Ion exchange reactions - reactions in aqueous solutions between electrolytes, flowing without changes in the oxidation states of the elements that form them

A necessary condition for the reaction between electrolytes (salts, acids and bases) is the formation of a low-dissociating substance (water, weak acid, ammonium hydroxide), a precipitate or a gas.

Consider the reaction that produces water. These reactions include all reactions between any acid and any base. For example, interaction nitric acid with potassium hydroxide:

HNO 3 + KOH \u003d 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 decompose into ions. Thus, it is possible to rewrite the equation above 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 - \u003d K + + NO 3 - + H 2 O (2)

As can be seen from equation (2), both before and after the reaction, there are NO 3 − and K + ions in the solution. In other words, in fact, nitrate ions and potassium ions did not participate in the reaction in any way. The reaction occurred only due to the combination of H + and OH − particles into water molecules. Thus, having algebraically reduced identical ions in equation (2):

H + + NO 3 - + K + + OH - \u003d K + + NO 3 - + H 2 O

we'll get:

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

Equations of the form (3) are called reduced ionic equations, of the form (2) — complete ionic equations, and of the form (1) — molecular reaction equations.

In fact, the ionic equation of the reaction maximally reflects its essence, exactly what makes it possible to proceed. It should be noted that many different reactions can correspond to one reduced ionic equation. Indeed, if we take, for example, not nitric acid, but hydrochloric acid, and use, say, barium hydroxide instead of potassium hydroxide, we have the following molecular reaction equation:

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 mainly in the form of ions. Water, as discussed above, is a weak electrolyte, that is, it exists in solution almost exclusively in the form of molecules. In this way, complete ionic equation this reaction will look like this:

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

We reduce 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 - \u003d H 2 O,

Received reduced ionic equation completely coincides with the reduced ionic equation of the interaction of nitric acid and potassium hydroxide.

When compiling ionic equations in the form of ions, only formulas are written:

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 (alkali hydroxides (ALH) and alkaline earth metals (ALHM))

3) soluble salts

In molecular form, the formulas are written:

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 alkaline metals and alkaline earth metals

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 . Sulphuric acid 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 in the form of ions in the ionic equation. Considering 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 reduced ionic equation:

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

Now let's look at the ion exchange reaction that results in the formation of a precipitate. For example, the interaction of two soluble salts:

All three salts - sodium carbonate, calcium chloride, sodium chloride and calcium carbonate (yes, yes, and he too) - are strong electrolytes and everything except calcium carbonate is 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 −

Reducing the same ions on the left and right in this equation, we get the abbreviated ionic:

CO 3 2- + Ca 2+ \u003d CaCO 3 ↓

The last equation displays the reason for the interaction of solutions of sodium carbonate and calcium chloride. Calcium ions and carbonate ions are combined into neutral calcium carbonate molecules, which, when combined with each other, give rise to small crystals of CaCO 3 precipitate of ionic structure.

Note important for passing the exam in chemistry In order for the reaction of salt1 with salt2 to proceed, in addition to the basic requirements for the occurrence of ionic reactions (gas, precipitate or water in the reaction products), one more requirement is imposed on such reactions - the initial salts must be soluble. That is, for example, CuS + Fe(NO 3) 2 ≠ FeS + Cu(NO 3) 2 no reaction, thoughFeS - could potentially precipitate, because. insoluble. The reason that the reaction does not go is the insolubility of one of the starting salts (CuS). And here, for example, Na 2 CO 3 + CaCl 2 \u003d CaCO 3 ↓ + 2NaCl proceeds, since calcium carbonate is insoluble and the original salts are soluble. The same applies to the interaction of salts with bases. In addition to the basic requirements for the occurrence of ion exchange reactions, in order for the salt to react with the base, the solubility of both of them is necessary. In this way: Cu(OH) 2 + Na 2 S - does not flow becauseCu(OH) 2 is insoluble, although the potential productCuS would be sediment. Here is the reaction betweenNaOH andCu(NO 3) 2 flows, so both starting materials are soluble and precipitateCu(OH) 2: 2NaOH + Cu(NO 3) 2 = Cu(OH) 2 ↓+ 2NaNO 3 Attention! In no case do not extend the requirement for the solubility of the starting substances beyond the reactions salt1 + salt2 and salt + base. For example, with acids, this requirement is not necessary. In particular, all soluble acids perfectly react with all carbonates, including insoluble ones. In other words: 1) Salt 1 + salt 2 - the reaction proceeds if the initial salts are soluble, and there is a precipitate in the products 2) Salt + metal hydroxide - the reaction proceeds if the starting substances are soluble and the products contain a cage or ammonium hydroxide.

Let us consider the third condition for the occurrence of ion exchange reactions - the formation of gas. Strictly speaking, only as a result of ion exchange, the formation of gas is possible only in rare cases, for example, in the formation of gaseous hydrogen sulfide:

K2S + 2HBr = 2KBr + H2S

In most other cases, the 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 within the framework of the exam 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 \u003d H 2 O + CO 2

NH 4 OH \u003d H 2 O + NH 3

H 2 SO 3 \u003d H 2 O + SO 2

In other words, if as a result of ion exchange, 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:

K2S + 2HBr = 2KBr + H2S

In ionic form, potassium sulfide and potassium bromide will be recorded, because. are soluble salts, as well as hydrobromic acid, tk. refers to strong acids. Hydrogen sulfide, being a poorly soluble and poorly dissociating gas into ions, will be written in molecular form:

2K + + S 2- + 2H + + 2Br - \u003d 2K + + 2Br - + H 2 S

Reducing the same ions, we get:

S 2- + 2H + = H 2 S

2) For the equation:

Na 2 CO 3 + H 2 SO 4 \u003d 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 low-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- \u003d 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 \u003d KNO 3 + H 2 O + NH 3

Molecules of water and ammonia will be recorded as a whole, and NH 4 NO 3 , KNO 3 and KOH will be recorded 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 - \u003d H 2 O + NH 3

For the equation:

Na 2 SO 3 + 2HCl \u003d 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

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

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

Determine the state of each substance that participates in the reaction. Often this can be judged by the condition of the problem. There are certain rules that help determine what state an element or connection is in.

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

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

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

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

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

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

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

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

Barium sulfate will precipitate in the form of a precipitate, in the crystals of which chemical bond between Ba 2+ and SO 2- 4 ions is stronger than the bond with water molecules that hydrate them. The bond between the K+ and Cl - ions only slightly exceeds the sum of their hydration energies, so the collision of these ions will not lead to the formation of a precipitate.

Therefore, the following conclusion can be drawn. Exchange reactions occur when such ions interact, the binding energy between which in the reaction product is much greater than the sum of their hydration energies.

Ion exchange reactions are described by ionic equations. Sparingly soluble, volatile and slightly dissociated compounds are written in molecular form. If during the interaction of electrolyte solutions none of the indicated types of compounds is formed, this means that practically no reactions occur.

Formation of sparingly soluble compounds

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

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

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

Only Ba 2+ and CO -2 ions reacted, the state of the remaining ions did not change, so the short ionic equation will take the form:

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

Formation of volatile substances

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

CaCO 3 + 2HCl \u003d CaCl 2 + H 2 O + CO 2

One of the reaction products - carbon dioxide CO 2 - was released from the reaction sphere in the form of a gas. The expanded ionic equation has the form:

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

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

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

Formation of a slightly dissociated compound

An example of such a reaction is any neutralization reaction, resulting in the formation of water - a slightly dissociated compound:

NaOH + HCl \u003d NaCl + H 2 O

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

OH- + H + \u003d H 2 O

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

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

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

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

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

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

Example 2. Make a molecular and ionic reaction equation between: a) iron (III) chloride and potassium hydroxide; b) potassium sulfate and zinc iodide.

a) We compose the molecular equation for the reaction between FeCl 3 and KOH:

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

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

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

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

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

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

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

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

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

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

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

1. Write down the formulas of the substances that have reacted, put an "equal" sign and write down the formulas of the formed substances. Set up coefficients.

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

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

4. Reduce identical ions before and after the equal sign in the equation. Get the reduced ionic equation.

5. Remember!

P - soluble substance;

M - poorly soluble substance;

TP - table of solubility.

Algorithm for compiling ion exchange reactions (RIO)

in molecular, full and short ionic form

Examples of compiling ion exchange reactions

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

AT this case the full ionic equation is the same as the reduced ionic equation.

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

In this case, the full ionic reaction equation coincides with the reduced one. This reaction proceeds to the end, 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 the reaction.

COMPLETE THE TASKS ON THE TOPIC "ION EXCHANGE REACTIONS"

Task number 1.
Determine whether the interaction between solutions of the following substances can be carried out, write down the reactions in molecular, full, short ionic form:
potassium hydroxide and ammonium chloride.

Solution

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

KOH + NH4 Cl = KCl + NH4 OH

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

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

We compose the full RIO ionic equation using TR (do not forget to write down the charge of the ion in the upper right corner):

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

We compose a short RIO ionic equation, deleting the same ions before and after the reaction:

Oh - +NH 4 + =NH 3 + H2O

We conclude:
The interaction between solutions of the following substances can be carried out, since the products of this RIO are gas (NH3) and a low-dissociating substance water (H2 O).

Task number 2

Given scheme:

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

Select substances, the interaction between which in aqueous solutions is expressed by the following abbreviated equations. Write the corresponding molecular and full ionic equations.

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

For example, acid - H 3 PO4 (p) and salt -K2 CO3 (p).

We compose the RIO molecular equation:

2H 3 PO4 (p) +3 K2 CO3 (p) -> 2K3 PO4 (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 PO4 (p) +3 K2 CO3 (p) -> 2K3 PO4 (p) + 3CO2 + 3H2 O

We compose the full RIO ionic equation:

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

We compose a short RIO ionic equation:

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

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

We conclude:

In the end, we got the desired reduced ionic equation, therefore, the task was completed correctly.

Task number 3

Write down the exchange reaction between sodium oxide and phosphoric acid in molecular, full 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 PO4 (p) -> 2Na3 PO4 (p) + 3H2 O (md)

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

2. We compose a complete ionic equation:

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

3. We cancel the same 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 a low-dissociating substance water is formed in the products.

TASKS FOR INDEPENDENT WORK

Task number 1

Reaction between sodium carbonate and sulfuric acid

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

Task number 2

ZnF 2 + Ca(OH)2 ->
K2 S+H3 PO4 ->

Task number 3

Check out the following experiment

Precipitation of barium sulfate

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

Task number 4

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

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

When completing the task, use the table of the solubility of substances in water. Remember about exceptions!

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

This suggests that such reactions are reduced to one process. We will obtain the equation of this process if we consider in more detail one of the above reactions, for example, the first one. We 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, see § 90):

Considering the resulting equation, we see that during the reaction, the ions and did not change. Therefore, we rewrite the equation again, excluding these ions from both sides of the equation. We get:

Thus, the reactions of neutralization of any strong acid with any strong base are reduced to the same process - to the formation of water molecules from hydrogen ions and hydroxide ions. It is clear that the thermal effects of these reactions must also be the same.

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

However, as we shall see below, water is a very weak electrolyte and dissociates only to a negligible degree. 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 the end.

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:

Similar reactions are also reduced 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 precipitate in molecular form:

As can be seen, the ions and do not undergo changes during the reaction. Therefore, we eliminate them and rewrite the equation again:

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 ions and in solution, so that the process expressed by the last equation is reversible:

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 formation from ions practically comes to an end.

The formation of a precipitate will always be observed when ions and are in a significant concentration in one solution. Therefore, with the help of silver ions, it is possible to detect the presence of ions in a solution and, conversely, with the help of chloride ions, the presence of silver ions; an ion can serve as a reactant for an ion, and an ion as a reactant for an ion.

In the future, we will widely use the ion-molecular form of writing the equations of 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 solubility in water of the most important salts is given in table. fifteen.

Table 15. Solubility of the most important salts in water

Ionic-molecular equations help to understand the features of reactions between electrolytes. Consider, as an example, several reactions involving weak acids and bases.

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 a strong acid is neutralized with a weak base, weak acid strong or weak base thermal effects are different. Let us write the ion-molecular equations for such reactions.

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

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

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

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

Here, in the form of ions, we must write the acid and the resulting salt, and in the form of molecules, ammonium hydroxide and water:

Ions do not undergo changes. Omitting them, we obtain the ion-molecular equation:

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

In this reaction, all substances, except for the resulting weak electrolytes. Therefore, the ion-molecular form of the equation has the form:

Comparing the obtained ion-molecular equations, we see that they are all different. Therefore, it is clear that the heats of the considered reactions are not the same.

As already mentioned, the reactions of neutralization of strong acids with strong bases, during which hydrogen ions and hydroxide ions combine into a water molecule, proceed almost to the end. Neutralization reactions, on the other hand, in which at least one of the starting substances is a weak electrolyte and in which molecules of weakly associated substances are present not only on the right, but also on the left side of the ion-molecular equation, do not proceed to the end.

They reach a state of equilibrium in which the salt coexists with the acid and base from which it is derived. Therefore, it is more correct to write the equations of such reactions as reversible reactions.