When compiling ionic equations, one should be guided by the fact that the formulas of low-dissociating, insoluble and gaseous substances are written in molecular form. If a substance precipitates, then, as you already know, an arrow pointing down (↓) is placed next to its formula, and if a gaseous substance is released during the reaction, then an upward arrow () 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 a white precipitate of barium sulfate BaSO 4 is formed as a result of the reaction. We write the molecular reaction equation:

Rice. 132.
Reaction between sodium sulfate and barium chloride

We rewrite this equation, depicting strong electrolytes as ions, and those leaving the reaction sphere as molecules:

We have thus written down the complete ionic reaction equation. 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), then we get the reduced ionic reaction equation:

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 BaSO 4 precipitate is formed. In this case, it does not matter at all which electrolytes included these ions before the reaction. A similar interaction can also be observed between K 2 SO 4 and Ba(NO 3) 2 , H 2 SO 4 and BaCl 2 .

Laboratory experiment No. 17
Interaction of 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 with a pipette. What are you watching? Write down the molecular and ionic equations of the reaction. According to the abbreviated ionic equation, offer several options for carrying out such a reaction with other electrolytes. Write down the molecular equations of the reactions performed.

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

Rice. 133.
Reaction between nitric acid and sodium hydroxide

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

NaOH + HNO 3 \u003d NaNO 3 + H 2 O.

The full 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 reaction equation is written as follows:

H + + OH - \u003d H 2 O.

It shows that the interaction of a strong acid and an 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 get 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.
The interaction of copper (II) hydroxide with acids:
a - sulfuric; b - salt; in - nitrogen

This will mean that in all cases a chemical reaction has taken 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, full and abbreviated ionic equations of the above reactions.

Laboratory experiment No. 18
Obtaining 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 watching? 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 of the reaction.

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

Consider ionic reactions that proceed with the formation of gas.

Pour 2 ml of sodium carbonate and potassium carbonate solutions 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 carbon dioxide released.

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

Let us write 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 solution. Thus, two conclusions can be drawn.

Keywords and phrases

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

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Questions and tasks

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 down 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's the same.
• For example, after balancing the equation Cr + NiCl 2 --> CrCl 3 + Ni, we get 2Cr + 3NiCl 2 --> 2CrCl 3 + 3Ni.

Determine the state of each substance that participates in the reaction. Often this can be judged by the condition 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. During dissociation, the compound decomposes 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. When doing this, remember that metals form positively charged cations, and non-metal atoms turn into negative anions. Determine the charges of the elements according to the periodic table. It is also necessary to balance all charges in neutral compounds.

Rewrite the equation so that all soluble compounds are separated into individual ions. Anything that dissociates or ionizes (such as strong acids) will split into two separate ions. In this case, the substance will remain in a dissolved state ( rr). Check that the equation is balanced.

• Solids, liquids, gases, weak acids and ionic compounds with low solubility will not change their state and will not separate into ions. Leave them as they are.
• Molecular compounds will simply dissipate in solution, and their state will change to dissolved ( rr). There are three molecular compounds that not go to state ( rr), this is CH 4( G), C 3 H 8( G) and C 8 H 18( well) .
• For the reaction under consideration, the complete ionic equation can be written in the following form: 2Cr ( tv) + 3Ni 2+ ( rr) + 6Cl - ( rr) --> 2Cr 3+ ( rr) + 6Cl - ( rr) + 3Ni ( tv) . If chlorine is not part of the compound, it breaks down into individual atoms, so we multiply the number of Cl ions by 6 on both sides of the equation.

Cancel the same ions on the left and right side of the equation. You can cross out only those ions that are completely identical on both sides of the equation (have the same charges, subscripts, and so on). Rewrite the equation without these ions.

• In our example, both sides of the equation contain 6 Cl - ions, which can be crossed out. Thus, we obtain a short ionic equation: 2Cr ( tv) + 3Ni 2+ ( rr) --> 2Cr 3+ ( rr) + 3Ni ( tv) .
• Check the result. The total charges of the left and right sides of the ionic equation must be equal.

Since electrolytes in solution are in the form of ions, the reactions between solutions of salts, bases and acids are reactions between ions, i.e. ionic reactions. Some of the ions, participating in the reaction, lead to the formation of new substances (slightly dissociating substances, precipitation, gases, water), while other ions, being present in the solution, do not give new substances, but remain in the solution. In order to show the interaction of which ions leads to the formation of new substances, molecular, complete and brief ionic equations are composed.

AT molecular equations All substances are represented as molecules. Complete ionic equations show the entire list of ions present in solution during a given reaction. Brief ionic equations are composed only of those ions, the interaction between which leads to the formation of new substances (slightly dissociating substances, precipitation, gases, water).

When compiling ionic reactions, it should be remembered that substances are slightly dissociated (weak electrolytes), slightly - and sparingly soluble (precipitating - “ H”, “M”, see appendix‚ table 4) and gaseous are written in the form of molecules. Strong electrolytes, almost completely dissociated, are in the form of ions. The sign “↓” after the formula of a substance indicates that this substance is removed from the reaction sphere in the form of a precipitate, and the sign “”, indicates the removal of a substance in the form of a gas.

The procedure for compiling ionic equations from known molecular equations consider the example of the reaction between solutions of Na 2 CO 3 and HCl.

1. The reaction equation is written in molecular form:

Na 2 CO 3 + 2HCl → 2NaCl + H 2 CO 3

2. The equation is rewritten in ionic form, while well-dissociating substances are written in the form of ions, and poorly dissociating substances (including water), gases or hardly soluble substances are written in the form of molecules. The coefficient before the formula of a substance in the molecular equation equally applies to each of the ions that make up the substance, and therefore it is taken out in the ionic equation before the ion:

2 Na + + CO 3 2- + 2H + + 2Cl -<=>2Na + + 2Cl - + CO 2 + H 2 O

3. From both parts of the equality, ions that occur in the left and right parts are excluded (reduced) (underlined by the corresponding dashes):

2 Na++ CO 3 2- + 2H + + 2Cl-<=> 2Na+ + 2Cl-+ CO 2 + H 2 O

4. The ionic equation is written in its final form (short ionic equation):

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

If in the course of the reaction and / or slightly dissociated, and / or hardly soluble, and / or gaseous substances, and / or water are formed, and such compounds are absent in the starting substances, then the reaction will be practically irreversible (→), and for it it is possible to compose a molecular, full and short ionic equation. If such substances exist both in the reactants‚ and in the products, then the reaction will be reversible (<=>):

molecular equation: CaCO 3 + 2HCl<=>CaCl 2 + H 2 O + CO 2

Full ionic equation: CaCO 3 + 2H + + 2Cl -<=>Ca 2+ + 2Cl - + H 2 O + CO 2

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 of great importance. Usually, the exchange in electrolyte solutions leads to the formation of a compound with a stronger chemical bond. So, during the interaction of solutions of salts of barium chloride BaCl 2 and potassium sulfate K 2 SO 4, four types of hydrated ions Ba 2 + (H 2 O) n, Cl - (H 2 O) m, K + (H 2 O) will be in the mixture p, SO 2 -4 (H 2 O) q, between which a reaction will occur according to the equation:

BaCl 2 + K 2 SO 4 \u003d BaSO 4 + 2 KCl

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 that hydrate them. The bond 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, the following conclusion can be drawn. Exchange reactions occur when such ions interact, 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 is formed, this means that practically no reactions occur.

Formation of sparingly soluble compounds

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

Na 2 CO 3 + BaCl 2 \u003d BaCO 3 + 2NaCl or in the form:

2Na + + CO 2- 3 + Ba 2+ + 2Cl - \u003d BaCO 3 + 2Na + + 2Cl -

Only 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+ \u003d BaCO 3

Formation of volatile substances

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

CaCO 3 + 2HCl \u003d 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 has the form:

CaCO 3 + 2H + + 2Cl - \u003d Ca 2+ + 2Cl - + H 2 O + CO 2

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

CaCO 3 + 2H + \u003d 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 \u003d NaCl + H 2 O

Na + + OH- + H + + Cl - \u003d Na + + Cl - + H 2 O

OH- + H + \u003d H 2 O

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

All three types of reactions go irreversibly, to the end.

If solutions of, for example, sodium chloride and calcium nitrate are drained, then, as the ionic equation shows, no reaction will occur, since neither a precipitate, nor a gas, nor a low-dissociating compound is formed:

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

We compose the ionic equation of the reaction, taking into account the solubility of the compounds:

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

Example 2. Make a molecular and ionic reaction equation 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:

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

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

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

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

K 2 SO 4 + ZnI 2 \u003d 2KI + ZnSO 4

From the solubility table it follows that the starting and obtained compounds are soluble, therefore the reaction is reversible, does not reach the end. Indeed, neither a precipitate, nor a gaseous compound, nor a slightly dissociated compound is formed here. Let us compose the complete ionic reaction equation:

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

Example 3. According to the ionic equation: Cu 2+ +S 2- -= CuS, draw up a molecular equation for the reaction.

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

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

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

When dissolved in water, not all substances have the ability to conduct electricity. 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) have. There are substances that have strongly 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 decompose into ions, but retain their molecular structure. Such substances do not conduct electricity and are called non-electrolytes.

Here are some regularities, guided by which it is possible to determine whether one or another compound belongs to strong or weak electrolytes:

1. acids . Among the most common strong acids are HCl, HBr, HI, HNO 3 , H 2 SO 4 , HClO 4 . Almost all other acids are weak electrolytes.
2. Foundations. 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. The exceptions are mainly salts of heavy metals.

Theory of electrolytic dissociation

Electrolytes, both strong and weak, and even very dilute ones, do not obey Raoult's law and . Having the ability to conduct electricity, 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 the same values ​​of a pure solvent. In 1887, S. Arrhenius, studying these deviations, came to the creation of a theory of electrolytic dissociation.

Electrolytic dissociation assumes that the electrolyte molecules in solution decompose into positively and negatively charged ions, which are called cations and anions, respectively.

The theory puts forward the following postulates:

1. In solutions, electrolytes decompose 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 also on the nature of the solvent, as well as electrolyte concentration and temperature.

Degree of dissociation α , shows how many molecules n decayed 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 characterizes dissociation constant, independent of the concentration of the solution. If we present the process of electrolyte dissociation in a 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 electrolyte C, so the expression for the dissociation constant can be converted:

K = α 2 C/(1-α)

For dilute solutions(1-α) =1, then

K = α 2 C

From here it's easy to find degree of dissociation

Ionic–molecular equations

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

HCl + NaOH = NaCl + HOH

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

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

After the "reduction" of identical ions in the left and right parts of the equation, we obtain reduced ionic equation:

H + + OH - = HOH

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

When compiling 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 precipitation process is reduced to the interaction of only Ag + and I - and the formation of water-insoluble AgI.

To find out whether the substance of interest to us is capable of solubility in water, it is necessary to use the insolubility table.

Let us consider the third type of reactions, as a result of which a volatile compound is formed. These are reactions of interaction of carbonates, sulfites or sulfides with acids. For example,

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

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

• Non-electrolyte formation. Water can act as a non-electrolyte.
• Sediment formation.
• Gas release.
• The formation of a weak electrolyte, such as acetic acid.
• Transfer of one or more electrons. This is realized in redox reactions.
• The formation or rupture of one or more

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