Abbreviated ion equation. Drawing up equations for ion exchange reactions

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

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

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

Electrolytic dissociation theory

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

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

The theory puts forward the following postulates:

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

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

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

α = n/N

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

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

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

A a B b ↔ aA — + bB +

K = a b /

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

K = α 2 C/(1-α)

For dilute solutions(1-α) =1, then

K = α2C

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

Ionic-molecular equations

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

HCl + NaOH = NaCl + HOH

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

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

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

H + + OH - = HOH

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

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

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

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

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

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

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

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

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

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

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

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

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

1. Write down the formulas of the substances that reacted, put an equal sign and write down the formulas of the formed substances. The coefficients are set.

2. Using the solubility table, write down in ionic form the formulas of substances (salts, acids, bases) designated in the solubility table by the letter “P” (highly soluble in water), with the exception of calcium hydroxide, which, although designated by the letter “M”, still in aqueous solution dissociates well into ions.

3. It must be remembered that metals, oxides of metals and non-metals, water, gaseous substances, and water-insoluble compounds indicated in the solubility table with the letter “H” do not decompose into ions. The formulas of these substances are written in molecular form. The complete ionic equation is obtained.

4. Abbreviate identical ions before and after the equal sign in the equation. The abbreviated ionic equation is obtained.

5. Remember!

P - soluble substance;

M - slightly soluble substance;

TP - solubility table.

Algorithm for composing ion exchange reactions (IER)

in molecular, full and short ionic form

Examples of composing ion exchange reactions

1. If, as a result of the reaction, a low-dissociating (ppm) substance is released - water.

IN in this case the full ionic equation is the same as the abbreviated ionic equation.

2. If, as a result of the reaction, a substance insoluble in water is released.

In this case, the full ionic equation of the reaction coincides with the abbreviated one. This reaction proceeds to completion, as evidenced by two facts at once: the formation of a substance insoluble in water and the release of water.

3. If a gaseous substance is released as a result of a reaction.

COMPLETE TASKS ON THE TOPIC "ION EXCHANGE REACTIONS"

Task No. 1.
Determine whether interaction can occur between solutions of the following substances, write down the reactions in molecular, complete, short ionic form:
potassium hydroxide and ammonium chloride.

Solution

We compose chemical formulas of substances by their names, using valencies and write RIO in molecular form (we check the solubility of substances using TR):

KOH + NH4 Cl = KCl + NH4 OH

since NH4 OH is an unstable substance and decomposes into water and NH3 gas, the RIO equation will take its final form

KOH (p) + NH4 Cl (p) = KCl (p) + NH3 + H2 O

We compose the complete ionic equation of RIO using TR (don’t forget to write down the charge of the ion in the upper right corner):

K+ + OH- + NH4 + + Cl- = K+ + Cl- + NH3 + H2 O

We create a short ionic equation for RIO, crossing out identical ions before and after the reaction:

OH - + NH 4 + = NH 3 + H2O

We conclude:
Interaction between solutions of the following substances can occur, since the products of this RIO are gas (NH3) and a poorly dissociating substance water (H2 O).

Task No. 2

The diagram is given:

2H + + CO 3 2- = H2 O+CO2

Select substances whose interaction in aqueous solutions is expressed by the following abbreviated equations. Write the corresponding molecular and total ionic equations.

Using TR we select reagents - water-soluble substances containing 2H ions + and CO3 2- .

For example, acid - H 3 P.O.4 (p) and salt -K2 CO3 (p).

We compose the molecular equation of RIO:

2H 3 P.O.4 (p) +3 K2 CO3 (p) -> 2K3 P.O.4 (p) + 3H2 CO3 (p)

Since carbonic acid is an unstable substance, it decomposes into carbon dioxide CO 2 and water H2 O, the equation will take the final form:

2H 3 P.O.4 (p) +3 K2 CO3 (p) -> 2K3 P.O.4 (p) + 3CO2 + 3H2 O

We compose the complete ionic equation of RIO:

6H + +2PO4 3- +6K+ + 3CO3 2- -> 6K+ +2PO4 3- + 3CO2 + 3H2 O

Let's create a short ionic equation for RIO:

6H + +3CO3 2- = 3CO2 + 3H2 O

2H + +CO3 2- = CO2 +H2 O

We conclude:

In the end, we received the desired abbreviated ionic equation, therefore, the task was completed correctly.

Task No. 3

Write down the exchange reaction between sodium oxide and phosphoric acid in molecular, total and short ionic form.

1. We compose a molecular equation; when compiling formulas, we take into account valencies (see TR)

3Na 2 O(ne) + 2H3 P.O.4 (p) -> 2Na3 P.O.4 (p) + 3H2 O (md)

where ne is a non-electrolyte, does not dissociate into ions,
MD is a low dissociating substance, we do not break it down into ions, water is a sign of the irreversibility of the reaction

2. We compose the complete ionic equation:

3Na 2 O+6H+ +2PO4 3- -> 6Na+ +2PO 4 3- + 3H2 O

3. We cancel identical ions and get a short ionic equation:

3Na 2 O+6H+ -> 6Na+ + 3H2 O
We reduce the coefficients by three and get:
Na2 O+2H+ -> 2Na+ +H2 O

This reaction is irreversible, i.e. goes to the end, since the low-dissociating substance water is formed in the products.

TASKS FOR INDEPENDENT WORK

Task No. 1

Reaction between sodium carbonate and sulfuric acid

Write an equation for the ion exchange reaction of sodium carbonate with sulfuric acid in molecular, total and short ionic form.

Task No. 2

ZnF 2 +Ca(OH)2 ->
K2 S+H3 P.O.4 ->

Task No. 3

Check out the next experiment

Barium sulfate precipitation

Write an equation for the ion exchange reaction of barium chloride with magnesium sulfate in molecular, total and short ionic form.

Task No. 4

Complete the reaction equations in molecular, complete and short ionic form:

Hg(NO 3 ) 2 +Na2 S ->
K2 SO3 + HCl ->

When completing the task, use the table of solubility of substances in water. Be aware of exceptions!

>> Chemistry: Ionic equations

Ionic equations

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

Typically, ionic reaction equations are obtained from molecular equations, but this happens 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.

11. Electrolytic dissociation. Ionic reaction equations

11.5. Ionic reaction equations

Since electrolytes in aqueous solutions break down into ions, it can be argued that reactions in aqueous solutions of electrolytes are reactions between ions. Such reactions can occur with a change in the oxidation state of atoms:

Fe 0  + 2 H + 1 Cl = Fe + 2 Cl 2 + H 0 2

and without change:

NaOH + HCl = NaCl + H2O

IN general case reactions between ions in solutions are called ionic, and if they are exchange reactions, then ion exchange reactions. Ion exchange reactions occur only when substances are formed that leave the reaction sphere in the form of: a) a weak electrolyte (for example, water, acetic acid); b) gas (CO 2, SO 2); c) sparingly soluble substance (precipitate). The formulas of sparingly soluble substances are determined from the solubility table (AgCl, BaSO 4, H 2 SiO 3, Mg(OH) 2, Cu(OH) 2, etc.). The formulas of gases and weak electrolytes need to be memorized. Note that weak electrolytes can be highly soluble in water: for example, CH 3 COOH, H 3 PO 4, HNO 2.

The essence of ion exchange reactions is reflected ionic reaction equations, which are obtained from molecular equations following the following rules:

1) the formulas of weak electrolytes, insoluble and slightly soluble substances, gases, oxides, hydroanions of weak acids (HS − , HSO 3 − , HCO 3 − , H 2 PO 4 − , HPO 4 2 − ; exception - HSO ion) are not written in the form of ions 4 – in a dilute solution); hydroxocations of weak bases (MgOH +, CuOH +); complex ions ( 3− , 2− , 2− );

2) the formulas of strong acids, alkalis, and water-soluble salts are represented in the form of ions. The formula Ca(OH) 2 is written as ions if lime water is used, but is not written as ions in the case of lime milk containing insoluble Ca(OH) 2 particles.

There are full ionic and abbreviated (short) ionic reaction equations. The abbreviated ionic equation is missing the ions present on both sides of the full ionic equation. Examples of writing molecular, full ionic and abbreviated ionic equations:

• NaHCO 3 + HCl = NaCl + H 2 O + CO 2 - molecular,

Na + + HCO 3 − + H + + Cl − = Na + + Cl − + H 2 O + CO 2   - complete ionic,

HCO 3 − + H + = H 2 O + CO 2   - abbreviated ionic;

• BaCl 2 + K 2 SO 4 = BaSO 4 ↓ + 2KCl - molecular,

Ba 2 + + 2 Cl − + 2 K + + SO 4 2 − = BaSO 4   ↓ + 2 K + + 2 Cl − - complete ionic,

Ba 2 + + SO 4 2 − = BaSO 4   ↓ - abbreviated ionic.

Sometimes the full ionic equation and the abbreviated ionic equation are the same:

Ba(OH) 2 + H 2 SO 4 = BaSO 4 ↓ + 2H 2 O

Ba 2+ + 2OH − + 2H + + SO 4 2 − = BaSO 4 ↓ + 2H 2 O,

and for some reactions the ionic equation cannot be compiled at all:

3Mg(OH) 2 + 3H 3 PO 4 = Mg 3 (PO 4) 2 ↓ + 6H 2 O

Example 11.5. Indicate a pair of ions that can be present in the full ion-molecular equation if it corresponds to the abbreviated ion-molecular equation

Ca 2 + + SO 4 2 − = CaSO 4 .

1) SO 3 2 − and H +; 3) CO 3 2 − and K + ; 2) HCO 3 − and K + ; 4) Cl− and Pb 2+.

Solution. The correct answer is 2):

Ca 2 + + 2 HCO 3 − + 2 K + + SO 4 2 − = CaSO 4   ↓ + 2 HCO 3 − + 2 K + (Ca(HCO 3) 2 salt is soluble) or Ca 2+ + SO 4 2 − = CaSO4.

For other cases we have:

1) CaSO 3 + 2H + + SO 4 2 − = CaSO 4 ↓ + H 2 O + SO 2 ;

3) CaCO 3 + 2K + + SO 4 2 − (reaction does not occur);

4) Ca 2+ + 2Cl − + PbSO 4 (reaction does not occur).

Answer: 2).

Substances (ions) that react with each other in an aqueous solution (i.e., the interaction between them is accompanied by the formation of a precipitate, gas or weak electrolyte) cannot coexist in an aqueous solution in significant quantities

Table 11.2

Examples of ion pairs that do not exist together in significant quantities in aqueous solution

Example 11.6. Indicate in this row: HSO 3 − , Na + , Cl − , CH 3 COO − , Zn 2+ - formulas of ions that cannot be present in significant quantities: a) in an acidic environment; b) in an alkaline environment.

Solution. a) In an acidic environment, i.e. together with H + ions, the anions HSO 3 − and CH 3 COO − cannot be present, since they react with hydrogen cations, forming a weak electrolyte or gas:

CH 3 COO − + H + ⇄ CH 3 COOH

HSO 3 − + H + ⇄ H 2 O + SO 2

b) HSO 3 − and Zn 2+ ions cannot be present in an alkaline medium, since they react with hydroxide ions to form either a weak electrolyte or a precipitate:

HSO 3 − + OH − ⇄ H 2 O + SO 3 2 −

Zn 2+ + 2OH– = Zn(OH) 2 ↓.

Answer: a) HSO 3 − and CH 3 COO −; b) HSO 3 − and Zn 2+.

Residues of acid salts of weak acids cannot be present in significant quantities in either an acidic or an alkaline medium, because in both cases a weak electrolyte is formed

The same can be said about the residues of basic salts containing a hydroxo group:

CuOH + + OH − = Cu(OH) 2 ↓