 − = CaSO4.

For other cases we have:

1) CaSO 3 + 2H + + SO 4 2 − = CaSO 4 ↓ + H 2 O + SO 2 ;

3) CaCO 3 + 2K + + SO 4 2 − (reaction does not occur);

4) Ca 2+ + 2Cl − + PbSO 4 (reaction does not occur).

Answer: 2).

Substances (ions) that react with each other in an aqueous solution (i.e., the interaction between them is accompanied by the formation of a precipitate, gas or weak electrolyte) cannot coexist in an aqueous solution in significant quantities

Table 11.2

Examples of ion pairs that do not exist together in significant quantities in aqueous solution

Example 11.6. Indicate in this series: HSO 3 − , Na + , Cl − , CH 3 COO − , Zn 2+ - formulas of ions that cannot be present in significant quantities: a) in acidic environment; b) in an alkaline environment.

Solution. a) In an acidic environment, i.e. together with H + ions, the anions HSO 3 − and CH 3 COO − cannot be present, since they react with hydrogen cations, forming a weak electrolyte or gas:

CH 3 COO − + H + ⇄ CH 3 COOH

HSO 3 − + H + ⇄ H 2 O + SO 2

b) HSO 3 − and Zn 2+ ions cannot be present in an alkaline medium, since they react with hydroxide ions to form either a weak electrolyte or a precipitate:

HSO 3 − + OH − ⇄ H 2 O + SO 3 2 −

Zn 2+ + 2OH– = Zn(OH) 2 ↓.

Answer: a) HSO 3 − and CH 3 COO −; b) HSO 3 − and Zn 2+.

Residues of acid salts of weak acids cannot be present in significant quantities in either an acidic or an alkaline medium, because in both cases a weak electrolyte is formed

The same can be said about the residues of basic salts containing a hydroxo group:

CuOH + + OH − = Cu(OH) 2 ↓

When any strong acid is neutralized by any strong base, for each mole of water formed, about the heat is released:

This suggests that such reactions are reduced to one process. We will obtain the equation for this process if we consider in more detail one of the given reactions, for example, the first. Let's rewrite its equation, writing strong electrolytes in ionic form, since they exist in solution in the form of ions, and weak electrolytes in molecular form, since they are in solution mainly in the form of molecules (water is a very weak electrolyte, see § 90):

Considering the resulting equation, we see that the ions did not undergo changes during the reaction. Therefore, we will rewrite the equation again, eliminating these ions from both sides of the equation. We get:

Thus, the reactions of neutralization of any strong acid with any strong base come down to the same process - the formation of water molecules from hydrogen ions and hydroxide ions. It is clear that the thermal effects of these reactions must also be the same.

Strictly speaking, the reaction of the formation of water from ions is reversible, which can be expressed by the equation

However, as we will see below, water is a very weak electrolyte and dissociates only to a negligible extent. In other words, the equilibrium between water molecules and ions is strongly shifted towards the formation of molecules. Therefore, in practice, the reaction of neutralization of a strong acid with a strong base proceeds to completion.

When mixing a solution of any silver salt with hydrochloric acid or with a solution of any of its salts, a characteristic white cheesy precipitate of silver chloride is always formed:

Such reactions also come down to one process. In order to obtain its ionic-molecular equation, we rewrite, for example, the equation of the first reaction, writing strong electrolytes, as in the previous example, in ionic form, and the substance in the sediment in molecular form:

As can be seen, the ions do not undergo changes during the reaction. Therefore, we exclude them and rewrite the equation again:

This is the ion-molecular equation of the process under consideration.

Here we must also keep in mind that the silver chloride precipitate is in equilibrium with the ions in solution, so that the process expressed by the last equation is reversible:

However, due to the low solubility of silver chloride, this equilibrium is very strongly shifted to the right. Therefore, we can assume that the reaction of formation from ions is almost completed.

The formation of a precipitate will always be observed when there are significant concentrations of and ions in one solution. Therefore, with the help of silver ions it is possible to detect the presence of ions in a solution and, conversely, with the help of chloride ions - the presence of silver ions; An ion can serve as a reactant on an ion, and an ion can serve as a reactant on an ion.

In the future, we will widely use the ionic-molecular form of writing equations for reactions involving electrolytes.

To draw up ion-molecular equations, you need to know which salts are soluble in water and which are practically insoluble. general characteristics The solubility of the most important salts in water is given in Table. 15.

Table 15. Solubility of the most important salts in water

Ionic-molecular equations help to understand the characteristics of reactions between electrolytes. Let us consider, as an example, several reactions that occur with the participation of weak acids and bases.

As already mentioned, the neutralization of any strong acid by any strong base is accompanied by the same thermal effect, since it comes down to the same process - the formation of water molecules from hydrogen ions and hydroxide ions.

However, when neutralizing a strong acid weak foundation, weak acid, strong or weak base, the thermal effects are different. Let's write ion-molecular equations for such reactions.

Neutralization of a weak acid (acetic acid) with a strong base (sodium hydroxide):

Here, the strong electrolytes are sodium hydroxide and the resulting salt, and the weak electrolytes are acid and water:

As can be seen, only sodium ions do not undergo changes during the reaction. Therefore, the ion-molecular equation has the form:

Neutralization of a strong acid (nitrogen) with a weak base (ammonium hydroxide):

Here we must write the acid and the resulting salt in the form of ions, and ammonium hydroxide and water in the form of molecules:

The ions do not undergo changes. Omitting them, we obtain the ionic-molecular equation:

Neutralization of a weak acid (acetic acid) with a weak base (ammonium hydroxide):

In this reaction, all substances except those formed are weak electrolytes. Therefore, the ion-molecular form of the equation looks like:

Comparing the obtained ion-molecular equations with each other, we see that they are all different. Therefore, it is clear that the heats of the reactions considered are also different.

As already indicated, the reactions of neutralization of strong acids with strong bases, during which hydrogen ions and hydroxide ions combine to form a water molecule, proceed almost to completion. Neutralization reactions, in which at least one of the starting substances is a weak electrolyte and in which molecules of weakly associated substances are present not only on the right, but also on the left side of the ion-molecular equation, do not proceed to completion.

They reach a state of equilibrium in which the salt coexists with the acid and base from which it was formed. Therefore, it is more correct to write the equations of such reactions as reversible reactions.