Drawing up equations for ion exchange reactions. Ionic reaction equations

>> Chemistry: Ionic equations

Ionic equations

As you already know from previous chemistry lessons, most chemical reactions occur 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.

Instructions

Consider an example of the formation of a sparingly soluble compound.

Na2SO4 + BaCl2 = BaSO4 + 2NaCl

Or an ionic version:

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

When solving ionic equations, the following rules must be observed:

Identical ions from both parts are excluded;

It should be remembered that the sum of the electric charges on the left side of the equation must be equal to the sum of the electric charges on the right side of the equation.

Write ionic equations for the interaction between aqueous solutions of the following substances: a) HCl and NaOH; b) AgNO3 and NaCl; c) K2CO3 and H2SO4; d) CH3COOH and NaOH.

Solution. Write down the equations of interaction of these substances in molecular form:

a) HCl + NaOH = NaCl + H2O

b) AgNO3 + NaCl = AgCl + NaNO3

c) K2CO3 + H2SO4 = K2SO4 + CO2 + H2O

d) CH3COOH + NaOH = CH3COONa + H2O

Note that the interaction of these substances is possible, because the result is the binding of ions with the formation of either weak (H2O), or sparingly soluble substance (AgCl), or gas (CO2).

By excluding identical ions from the left and right sides of the equality (in the case of option a) - ions and , in case b) - sodium ions and -ions, in case c) - potassium ions and sulfate ions), d) - sodium ions, you get solving these ionic equations:

a) H+ + OH- = H2O

b) Ag+ + Cl- = AgCl

c) CO32- + 2H+ = CO2 + H2O

d) CH3COOH + OH- = CH3COO- + H2O

Quite often in independent and tests There are tasks that involve solving reaction equations. However, without some knowledge, skills and abilities, even the simplest chemical equations don't write.

Instructions

First of all, you need to study the basic organic and inorganic compounds. As a last resort, you can have a suitable cheat sheet in front of you that can help during the task. After training they will still be remembered necessary knowledge and skills.

The basic material is covering, as well as methods for obtaining each compound. They are usually presented in the form general schemes, for example: 1. + base = salt + water
2. acid oxide + base = salt + water
3. basic oxide + acid = salt + water
4. metal + (diluted) acid = salt + hydrogen
5. soluble salt + soluble salt = insoluble salt + soluble salt
6. soluble salt + = insoluble base + soluble salt
Having before your eyes a table of salt solubility, and, as well as cheat sheets, you can decide on them equations reactions. It is only important to have full list such schemes, as well as information about the formulas and names of various classes of organic and inorganic compounds.

After the equation itself is completed, it is necessary to check the correctness of the spelling of the chemical formulas. Acids, salts and bases are easily checked using the solubility table, which shows the charges of the acidic residues and metal ions. It is important to remember that any one must be generally electrically neutral, that is, the number of positive charges must coincide with the number of negative ones. In this case, it is necessary to take into account the indices, which are multiplied by the corresponding charges.

If this stage has been passed and you are confident in the correctness of the spelling equations chemical reactions, then you can now safely set the coefficients. Chemical equation represents a conditional record reactions using chemical symbols, indices and coefficients. At this stage of the task, you must adhere to the rules: The coefficient is placed in front of the chemical formula and applies to all elements that make up the substance.
The index is placed after chemical element slightly below, and refers only to the chemical element to the left of it.
If a group (for example, an acid residue or a hydroxyl group) is in brackets, then you need to understand that two adjacent indices (before and after the bracket) are multiplied.
When counting the atoms of a chemical element, the coefficient is multiplied (not added!) by the index.

Next, the amount of each chemical element is calculated so that the total number of elements included in the starting substances coincides with the number of atoms included in the compounds formed in the products reactions. By analyzing and applying the above rules, you can learn to solve equations reactions included in chains of substances.

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

IN molecular equations All substances are presented in the form of molecules. Complete ionic equations show the entire list of ions present in the 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 (lowly dissociating substances, sediments, gases, water).

When composing ionic reactions, it should be remembered that substances are slightly dissociated (weak electrolytes), slightly and poorly soluble (precipitated - “ N”, “M”, see appendix, table 4) and gaseous ones 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 that the substance is removed in the form of a gas.

The procedure for composing ionic equations using known molecular equations Let's look at 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, with well-dissociating substances written in the form of ions, and poorly dissociating substances (including water), gases or sparingly soluble substances - in the form of molecules. The coefficient in front of the formula of a substance in a molecular equation applies equally to each of the ions that make up the substance, and therefore it is placed in front of the ion in the ionic equation:

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

3. From both sides of the equality, ions found in the left and right parts(underlined with appropriate dashes):

2Na++ 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 during the reaction, and/or slightly dissociated, and/or sparingly 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, complete and brief ionic equation. If such substances are present both in the reagents and in the products, then the reaction will be reversible (<=>):

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

Complete ionic equation: CaCO 3 + 2H + + 2Cl –<=>Ca 2+ + 2Cl – + H 2 O + CO 2

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, we will analyze many examples different levels difficulties.

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 aquatic 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 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. Those demanding readers who are not satisfied with the vague term “all other substances” and who, following the example of the hero famous film, demand to "announce 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 everything organic compounds(exception is 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 small 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 aqueous solution iron(III) nitrate.

2.6 Ionic-molecular equations

When any strong acid is neutralized by any strong base, about 57.6 kJ of heat is released for each mole of water formed:

HCl + NaOH = NaCl + H 2 O + 57.53 kJ

HNO 3 + KOH = KNO 3 + H 2 O +57.61 kJ

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):

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

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

H + + OH - = H 2 O

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's clear that thermal effects these reactions should also be the same.

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

H + + OH - ↔ H 2 O

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:

AgNO 3 + HC1 = AgCl↓ + HNO 3

Ag 2 SO 4 + CuCl 2 = 2AgCl↓ + CuSO 4

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:

Ag + + NO 3 - + H + + C1 - = AgCl↓+ H + + NO 3 -

As can be seen, the H + and NO 3 - ions do not undergo changes during the reaction. Therefore, we exclude them and rewrite the equation again:

Ag + + С1 - = AgCl↓

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

Here it must also be borne in mind that the silver chloride precipitate is in equilibrium with the Ag + and C1 - ions in solution, so that the process expressed by the last equation is reversible:

Ag + + C1 - ↔ AgCl↓

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 the formation of AgCl from ions is almost complete.

The formation of an AgCl precipitate will always be observed when there are significant concentrations of Ag + and C1 - ions in the same solution. Therefore, using silver ions, you can detect the presence of C1 - ions in a solution and, conversely, using chloride ions - the presence of silver ions; the C1 - ion can serve as a reagent for the Ag + ion, and the Ag + ion can serve as a reagent for the C1 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 2.

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

Table 2. Solubility of the most important salts in water

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):

CH 3 COOH + NaOH = CH 3 COONa + H 2 O

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

CH 3 COOH + Na + + OH - = CH 3 COO - + Na + + H 2 O

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

CH 3 COOH + OH - = CH 3 COO - + H 2 O

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

HNO 3 + NH 4 OH = NH 4 NO 3 + H 2 O

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:

H + + NO 3 - + NH 4 OH = NH 4 - + NH 3 - + H 2 O

NO 3 - ions do not undergo changes. Omitting them, we obtain the ionic-molecular equation:

H + + NH 4 OH= NH 4 + + H 2 O

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

CH 3 COOH + NH 4 OH = CH 3 COONH 4 + H 2 O

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

CH 3 COOH + NH 4 OH = CH 3 COO - + NH 4 + + H 2 O

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.

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 dissociating substances are present not only on the right side, but also on the left side ion-molecular equation, do not proceed completely. 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:

CH 3 COOH + OH - ↔ CH 3 COO - + H 2 O

H + + NH 4 OH↔ NH 4 + + H 2 O

CH 3 COOH + NH 4 OH ↔ CH 3 COO - + NH 4 + + H 2 O

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