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 identical 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

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. Thus, 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 proceed 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. Thus:

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

Most chemical reactions take place in solutions. Electrolyte solutions contain ions, in connection with this, reactions in electrolyte solutions are actually reduced to reactions between ions. Reactions between ions are called ionic reactions, and the equations of such reactions are called ionic equations. When compiling ionic equations, one should be guided by the fact that the formulas of low-dissociating, insoluble and gaseous substances are written in molecular form.

A white substance precipitates, then an arrow pointing down is placed next to its formula, and if a gaseous substance is released during the reaction, then an arrow pointing up is placed next to its formula.

We rewrite this equation, depicting strong electrolytes as ions, and those leaving the reaction sphere as molecules:

We have thus written down the complete ionic reaction equation.

If we exclude identical ions from both parts of the equation, that is, those that do not participate in the reaction in the left and right often equations), then we obtain a reduced ionic reaction equation:

Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, abbreviated ionic equations are general equations that characterize the essence of a chemical reaction, show which ions react and which substance is formed as a result.

Ion exchange reactions proceed to completion when either a precipitate or a low-dissociating substance, such as water, is formed. By adding an excess of nitric acid solution to a solution of sodium hydroxide dyed crimson with phenolphthalein, the solution will become colorless, which will serve as a signal for a chemical reaction:

It shows that the interaction of a strong acid and alkali is reduced to the interaction of H + ions and OH - ions, as a result of which a slightly dissociating substance is formed - water.

The indicated reaction of the interaction of a strong acid with an alkali is commonly called the neutralization reaction. This is a special case of an exchange reaction.

Such an exchange reaction can occur not only between acids and alkalis, but also between acids and insoluble bases. For example, if you get a blue precipitate of insoluble copper (II) hydroxide by reacting copper II sulfate with alkali:

and then divide the resulting precipitate into three parts and add a solution of sulfuric acid to the precipitate in the first test tube, a solution of hydrochloric acid to the precipitate in the second test tube, and a solution of nitric acid to the precipitate in the third test tube, then the precipitate will dissolve in all three test tubes. This will mean that in all cases a chemical reaction has taken place, the essence of which is reflected using the same ionic equation.

To make sure of this, write down the molecular, full and abbreviated ionic equations of the given reactions.

Consider ionic reactions that proceed with the formation of gas. Pour 2 ml of sodium carbonate and potassium carbonate solutions into two test tubes. Next, pour a solution of hydrochloric acid into the first, and nitric acid into the second. In both cases, we will notice a characteristic "boiling" due to the carbon dioxide released. We write the reaction equations for the first case:

Reactions occurring in electrolyte solutions are described using ionic equations. These reactions are called ion exchange reactions, since electrolytes exchange their ions in solutions. Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, two conclusions can be drawn. 1. Reactions in aqueous electrolyte solutions are reactions between ions, and therefore are depicted as ionic equations. Οʜᴎ are simpler than molecular ones and are more general.

2. Ion exchange reactions in electrolyte solutions proceed almost irreversibly only if, as a result, a precipitate, a gas, or a low-dissociating substance is formed.

7. Complex compounds

Ionic equations - concept and types. Classification and features of the category "Ionic Equations" 2017, 2018.

Ionic equations are an integral part of chemistry. They represent only those components that change during a chemical reaction. Most often, ionic equations are used to describe redox reactions, exchange reactions, and neutralization reactions. To write an ionic equation, there are three main steps: balance the molecular equation of a chemical reaction, convert it to a full ionic equation (that is, write down the components as they exist in solution), and finally write down a short ionic equation.

Steps

Part 1

Understand the difference between molecular and ionic compounds . To write the ionic equation, the first step is to determine the ionic compounds involved in the reaction. Ionic substances are those substances that dissociate (decompose) into charged ions in aqueous solutions. Molecular compounds do not break down into ions. They are made up of two non-metallic elements and are sometimes referred to as covalent compounds.

Determine the solubility of the compound. Not all ionic compounds dissolve in aqueous solutions, that is, not all of them dissociate into individual ions. Before you start writing the equation, you should find the solubility of each compound. The following are brief rules for solubility. More details and exceptions to the rules can be found in the solubility table.

• Follow the rules in the order they are listed below:
• all Na + , K + and NH 4 + salts dissolve;
• all salts of NO 3 - , C 2 H 3 O 2 - , ClO 3 - and ClO 4 - are soluble;
• all salts Ag + , Pb 2+ and Hg 2 2+ are insoluble;
• all salts Cl - , Br - and I - dissolve;
• salts of CO 3 2-, O 2-, S 2-, OH -, PO 4 3-, CrO 4 2-, Cr 2 O 7 2- and SO 3 2- are insoluble (with some exceptions);
• salts of SO 4 are 2- soluble (with some exceptions).

1. Identify the cation and anion of the compound. Cations are positively charged ions (usually metals). Anions have a negative charge, usually non-metal ions. Some non-metals can form not only anions, but also cations, while metal atoms always act as cations.

   • For example, in the compound NaCl (common salt), Na is a positively charged cation because it is a metal, and Cl is a negatively charged anion because it is a non-metal.

2. Determine the polyatomic (complex) ions involved in the reaction. Such ions are charged molecules, between the atoms of which there is such a strong bond that they do not dissociate during chemical reactions. It is necessary to identify polyatomic ions, since they have their own charge and do not break up into individual atoms. Polyatomic ions can have both positive and negative charges.

   Part 2

   Writing Ionic Equations

   1. 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.

   2. 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.

   3. 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 that not go to state ( rr), this is CH 4( G), C 3 H 8( G) and C 8 H 18( well) .
      • 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.

   4. 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 sides of the ionic equation must be equal.

The essence of the exchange reactions occurring in solutions reflect ionic(ionic-molecular) reaction equations. Such reactions are generally written in the form of three equations: a) molecular; b) full ionic; in) abbreviated ionic. For example, when sodium carbonate reacts with hydrochloric acid, all three equations look like this:

molecular

Na 2 CO 3 + 2 HCl  2 NaCl + H 2 O + CO 2 ,

completeionic

2 Na + + +2 H + + 2 Cl -  2 Na + + 2 Cl - + H 2 O + CO 2 .

abbreviated ionic

2H++
 H 2 O + CO 2 .

The abbreviated ionic equation does not contain those ions that remained unchanged before and after the reaction.

When writing ionic equations, it is customary to adhere to the following rules.

Do not write in the form of ions in both the left and right parts of the equation of the formula:

a) weak electrolytes, i.e. substances that in aqueous solutions only partially decompose into ions. Weak electrolytes include: water, acids (H 2 CO 3, H 2 SiO 3, H 2 S, CH 3 COOH, H 3 PO 4, H 2 SO 3, HF, HNO 2, HClO, HClO 2, H 2 SO 4(conc.)), bases, with the exception of hydroxides of alkali and alkaline earth metals (NH 4 OH, Cu(OH) 2, Al(OH) 3, Fe(OH) 2, etc.);

b) substances insoluble and poorly soluble in water, which are established according to the table of solubility of acids, bases and salts;

c) gases: CO 2, SO 2, NH 3, etc.;

d) oxides: Al 2 O 3 , CuO, FeO, P 2 O 5 etc.;

e) hydrogen-containing residues of weak acids:
,
,
, HS – ,
etc.;

f) residues of weak bases containing hydroxo groups: CuOH + , MgOH + , AlOH 2+ ,
.

In the form of ions, the formulas are written:

a) strong acids: HCl, HNO 3 , HBr, HI, HClO 3 , HClO 4 , HMnO 4 , H 2 SO 4 ;

b) alkalis (hydroxides of alkali and alkaline earth metals): LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2 , Sr(OH) 2 , Ba(OH) 2 ;

c) water-soluble salts: NaCl, K 2 SO 4, Сu (NO 3) 2, etc. The formulas of soluble complex salts are also presented in the form of ions:

K  K + + – .

EXPERIMENTAL Experiment 1. Obtaining and chemical properties of oxides

a) Obtaining the basic oxide

In a metal burning spoon, put some magnesium shavings and heat in the flame of an alcohol lamp until the magnesium ignites.

Carefully! Magnesium burns very brightly. Write the reaction equation. Note the color of the oxide. Save the resulting oxide for the next experiment.

b) Interaction of basic oxide with water

Transfer the oxide obtained in the previous experiment to a test tube and add 1-2 ml of water and 2-3 drops of phenolphthalein. How has the color changed? Write the reaction equation for the interaction of magnesium oxide with water.

in) Obtaining acid oxide

Put a piece of chalk or marble into a test tube and add 1-2 ml of hydrochloric acid solution. What is observed? Get carbon dioxide in the Kipp apparatus, in which a similar reaction of hydrochloric acid with marble takes place. Write the reaction equation in molecular and ion-molecular forms. Make a conclusion about the stability of carbonic acid.

G) Reaction of acid oxide with water and bases

Pass a current of carbon dioxide from the Kipp apparatus into a test tube with water. Add 2-3 drops of methyl red indicator solution to the contents of the tube. Note the color change and explain the reason. Write the reaction equation for the interaction of carbon dioxide with water.

Pass a current of carbon dioxide into a test tube with freshly prepared lime water (a saturated solution of calcium hydroxide). What causes the cloudiness of the solution? What kind of salt is formed? Continue to pass excess carbon dioxide through the solution until the precipitate is completely dissolved. What kind of salt is formed? Make a reaction equation in the molecular and ionic-molecular forms of the formation of the average salt CaCO 3 and the interaction of the average salt with an excess of carbonic acid. Save the resulting solution for experiment 4, c).

e) Properties of amphoteric oxides

Place one zinc oxide microspatula into two test tubes. Add 10-15 drops to the first tube2 M hydrochloric acid solution, in another - the same amount of concentrated alkali solution. Gently shake the contents of the tubes until the precipitates in both tubes are dissolved. Write reaction equations in molecular and ion-molecular forms. Make a conclusion about the nature of the taken oxide.