How to write a short ionic equation. Drawing up equations for ion exchange reactions

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 sides are excluded (reduced):

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, 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).

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, abbreviated ionic equations are equations in general form that 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 - with 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.

Work with computer

1. Refer to the electronic application. Study the lesson material and complete the assigned tasks.
2. Find email addresses on the Internet that can serve as additional sources that reveal the content of keywords and phrases in the paragraph. Offer your help to the teacher in preparing a new lesson - make a report on the key words and phrases of the next paragraph.

Questions and tasks

Topic: Chemical bond. Electrolytic dissociation

Lesson: Writing Equations for Ion Exchange Reactions

Let's create an equation for the reaction between iron (III) hydroxide and nitric acid.

Fe(OH) 3 + 3HNO 3 = Fe(NO 3) 3 + 3H 2 O

(Iron (III) hydroxide is an insoluble base, therefore it is not subjected to. Water is a poorly dissociated substance; it is practically not dissociated into ions in solution.)

Fe(OH) 3 + 3H + + 3NO 3 - = Fe 3+ + 3NO 3 - + 3H 2 O

Cross out the same number of nitrate anions on the left and right and write the abbreviated ionic equation:

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

This reaction proceeds to completion, because a slightly dissociable substance is formed - water.

Let's write an equation for the reaction between sodium carbonate and magnesium nitrate.

Na 2 CO 3 + Mg(NO 3) 2 = 2NaNO 3 + MgCO 3 ↓

Let's write this equation in ionic form:

(Magnesium carbonate is insoluble in water and therefore does not break down into ions.)

2Na + + CO 3 2- + Mg 2+ + 2NO 3 - = 2Na + + 2NO 3 - + MgCO 3 ↓

Let's cross out the same number of nitrate anions and sodium cations on the left and right, and write the abbreviated ionic equation:

CO 3 2- + Mg 2+ = MgCO 3 ↓

This reaction proceeds to completion, because a precipitate is formed - magnesium carbonate.

Let's write an equation for the reaction between sodium carbonate and nitric acid.

Na 2 CO 3 + 2HNO 3 = 2NaNO 3 + CO 2 + H 2 O

(Carbon dioxide and water are products of the decomposition of the resulting weak carbonic acid.)

2Na + + CO 3 2- + 2H + + 2NO 3 - = 2Na + + 2NO 3 - + CO 2 + H 2 O

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

This reaction proceeds to completion, because As a result, gas is released and water is formed.

Let's create two molecular reaction equations, which correspond to the following abbreviated ionic equation: Ca 2+ + CO 3 2- = CaCO 3 .

The abbreviated ionic equation shows the essence of the ion exchange reaction. In this case, we can say that to obtain calcium carbonate, it is necessary that the composition of the first substance include calcium cations, and the composition of the second - carbonate anions. Let's create molecular equations for reactions that satisfy this condition:

CaCl 2 + K 2 CO 3 = CaCO 3 ↓ + 2KCl

Ca(NO 3) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaNO 3

1. Orzhekovsky P.A. Chemistry: 9th grade: textbook. for general education establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. - M.: AST: Astrel, 2007. (§17)

2. Orzhekovsky P.A. Chemistry: 9th grade: general education. establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013. (§9)

3. Rudzitis G.E. Chemistry: inorganic. chemistry. Organ. chemistry: textbook. for 9th grade. / G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009.

4. Khomchenko I.D. Collection of problems and exercises in chemistry for high school. - M.: RIA “New Wave”: Publisher Umerenkov, 2008.

5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed. V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.

Additional web resources

1. A unified collection of digital educational resources (video experiences on the topic): ().

2. Electronic version of the journal “Chemistry and Life”: ().

Homework

1. In the table, mark with a plus sign the pairs of substances between which ion exchange reactions are possible and proceed to completion. Write reaction equations in molecular, full and reduced ionic form.

2. p. 67 No. 10,13 from the textbook P.A. Orzhekovsky “Chemistry: 9th grade” / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013.

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. in 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.

In fact, the ionic equation of a reaction maximally reflects its essence, 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.

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