Complete reaction equations. Electrolyte solutions. Ionic-molecular equations

In electrolyte solutions, reactions occur between hydrated ions, which is why they are called ionic reactions. In their direction, the nature and strength of the chemical bond in the reaction products are important. Typically, exchange in electrolyte solutions results in the formation of a compound with a stronger chemical bond. Thus, when solutions of barium chloride salts BaCl 2 and potassium sulfate K 2 SO 4 interact, the mixture will contain four types of hydrated ions Ba 2 + (H 2 O)n, Cl - (H 2 O)m, K + (H 2 O) p, SO 2 -4 (H 2 O)q, between which the reaction will occur according to the equation:

BaCl 2 +K 2 SO 4 =BaSO 4 +2КCl

Barium sulfate will precipitate in the form of a precipitate, in the crystals of which the chemical bond between the Ba 2+ and SO 2- 4 ions is stronger than the bond with the water molecules hydrating them. The connection between the K+ and Cl - ions only slightly exceeds the sum of their hydration energies, so the collision of these ions will not lead to the formation of a precipitate.

Therefore, we can draw the following conclusion. Exchange reactions occur during the interaction of such ions, the binding energy between which in the reaction product is much greater than the sum of their hydration energies.

Ion exchange reactions are described by ionic equations. Sparingly soluble, volatile and slightly dissociated compounds are written in molecular form. If, during the interaction of electrolyte solutions, none of the indicated types of compounds are formed, this means that practically no reaction occurs.

Formation of sparingly soluble compounds

For example, the interaction between sodium carbonate and barium chloride in the form of a molecular equation will be written as follows:

Na 2 CO 3 + BaCl 2 = BaCO 3 + 2NaCl or in the form:

2Na + +CO 2- 3 +Ba 2+ +2Сl - = BaCO 3 + 2Na + +2Сl -

Only the Ba 2+ and CO -2 ions reacted, the state of the remaining ions did not change, so the short ionic equation will take the form:

CO 2- 3 +Ba 2+ =BaCO 3

Formation of Volatile Substances

The molecular equation for the interaction of calcium carbonate and hydrochloric acid will be written as follows:

CaCO 3 +2HCl=CaCl 2 +H 2 O+CO 2

One of the reaction products - carbon dioxide CO 2 - was released from the reaction sphere in the form of a gas. The expanded ionic equation is:

CaCO 3 +2H + +2Cl - = Ca 2+ +2Cl - +H 2 O+CO 2

The result of the reaction is described by the following short ionic equation:

CaCO 3 +2H + =Ca 2+ +H 2 O+CO 2

Formation of a slightly dissociated compound

An example of such a reaction is any neutralization reaction, resulting in the formation of water, a slightly dissociated compound:

NaOH+HCl=NaCl+H 2 O

Na + +OH-+H + +Cl - = Na + +Cl - +H 2 O

OH-+H+=H 2 O

From the brief ionic equation it follows that the process is expressed in the interaction of H+ and OH- ions.

All three types of reactions proceed irreversibly to completion.

If you merge solutions of, for example, sodium chloride and calcium nitrate, then, as the ionic equation shows, no reaction will occur, since no precipitate, no gas, or low-dissociating compound is formed:

Using the solubility table, we establish that AgNO 3, KCl, KNO 3 are soluble compounds, AgCl is an insoluble substance.

We create an ionic equation for the reaction taking into account the solubility of the compounds:

A brief ionic equation reveals the essence of the chemical transformation taking place. It can be seen that only Ag+ and Cl - ions actually took part in the reaction. The remaining ions remained unchanged.

Example 2. Make up a molecular and ionic equation for the reaction between: a) iron (III) chloride and potassium hydroxide; b) potassium sulfate and zinc iodide.

a) We compose the molecular equation for the reaction between FeCl 3 and KOH:

Using the solubility table, we establish that of the resulting compounds, only iron hydroxide Fe(OH) 3 is insoluble. We compose the ionic equation of the reaction:

The ionic equation shows that the coefficients of 3 in the molecular equation apply equally to ions. This is a general rule for writing ionic equations. Let us represent the reaction equation in short ionic form:

This equation shows that only Fe3+ and OH- ions took part in the reaction.

b) Let's create a molecular equation for the second reaction:

K 2 SO 4 + ZnI 2 = 2KI + ZnSO 4

From the solubility table it follows that the starting and resulting compounds are soluble, therefore the reaction is reversible and does not reach completion. Indeed, no precipitate, no gaseous compound, or slightly dissociated compound is formed here. Let's create a complete ionic equation for the reaction:

2K + +SO 2- 4 +Zn 2+ +2I - + 2K + + 2I - +Zn 2+ +SO 2- 4

Example 3. Using the ionic equation: Cu 2+ +S 2- -= CuS, create a molecular equation for the reaction.

The ionic equation shows that on the left side of the equation there must be molecules of compounds containing Cu 2+ and S 2- ions. These substances must be soluble in water.

According to the solubility table, we will select two soluble compounds, which include the Cu 2+ cation and the S 2- anion. Let's create a molecular equation for the reaction between these compounds:

CuSO 4 +Na 2 S CuS+Na 2 SO 4

Quite often, schoolchildren and students have to compose the so-called. ionic reaction equations. In particular, task 31, proposed at the Unified State Exam in Chemistry, is devoted to this topic. In this article we will discuss in detail the algorithm for writing short and complete ionic equations, and will analyze many examples of different levels of complexity.

Why are ionic equations needed?

Let me remind you that when many substances are dissolved in water (and not only in water!), a dissociation process occurs - the substances break up into ions. For example, HCl molecules in an aqueous environment dissociate into hydrogen cations (H +, more precisely, H 3 O +) and chlorine anions (Cl -). Sodium bromide (NaBr) is found in an aqueous solution not in the form of molecules, but in the form of hydrated Na + and Br - ions (by the way, solid sodium bromide also contains ions).

When writing “ordinary” (molecular) equations, we do not take into account that it is not molecules that react, but ions. Here, for example, is what the equation for the reaction between hydrochloric acid and sodium hydroxide looks like:

HCl + NaOH = NaCl + H 2 O. (1)

Of course, this diagram does not describe the process entirely correctly. As we have already said, in an aqueous solution there are practically no HCl molecules, but there are H + and Cl - ions. The same is true with NaOH. It would be more correct to write the following:

H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O. (2)

That's what it is complete ionic equation. Instead of “virtual” molecules, we see particles that are actually present in the solution (cations and anions). We will not dwell on the question of why we wrote H 2 O in molecular form. This will be explained a little later. As you can see, there is nothing complicated: we replaced the molecules with ions that are formed during their dissociation.

However, even the complete ionic equation is not perfect. Indeed, take a closer look: both the left and right sides of equation (2) contain the same particles - Na + cations and Cl - anions. These ions do not change during the reaction. Why then are they needed at all? Let's remove them and get Brief ionic equation:

H + + OH - = H 2 O. (3)

As you can see, it all comes down to the interaction of H + and OH - ions with the formation of water (neutralization reaction).

All complete and brief ionic equations are written down. If we had solved problem 31 on the Unified State Exam in chemistry, we would have received the maximum score for it - 2 points.

So, once again about the terminology:

• HCl + NaOH = NaCl + H 2 O - molecular equation ("ordinary" equation, schematically reflecting the essence of the reaction);
• H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O - complete ionic equation (real particles in solution are visible);
• H + + OH - = H 2 O - a short ionic equation (we removed all the "garbage" - particles that do not participate in the process).

Algorithm for writing ionic equations

1. Let's create a molecular equation for the reaction.
2. All particles that dissociate in solution to a noticeable extent are written in the form of ions; substances that are not prone to dissociation are left “in the form of molecules.”
3. We remove the so-called from the two parts of the equation. observer ions, i.e. particles that do not participate in the process.
4. We check the coefficients and get the final answer - a short ionic equation.

Example 1. Write complete and short ionic equations describing the interaction of aqueous solutions of barium chloride and sodium sulfate.

Solution. We will act in accordance with the proposed algorithm. Let's first create a molecular equation. Barium chloride and sodium sulfate are two salts. Let's look at the section of the reference book "Properties of inorganic compounds". We see that salts can interact with each other if a precipitate is formed during the reaction. Let's check:

Exercise 2. Complete the equations for the following reactions:

1. KOH + H2SO4 =
2. H 3 PO 4 + Na 2 O=
3. Ba(OH) 2 + CO 2 =
4. NaOH + CuBr 2 =
5. K 2 S + Hg(NO 3) 2 =
6. Zn + FeCl 2 =

Exercise 3. Write the molecular equations for the reactions (in aqueous solution) between: a) sodium carbonate and nitric acid, b) nickel (II) chloride and sodium hydroxide, c) phosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope that you have no problems completing these three tasks. If this is not the case, you need to return to the topic "Chemical properties of the main classes of inorganic compounds."

How to turn a molecular equation into a complete ionic equation

The fun begins. We must understand which substances should be written as ions and which should be left in “molecular form”. You will have to remember the following.

In the form of ions write:

• soluble salts (I emphasize, only salts that are highly soluble in water);
• alkalis (let me remind you that alkalis are bases that are soluble in water, but not NH 4 OH);
• strong acids (H 2 SO 4, HNO 3, HCl, HBr, HI, HClO 4, HClO 3, H 2 SeO 4, ...).

As you can see, remembering this list is not at all difficult: it includes strong acids and bases and all soluble salts. By the way, for particularly vigilant young chemists who may be outraged by the fact that strong electrolytes (insoluble salts) are not included in this list, I can tell you the following: NOT including insoluble salts in this list does not at all deny the fact that they are strong electrolytes.

All other substances must be present in the ionic equations in the form of molecules. For those demanding readers who are not satisfied with the vague term “all other substances” and who, following the example of the hero of the famous film, demand “announcement of the full list”, I give the following information.

In the form of molecules write:

• all insoluble salts;
• all weak bases (including insoluble hydroxides, NH 4 OH and similar substances);
• all weak acids (H 2 CO 3, HNO 2, H 2 S, H 2 SiO 3, HCN, HClO, almost all organic acids...);
• in general, all weak electrolytes (including water!!!);
• oxides (all types);
• all gaseous compounds (in particular, H 2, CO 2, SO 2, H 2 S, CO);
• simple substances (metals and non-metals);
• almost all organic compounds (with the exception of water-soluble salts of organic acids).

Phew, looks like I haven't forgotten anything! Although it’s easier, in my opinion, to remember list No. 1. Of the fundamentally important things in list No. 2, I’ll once again mention water.

Example 2. Write a complete ionic equation describing the interaction of copper (II) hydroxide and hydrochloric acid.

Solution. Let's start, naturally, with the molecular equation. Copper(II) hydroxide is an insoluble base. All insoluble bases react with strong acids to form salt and water:

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

Now let’s find out which substances should be written down as ions and which ones as molecules. The lists above will help us. Copper(II) hydroxide is an insoluble base (see solubility table), a weak electrolyte. Insoluble bases are written in molecular form. HCl is a strong acid; in solution it almost completely dissociates into ions. CuCl 2 is a soluble salt. We write it in ionic form. Water - only in the form of molecules! We get the complete ionic equation:

Сu(OH) 2 + 2H + + 2Cl - = Cu 2+ + 2Cl - + 2H 2 O.

Example 3. Write a complete ionic equation for the reaction of carbon dioxide with an aqueous solution of NaOH.

Solution. Carbon dioxide is a typical acidic oxide, NaOH is an alkali. When acidic oxides interact with aqueous solutions of alkalis, salt and water are formed. Let’s create a molecular equation for the reaction (don’t forget about the coefficients, by the way):

CO 2 + 2NaOH = Na 2 CO 3 + H 2 O.

CO 2 - oxide, gaseous compound; maintaining molecular shape. NaOH - strong base (alkali); We write it in the form of ions. Na 2 CO 3 - soluble salt; we write in the form of ions. Water is a weak electrolyte and practically does not dissociate; leave in molecular form. We get the following:

CO 2 + 2Na + + 2OH - = Na 2+ + CO 3 2- + H 2 O.

Example 4. Sodium sulfide in aqueous solution reacts with zinc chloride to form a precipitate. Write a complete ionic equation for this reaction.

Solution. Sodium sulfide and zinc chloride are salts. When these salts interact, a precipitate of zinc sulfide precipitates:

Na 2 S + ZnCl 2 = ZnS↓ + 2NaCl.

I will immediately write down the complete ionic equation, and you will analyze it yourself:

2Na + + S 2- + Zn 2+ + 2Cl - = ZnS↓ + 2Na + + 2Cl - .

I offer you several tasks for independent work and a short test.

Exercise 4. Write molecular and complete ionic equations for the following reactions:

1. NaOH + HNO3 =
2. H2SO4 + MgO =
3. Ca(NO 3) 2 + Na 3 PO 4 =
4. CoBr 2 + Ca(OH) 2 =

Exercise 5. Write complete ionic equations describing the interaction of: a) nitric oxide (V) with an aqueous solution of barium hydroxide, b) a solution of cesium hydroxide with hydroiodic acid, c) aqueous solutions of copper sulfate and potassium sulfide, d) calcium hydroxide and an aqueous solution of iron nitrate ( III).

Balance the complete molecular equation. Before writing the ionic equation, the original molecular equation must be balanced. To do this, it is necessary to place the appropriate coefficients in front of the compounds, so that the number of atoms of each element on the left side is equal to their number on the right side of the equation.

• Write the number of atoms of each element on both sides of the equation.
• Add coefficients in front of the elements (except oxygen and hydrogen) so that the number of atoms of each element on the left and right sides of the equation is the same.
• Balance the hydrogen atoms.
• Balance the oxygen atoms.
• Count the number of atoms of each element on both sides of the equation and make sure it is the same.
• For example, after balancing the equation Cr + NiCl 2 --> CrCl 3 + Ni, we get 2Cr + 3NiCl 2 --> 2CrCl 3 + 3Ni.

Determine what state each substance that participates in the reaction is in. This can often be judged by the conditions of the problem. There are certain rules that help determine what state an element or connection is in.

Determine which compounds dissociate (separate into cations and anions) in solution. Upon dissociation, a compound breaks down into positive (cation) and negative (anion) components. These components will then enter the ionic equation of the chemical reaction.

Calculate the charge of each dissociated ion. Remember that metals form positively charged cations, and non-metal atoms turn into negative anions. Determine the charges of elements using the periodic table. It is also necessary to balance all charges in neutral compounds.

When dissolved in water, not all substances have the ability to conduct electric current. Those compounds, water solutions which are capable of conducting electric current are called electrolytes. Electrolytes conduct current due to the so-called ionic conductivity, which many compounds with an ionic structure (salts, acids, bases) possess. There are substances that have highly polar bonds, but in solution they undergo incomplete ionization (for example, mercury chloride II) - these are weak electrolytes. Many organic compounds (carbohydrates, alcohols) dissolved in water do not disintegrate into ions, but retain their molecular structure. Such substances do not conduct electric current and are called non-electrolytes.

Here are some principles that can be used to determine whether a particular compound is a strong or weak electrolyte:

1. Acids . The most common strong acids include HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4. Almost all other acids are weak electrolytes.
2. Grounds. The most common strong bases are hydroxides of alkali and alkaline earth metals (excluding Be). Weak electrolyte – NH 3.
3. Salt. Most common salts, ionic compounds, are strong electrolytes. Exceptions are mainly salts of heavy metals.

Electrolytic dissociation theory

Electrolytes, both strong and weak and even very diluted, do not obey Raoult's law And . Having the ability to conduct electrically, the vapor pressure of the solvent and the melting point of electrolyte solutions will be lower, and the boiling point will be higher compared to similar values ​​of a pure solvent. In 1887, S. Arrhenius, studying these deviations, came to the creation of the theory of electrolytic dissociation.

Electrolytic dissociation suggests that electrolyte molecules in solution break down into positively and negatively charged ions, which are called cations and anions, respectively.

The theory puts forward the following postulates:

1. In solutions, electrolytes break down into ions, i.e. dissociate. The more dilute the electrolyte solution, the greater its degree of dissociation.
2. Dissociation is a reversible and equilibrium phenomenon.
3. Solvent molecules interact infinitely weakly (i.e., solutions are close to ideal).

Different electrolytes have different degrees of dissociation, which depends not only on the nature of the electrolyte itself, but the nature of the solvent, as well as the concentration of the electrolyte and temperature.

Degree of dissociation α , shows how many molecules n disintegrated into ions, compared to the total number of dissolved molecules N:

α = n/N

In the absence of dissociation α = 0, with complete dissociation of the electrolyte α = 1.

From the point of view of the degree of dissociation, according to strength, electrolytes are divided into strong (α > 0.7), medium strength (0.3 > α > 0.7), weak (α< 0,3).

More precisely, the process of electrolyte dissociation is characterized by dissociation constant, independent of the concentration of the solution. If we imagine the process of electrolyte dissociation in general form:

A a B b ↔ aA — + bB +

K = a b /

For weak electrolytes the concentration of each ion is equal to the product of α by the total concentration of the electrolyte C, so the expression for the dissociation constant can be transformed:

K = α 2 C/(1-α)

For dilute solutions(1-α) =1, then

K = α2C

It's not hard to find from here degree of dissociation

Ionic-molecular equations

Consider an example of neutralization of a strong acid with a strong base, for example:

HCl + NaOH = NaCl + HOH

The process is presented as molecular equation. It is known that both the starting substances and the reaction products in solution are completely ionized. Therefore, let us represent the process in the form complete ionic equation:

H + + Cl - + Na + + OH - = Na + + Cl - + HOH

After “reducing” identical ions on the left and right sides of the equation, we get abbreviated ionic equation:

H + + OH - = HOH

We see that the neutralization process comes down to the combination of H + and OH - and the formation of water.

When composing ionic equations, it should be remembered that only strong electrolytes are written in ionic form. Weak electrolytes, solids, and gases are written in their molecular form.

The deposition process is reduced to the interaction of only Ag + and I - and the formation of water-insoluble AgI.

To find out whether the substance we are interested in is able to dissolve in water, we need to use the insolubility table.

Let's consider the third type of reaction, which results in the formation of a volatile compound. These are reactions involving carbonates, sulfites or sulfides with acids. For example,

When mixing some solutions of ionic compounds, interactions between them may not occur, for example

So, to summarize, we note that chemical transformations observed when one of the following conditions is met:

• Non-electrolyte formation. Water can act as a non-electrolyte.
• Formation of sediment.
• Gas release.
• Formation of a weak electrolyte for example acetic acid.
• Transfer of one or more electrons. This is realized in redox reactions.
• Formation or rupture of one or more.

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. The general characteristics of the solubility of the most important salts in water are 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 with a weak base, or a weak acid with a 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.