Quite often, schoolchildren and students have to make up the so-called. ionic equations reactions. In particular, problem 31, proposed at the Unified State Examination 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 ionic equations are needed

Let me remind you that when many substances are dissolved in water (and not only in water!) A process of dissociation occurs - 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 in an aqueous solution not in the form of molecules, but in the form of hydrated Na + and Br - ions (by the way, ions are also present in solid sodium bromide).

When writing the "ordinary" (molecular) equations, we do not take into account that not molecules enter into the reaction, but ions. For example, here is the equation for the reaction between hydrochloric acid and sodium hydroxide:

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

Of course, this diagram does not quite correctly describe the process. As we have already said, there are practically no HCl molecules in an aqueous solution, but there are H + and Cl - ions. The same is true for NaOH. It would be better 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 why we have written H 2 O in molecular form. This will be explained a little later. As you can see, there is nothing complicated: we have replaced the molecules with ions, which are formed during their dissociation.

However, even the complete ionic equation is not perfect. Indeed, take a closer look: both in the left and in the right parts of equation (2) there are identical 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 short 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 short ionic equations are written down. If we solved problem 31 at the exam in chemistry, we would get the maximum mark for it - 2 points.

So, once again about the terminology:

• HCl + NaOH = NaCl + H 2 O - molecular equation ("usual" 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. We compose the molecular equation of the reaction.
2. All particles that dissociate in solution to a noticeable degree are written as ions; substances that are not prone to dissociation, we leave "in the form of molecules."
3. We remove from the two parts of the equation the so-called. 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 a complete and short ionic equation describing the interaction of aqueous solutions of barium chloride and sodium sulfate.

Solution. We will act in accordance with the proposed algorithm. Let's set up the molecular equation first. 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 forms during the reaction. Let's check:

Exercise 2. Complete the equations for the following reactions:

1. KOH + H 2 SO 4 \u003d
2. H 3 PO 4 + Na 2 O \u003d
3. Ba(OH) 2 + CO 2 =
4. NaOH + CuBr 2 =
5. K 2 S + Hg (NO 3) 2 \u003d
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) orthophosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope that you had 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 most interesting begins. We must understand which substances should be written as ions and which should be left in "molecular form". You have to remember the following.

In the form of ions write:

• soluble salts (I emphasize that only salts are highly soluble in water);
• alkalis (let me remind you that water-soluble bases are called alkalis, 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, this list is easy to remember: it includes strong acids and bases and all soluble salts. By the way, to especially 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 reject 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 of a famous film, demand "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 all organic compounds(exception - water-soluble salts of organic acids).

Phew, I don't think I forgot anything! Although it is easier, in my opinion, to remember list No. 1. Of the fundamentally important in list No. 2, I will once again note the water.

Let's train!

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

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

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

And now we find out which substances to write in the form of ions, and which - in the form of 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- strong acid, in solution almost completely dissociates into ions. CuCl 2 is a soluble salt. We write in ionic form. Water - only in the form of molecules! We get the full ionic equation:

Cu (OH) 2 + 2H + + 2Cl - \u003d 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. We compose the molecular reaction equation (do not forget, by the way, about the coefficients):

CO 2 + 2NaOH \u003d Na 2 CO 3 + H 2 O.

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

CO 2 + 2Na + + 2OH - \u003d Na 2+ + CO 3 2- + H 2 O.

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

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

Na 2 S + ZnCl 2 \u003d ZnS ↓ + 2NaCl.

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

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

Here are some tasks for you to independent work and a little test.

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

1. NaOH + HNO3 =
2. H 2 SO 4 + 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.

Reactions between various kinds of chemicals and elements are one of the main subjects of study in chemistry. To understand how to draw up a reaction equation and use them for your own purposes, you need a fairly deep understanding of all the patterns in the interaction of substances, as well as processes with chemical reactions.

Writing Equations

One way to express a chemical reaction is a chemical equation. It contains the formula of the starting substance and product, the coefficients that show how many molecules each substance has. All known chemical reactions are divided into four types: substitution, combination, exchange and decomposition. Among them are: redox, exogenous, ionic, reversible, irreversible, etc.

Learn more about how to write equations chemical reactions:

1. It is necessary to determine the name of the substances interacting with each other in the reaction. We write them on the left side of our equation. As an example, consider the chemical reaction that took place between sulfuric acid and aluminum. We have the reagents on the left: H2SO4 + Al. Next, write the equal sign. In chemistry, you can see an arrow sign that points to the right, or two opposite arrows that mean "reversibility." The result of the interaction of metal and acid is salt and hydrogen. Write the products obtained after the reaction after the “equal” sign, that is, on the right. H2SO4+Al= H2+Al2(SO4)3. So, we can see the reaction scheme.
2. For compiling chemical equation you need to find the coefficients. Let's go back to the previous diagram. Let's look at the left side of it. Sulfuric acid contains hydrogen, oxygen and sulfur atoms in an approximate ratio of 2:4:1. On the right side there are 3 sulfur atoms and 12 oxygen atoms in the salt. There are two hydrogen atoms in a gas molecule. On the left side, the ratio of these elements is 2:3:12
3. To equalize the number of oxygen and sulfur atoms that are in the composition of aluminum (III) sulfate, it is necessary to put a factor of 3 in front of the acid on the left side of the equation. Now we have 6 hydrogen atoms on the left side. In order to equalize the number of elements of hydrogen, you need to put 3 in front of hydrogen on the right side of the equation.
4. Now it remains only to equalize the amount of aluminum. Since the composition of the salt includes two metal atoms, then on the left side in front of aluminum we set the coefficient 2. As a result, we will get the reaction equation of this scheme: 2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2

Understanding the basic principles of how to write a reaction equation chemical substances, in the future it will not cause much difficulty to write down any, even the most exotic, from the point of view of chemistry, reaction.

Write down the chemical equation. As an example, consider the following reaction:

• C 3 H 8 + O 2 –> H 2 O + CO 2
• This reaction describes the combustion of propane (C 3 H 8) in the presence of oxygen to form water and carbon dioxide (carbon dioxide).

Write down the number of atoms of each element. Do this for both sides of the equation. Notice the subscripts next to each element to determine the total number of atoms. Write down the symbol for each element in the equation and note the corresponding number of atoms.

• For example, on the right side of the equation under consideration, as a result of addition, we get 3 oxygen atoms.
• On the left side we have 3 carbon atoms (C 3), 8 hydrogen atoms (H 8) and 2 oxygen atoms (O 2).
• On the right side we have 1 carbon atom (C), 2 hydrogen atoms (H 2) and 3 oxygen atoms (O + O 2).

Leave hydrogen and oxygen for later, as they are part of several compounds on the left and right side. Hydrogen and oxygen are part of several molecules, so it's best to balance them last.

• Before balancing hydrogen and oxygen, you will have to count the atoms again, as additional factors may be needed to balance other elements.

Start with the least frequently occurring element. If you need to balance several elements, choose one that is part of one molecule of reactants and one molecule of reaction products. So the first thing to do is to balance the carbon.

For balance, add a factor before the single carbon atom. Place a factor in front of the single carbon on the right side of the equation to balance it with the 3 carbons on the left side.

• C 3 H 8 + O 2 –> H 2 O + 3 CO 2
• The factor 3 in front of the carbon on the right side of the equation indicates that there are three carbon atoms, which correspond to the three carbon atoms included in the propane molecule on the left side.
• In a chemical equation, you can change the coefficients in front of atoms and molecules, but the subscripts must remain unchanged.

Then balance the hydrogen atoms. After you equalized the number of carbon atoms on the left and right side, hydrogen and oxygen remained unbalanced. The left side of the equation contains 8 hydrogen atoms, the same number should be on the right side. Achieve this with a ratio.

• C 3 H 8 + O 2 –> 4 H 2 O + 3CO 2
• We've added a factor of 4 on the right side because the subscript shows we already have two hydrogens.
• If you multiply the factor 4 by the subscript 2, you get 8.
• As a result, 10 oxygen atoms are obtained on the right side: 3x2=6 atoms in three 3CO 2 molecules and four more atoms in four water molecules.

Part I

1. Lomonosov-Lavoisier law - the law of conservation of the mass of substances:

2. The equations of a chemical reaction are conditional notation of a chemical reaction using chemical formulas and mathematical symbols.

3. The chemical equation must comply with the law conservation of the mass of substances, which is achieved by arranging the coefficients in the reaction equation.

4. What does the chemical equation show?
1) What substances react.
2) What substances are formed as a result.
3) Quantitative ratios of substances in the reaction, i.e., the amount of reacting and formed substances in the reaction.
4) Type of chemical reaction.

5. Rules for arranging the coefficients in the scheme of a chemical reaction on the example of the interaction of barium hydroxide and phosphoric acid with the formation of barium phosphate and water.
a) Write down the reaction scheme, i.e. the formulas of the reacting and formed substances:

b) begin to equalize the reaction scheme with the salt formula (if available). At the same time, remember that several complex ions in the composition of a base or salt are indicated by brackets, and their number is indicated by indices outside the brackets:

c) equalize hydrogen in the penultimate turn:

d) equalize oxygen last - this is an indicator of the correct placement of the coefficients.
Before the formula of a simple substance, it is possible to write a fractional coefficient, after which the equation must be rewritten with doubled coefficients.

Part II

1. Make up the reaction equations, the schemes of which are:

2. Write the equations of chemical reactions:

3. Establish a correspondence between the scheme and the sum of the coefficients in the chemical reaction.

4. Establish a correspondence between the starting materials and the reaction products.

5. What does the equation of the following chemical reaction show:

1) Copper hydroxide and hydrochloric acid have reacted;
2) Formed as a result of the reaction of salt and water;
3) Coefficients before starting substances 1 and 2.

6. Using the following diagram, write an equation for a chemical reaction using a doubling of the fractional coefficient:

7. Chemical reaction equation:
4P+5O2=2P2O5
shows the amount of substance of the starting substances and products, their mass or volume:
1) phosphorus - 4 mol or 124 g;
2) phosphorus (V) oxide - 2 mol, 284 g;
3) oxygen - 5 mol or 160 l.

The calculator below is designed to equalize chemical reactions.

As you know, there are several methods for equalizing chemical reactions:

• Coefficient selection method
• mathematical method
• Garcia Method
• Electronic balance method
• Electron-ion balance method (half-reaction method)

The last two are used for redox reactions.

This calculator uses mathematical method- as a rule, in the case of complex chemical equations, it is quite laborious for manual calculations, but it works great if the computer calculates everything for you.

The mathematical method is based on the law of conservation of mass. The law of conservation of mass states that the amount of substance of each element before the reaction is equal to the amount of substance of each element after the reaction. Thus, the left and right sides of a chemical equation must have the same number of atoms of one or another element. This makes it possible to balance the equations of any reactions (including redox ones). To do this, write the reaction equation in general view, based on the material balance (equality of masses of a certain chemical element in the initial and obtained substances) to compose a system of mathematical equations and solve it.

Let's look at this method with an example:

Let the chemical reaction be given:

Denote the unknown coefficients:

Let's compose the equations for the number of atoms of each element involved in the chemical reaction:
For Fe:
For Cl:
For Na:
For P:
For O:

We write them in the form of a general system:

In this case, we have five equations for four unknowns, and the fifth can be obtained by multiplying the fourth by four, so that it can be safely discarded.

Let us rewrite this system of linear algebraic equations in matrix form:

This system can be solved by the Gauss method. Actually, it will not always be so lucky that the number of equations will coincide with the number of unknowns. However, the beauty of the Gauss method is that it just allows you to solve systems with any number of equations and unknowns. Especially for this, a calculator was written Solving a system of linear equations by the Gauss method with finding a general solution, which is used when equalizing chemical reactions.
That is, the calculator below parses the reaction formula, compiles the SLAE and passes it to the calculator from the link above, which solves the SLAE using the Gauss method. The solution is then used to display the balanced equation.

Chemical elements should be written as they are written in the periodic table, i.e., take into account capital and small letters (Na3PO4 - correct, na3po4 - incorrect).