H2o decompose into ions. How to write ionic equations. How to turn a molecular equation into a complete ionic equation

Topic: Chemical bond. Electrolytic dissociation

Lesson: Writing Equations for Ion Exchange Reactions

Let's make 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 exposed. Water is a poorly dissociated substance, it is practically undissociated into ions in solution.)

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

We cross out the same number of nitrate anions on the left and right, we write the abbreviated ionic equation:

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

This reaction proceeds to the end, because a poorly dissociated substance, water, is formed.

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

Na 2 CO 3 + Mg (NO 3) 2 \u003d 2NaNO 3 + MgCO 3 ↓

We 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 ↓

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

CO 3 2- + Mg 2+ \u003d MgCO 3 ↓

This reaction proceeds to the end, 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 \u003d 2NaNO 3 + CO 2 + H 2 O

(Carbon dioxide and water are decomposition products 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 the end, because as a result, gas is released and water is formed.

Let's make 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 in order to obtain calcium carbonate, it is necessary that the first substance contains calcium cations, and the second one contains carbonate anions. Let us compose the molecular equations of reactions that satisfy this condition:

CaCl 2 + K 2 CO 3 \u003d CaCO 3 ↓ + 2KCl

Ca(NO 3) 2 + Na 2 CO 3 \u003d CaCO 3 ↓ + 2NaNO 3

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

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

3. Rudzitis G.E. Chemistry: inorgan. chemistry. Organ. chemistry: textbook. for 9 cells. / G.E. Rudzitis, F.G. Feldman. - M .: Education, JSC "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, leading. scientific ed. I. Leenson. - M.: Avanta +, 2003.

Additional web resources

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

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

Homework

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

2. with. 67 Nos. 10,13 from 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 proceed 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, the interaction of 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 instead of potassium hydroxide use, say, barium 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 ones)).

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 . 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 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.

An important note 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

the reaction does not go, although FeS - could potentially give a 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

because Cu(OH) 2 is insoluble, although the potential CuS product would be a precipitate.

But the reaction between NaOH and Cu (NO 3) 2 proceeds, so both starting materials are soluble and precipitate Cu (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 react perfectly with all carbonates, including insoluble ones.

In other words:

1) Salt1 + salt2 - 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 there is a precipitate or ammonium hydroxide in the products.

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 carbonic acid, ammonium hydroxide, or sulfurous acid is formed as a result of ion exchange, 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

Instruction

Before proceeding with ionic equations, you need to learn some rules. Water-insoluble, gaseous and low-dissociating substances (for example, water) do not decompose into ions, which means that you write them down in molecular form. This also includes weak electrolytes such as H2S, H2CO3, H2SO3, NH4OH. The solubility of compounds can be found in the solubility table, which is an approved reference material for all types of control. All the charges that are inherent in cations and anions are also indicated there. To fully complete the task, it is necessary to write the molecular, complete and ionic reduced equations.

Example No. 1. Neutralization reaction between sulfuric acid and potassium hydroxide, consider it from the point of view of TED (electrolytic dissociation theory). First, write down the reaction equation in molecular form and .H2SO4 + 2KOH = K2SO4 + 2H2O Analyze the resulting substances for their solubility and dissociation. All compounds are soluble in water, and therefore into ions. The only exception is water, which does not decompose into ions, therefore, it will remain in molecular form. Write the ionic complete equation, find the same ions on the left and right sides and. To reduce the same ions, cross them out. H2O

Example No. 2. Write the exchange reaction between copper chloride and sodium hydroxide, consider it from the point of view of TED. Write the reaction equation in molecular form and arrange the coefficients. As a result, the formed copper hydroxide precipitated blue. CuCl2 + 2NaOH \u003d Cu (OH) 2 ↓ + 2NaCl Analyze all substances for their solubility in water - everything is soluble, except for copper hydroxide, which will not dissociate into ions. Write down the ionic complete equation, underline and reduce the same ions: Cu2+ +2Cl- + 2Na+ +2OH- = Cu(OH) 2↓+2Na+ +2Cl- The ionic reduced equation remains: Cu2+ +2OH- = Cu(OH) 2↓

Example No. 3. Write the exchange reaction between sodium carbonate and hydrochloric acid, consider it from the point of view of TED. Write the reaction equation in molecular form and arrange the coefficients. As a result of the reaction, sodium chloride is formed and the gaseous substance CO2 (carbon dioxide or carbon monoxide (IV)) is released. It is formed due to the decomposition of weak carbonic acid, which decomposes into oxide and water. Na2CO3 + 2HCl = 2NaCl + CO2+H2O Analyze all substances for their water solubility and dissociation. Carbon dioxide leaves the system as a gaseous compound, water is a low-dissociating substance. All other substances break down into ions. Write down the ionic complete equation, underline and reduce the same ions: 2Na + + CO3 2- + 2H + + 2Cl- \u003d 2Na + + 2Cl- + CO2 + H2O The ionic reduced equation remains: CO3 2- + 2H + = CO2 + H2O

Definition

The reactions that take place between ions in electrolyte solutions are called ion exchange reactions(RIO).

In the course of RIO, there is no change in the oxidation states of the elements, therefore, RIO are not redox.

The criterion for the irreversibility of ion exchange reactions is the formation of a weak electrolyte.

Berthollet's rule

Ion exchange reactions proceed almost irreversibly if one of the resulting reaction products "leaves" the reaction sphere in the form:

• gas,
• draft
• or a weakly dissociating electrolyte (eg water).

If there are no ions in the solution that form a weak electrolyte, the reaction is reversible and in this case its equation is not written, putting the sign "$\ne$"

To write ionic equations, molecular (1), full ionic (2) and short ionic forms of equations (3.4) are used:

$2KOH + H_2SO_4 = K_2SO_4 + 2H_2O \hspace(3cm) (1)$

$2K^+ +2OH^- + 2H^+ + SO_4^(2-) = 2K^+ + SO_4^(2-) +2H_2O \hspace(0.2cm) (2)$

$2OH^- + 2H^+ = 2H_2O \hspace(5cm) (3)$

$OH^- + H^+ = H_2O \hspace(5.5cm) (4)$

Please note that in short ionic equation, the coefficients should be minimal. Therefore, in equation (3), all coefficients are canceled by 2, and the resulting equation (4) is considered to be a short ionic equation.

When compiling the RIO, it should be remembered that

• water, metals, oxides, gases, precipitation do not decompose into ions and are written in all equations in molecular form;
• $H_2SO_3$, $H_2CO_3$, $NH_4OH$, $AgOH$ are unstable and, upon formation, decompose almost instantly:

  $H_2SO_3 = H_2O + SO_2 \uparrow$

  $H_2CO_3 = H_2O + CO_2 \uparrow$

  $NH_4OH = H_2O + NH_3 \uparrow$

  $2AgOH = Ag_2O \downarrow + H_2O$

Algorithm for compiling ion exchange reactions

1. Write down the molecular equation and arrange the coefficients. When writing the chemical formulas of the reaction products, it is important to remember that the sum of the charges in the molecule must be equal to zero.
2. Compose a complete ionic equation, which takes into account the result of the dissociation of both the starting materials and the products of the exchange reaction. In the form of ions, all soluble compounds are recorded (indicated in the solubility table by the letter “P” (highly soluble in water), the exception is calcium hydroxide). Formulas of insoluble substances, gases, oxides, water are written in molecular form. Counting up total reaction coefficient, for which all the coefficients on the right and left sides of the equation are added.
3. To obtain an abbreviated ionic form of the equation, they give similar ones, that is, they reduce the same ions before and after the equal sign in the equation. The coefficients should be minimal, and the sums of the charges on the left and right sides of the equation should be the same. The total coefficient is calculated in an abbreviated form (similar to the full form).
4. The abbreviated ionic form of the equation reflects the essence of the past chemical reaction.

Interaction of basic oxides with acids. Write down the molecular, short and complete ionic equations for the interaction of calcium oxide and hydrochloric acid. Calculate the total coefficients in full and abbreviated form.

Solution

1. Molecular equation:

$CaO + 2HCl = CaCl_2 + H_2O$

2. Full ionic equation:

$CaO + 2H^+ + \underline(2Cl^-) = Ca^(2+) + \underline(2Cl^-) + H_2O$

The sum of the coefficients is (1+2+2+1+2+1)=9.

3. Reduced ionic equation:

$CaO + 2H^+ = Ca^(2+) + H_2O$

The total coefficient is (1+2+1+1)=5.

4. A brief ionic equation shows that when calcium oxide interacts with strong acids ($H^+$), the reaction proceeds almost irreversibly, resulting in the formation of a soluble calcium salt and a low-dissociating substance (water)

The interaction of salts with acids. Write down the molecular, short and complete ionic equations for the interaction of potassium carbonate and nitric acid. Calculate the total coefficients in full and abbreviated form.

Solution

1. Molecular equation:

$K_2CO_3 + 2HNO_3 = 2KNO_3 + CO_2\uparrow + H_2O$

2. Full ionic equation:

$\underline(2K^+) + CO_3^(2-) + 2H^+ + \underline(2NO_3^-) = \underline(2K^+) + \underline(2NO_3^-) + CO_2\uparrow + H_2O$

The sum of the coefficients is (2+1+2+2+2+2+1+1)=13.

3. Brief ionic equation:

$ CO_3^(2-) + 2H^+ = CO_2\uparrow + H_2O$

The sum of the coefficients is (1+2+1+1)=5.

4. A brief ionic equation shows that when soluble carbonates (alkali metals) interact with strong acids ($H^+$), the reaction proceeds almost irreversibly, as a result of which carbon dioxide ($CO_2\uparrow$) and a low-dissociating substance (water )

Quite often, schoolchildren and students have to make up the so-called. ionic reaction equations. 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 of different levels of complexity.

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 an aqueous medium 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. Here, for example, 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) phosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope that you did not have any problems completing these three tasks. If this is not the case, it is necessary 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 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. To 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 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, 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.

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 is a strong acid, in solution it 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 - .

I offer you several tasks for independent work and a small test.

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

1. NaOH + HNO3 =
2. H 2 SO 4 + MgO \u003d
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).