A neutralization reaction is called a reaction. Neutralization reaction: definition, examples, application. How is this test carried out?

Acid-base reactions include neutralization reactions

Neutralization reaction called the reaction of an acid and a base to form salt and water.

For example, when potassium hydroxide is added to hydrochloric acid, the reaction occurs:

KOH + HCl = KCL + H 2 O OH - + H +

The neutralization reaction proceeds irreversibly only when a strong acid interacts with a strong base, because in this case, the only weak electrolyte in the reaction mixture is the reaction product - water. If the acid and base are taken in strictly stoichiometric quantities, then the medium in the resulting salt solution will be neutral.

The neutralization reaction proceeds differently with the participation of weak acids (HNO 2, CH 3 COOH, H 2 SO 3) or weak bases (NH 3 *H 2 O, Mg(OH) 2, Fe(OH) 2).

HNO 2 + KOH ↔ KNO 2 + H 2 O

HNO 2 + K + + OH - ↔ K + + NO - 2 + H 2 O

HNO 2 + OH - ↔ NO 2 - + H 2 O

According to the abbreviated ionic-molecular reaction equation, it is clear that in the reaction system there are weak electrolytes not only among the reaction products (H 2 O), but also among the starting substances (HNO 2), which indicates the reversibility of the reaction. However, since water is the weakest electrolyte, the reaction is spontaneously strongly shifted to the right, towards the formation of salt.

Let's look at a few examples.

Example 1. Select from the listed acids and bases: HNO 2, HNO 3, H 2 SO 3, Ba(OH) 2, LiOH, Mn(OH) 2 - those whose pairwise interactions correspond to the neutralization reaction proceeding according to the equation: H + + OH - = H 2 O. write molecular equations for possible reactions.

Answer. This process corresponds to the interaction of a strong acid with a strong base. Among the listed compounds are strong acid –HNO 3, strong bases – Ba(OH) 2 and LiOH. The equations for possible reactions are as follows:

2HNO 3 + Ba(OH) 2 = Ba(NO 3) 2 + 2H 2 O

HNO 3 + LiOH = LiNO 3 + H 2 O

Example 2. The solution contains a mixture of HCl and CH 3 COOH. What reactions and in what sequence occur when this solution is neutralized with potassium hydroxide?

Answer. The acids contained in the solution belong to different types of electrolytes: HCl is a strong electrolyte, CH 3 COOH is a weak one. Due to the suppression of the dissociation of a weak electrolyte by a strong one, the neutralization of these acids with the gradual addition of alkali proceeds sequentially: first, OH - ions interact with free H + ions, i.e. with a strong acid, and then weak acid molecules are involved in the process. Thus, the reaction occurs first with HCl, and then with CH 3 COOH:

1) HCl + KOH = KCl + H 2 O H + + OH - = H 2 O

2) CH 3 COOH + KOH = CH 3 COOK + H 2 O CH 3 COOH + OH - = CH 3 COO - + H 2 O

Example 3. Indicate the qualitative and quantitative composition of the solution obtained by adding 3.36 g of KOH to 500 ml of H 3 PO 4 solution with a molar concentration of 0.1 mol/l

Given:

ϑ (solution H3PO4) = 500 ml = 0.5 l H 3 PO 4 with KOH can form three different salts.

c(H 3 PO 4) = 0.1 mol/l Let us write down the equations for the formation reactions of each of

m(KOH) = 3.36 g of possible salts and note the stoichiometric

M(KOH) = 56 g/mol molar ratio of reagents:

Composition of the solution? n(H 3 PO 4) n(KOH)

N 3 PO 4 + KOH = KN 2 PO 4 + H 2 O 1: 1

N 3 PO 4 + 2KON = K 2 NPO 4 + 2H 2 O 1: 2

H 3 PO 4 + 3KON = K 3 PO 4 + 3H 2 O

Let us determine the amounts of reagents according to the problem data and their molar ratio:

n (H 3 PO 4) = c(H 3 PO 4)* ϑ (solution H3PO4) = 0.1 mol/l * 0.5 l = 0.05 mol

n(KOH) = m(KOH)/ M(KOH) = 3.36 g/56 g/mol = 0.06 mol

n (H 3 PO 4) : n(KOH) = 0.05: 0.06 = 5: 6 = 1: 1.2

Comparing this ratio with the molar ratios of the reagents in possible reactions, we conclude that a mixture of KH 2 PO 4 and K 2 HPO 4 is formed in the solution, since there is more alkali than is required for the formation of the first salt, but less than is required for the formation of the next .

In accordance with the excess of KOH, according to the first equation, all the acid will turn into KH 2 PO 4, with n (KH 2 PO 4) = n (H 3 PO 4) = 0.05 mol.

The number of moles of KOH consumed in this reaction, n 1 (KOH) = n (H 3 PO 4) = 0.05 mol, will remain unconsumed 0.06 - 0.05 = 0.01 (mol). This amount of KOH will interact with KN 2 PO 4 according to the equation:

KN 2 RO 4 + KON = K 2 NRO 4 + H 2 O

It is obvious that 0.01 mol of KOH will convert 0.01 KN 2 PO 4 into 0.01 mol K 2 HPO 4, while 0.05 - 0.01 = 0.04 (mol) K 2 HPO 4 will remain in the solution .

Answer: 0.04 mol KN 2 PO 4 and 0.01 mol K 2 HPO 4

The lesson is devoted to the study of the reaction between substances with opposite properties - acids and bases. Such reactions are called neutralization reactions. During the lesson, you will learn to use the formula of a salt to form its name, and to write down its formula using the name of a salt.

Topic: Classes of inorganic substances

Lesson: Neutralization reaction

If you mix equal amounts of hydrochloric acid and sodium hydroxide, a solution is formed in which the medium will be neutral, i.e. there will be no acid or alkali present in it. Let us write the equation for the reaction between hydrochloric acid and sodium hydroxide if the result is sodium chloride and water.

When 1 mole of hydrogen chloride (HCl) and 1 mole of sodium hydroxide (NaOH) react, 1 mole of sodium chloride (NaCl) and 1 mole of water (H 2 O) are formed. Please note that during this reaction, two complex substances exchange their constituent parts and two new complex substances are formed:

NaOH+HCl=NaCl+H2O

Reactions during which two complex substances exchange their constituent parts are called exchange reactions.

A special case of an exchange reaction is a neutralization reaction.

A neutralization reaction is the interaction of an acid with a base.

Neutralization reaction scheme: BASE + ACID = SALT + WATER

Bases that are insoluble in water can also dissolve in acid solutions. As a result of these reactions, salts and water are formed. Reaction equation for the interaction of copper (II) hydroxide with sulfuric acid:

Cu(OH) 2 +H 2 SO 4 = CuSO 4 + 2H 2 O

A substance with the chemical formula CuSO 4 belongs to the class of salts. We compiled the formula for this salt, knowing that the valence of copper in this process is equal to II, and the valence of SO 4 is also equal to II. But what should we call this substance?

The name of a salt consists of two words: the first word is the name of the acid residue (these names are given in the table in the textbook, you need to learn them), and the second word is the name of the metal. If the valence of a metal is variable, it is indicated in parentheses.

So, a substance with the chemical formula CuSO 4 is called copper(II) sulfate.

NaNO 3 – sodium nitrate;

K 3 PO 4 – potassium phosphate (orthophosphate).

Now, let’s do the opposite task: create a formula for a salt based on its name. Let's make the formulas of the following salts: sodium sulfate; magnesium carbonate; calcium nitrate.

To correctly compose the formula of a salt, we first write down the symbol of the metal and the formula of the acid residue, and indicate their valencies at the top. Let's find the LCM of the valence values. By dividing the NOC by each valence value, we find the number of metal atoms and the number of acid residues.

Please note that if the acidic residue consists of a group of atoms, then when writing the formula of the salt, the formula of the acidic residue is written in parentheses, and the number of acidic residues is indicated behind the brackets by the corresponding index.

1. Collection of problems and exercises in chemistry: 8th grade: for textbooks. P.A. Orzhekovsky and others. “Chemistry. 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. – M.: AST: Astrel, 2006. (p. 106)

2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 107-108)

3. Chemistry. 8th grade. Textbook for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. – M.: Astrel, 2013. (§33)

4. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§39)

5. Chemistry: inorg. chemistry: textbook. for 8th grade. general education institutions / G.E. Rudzitis, F.G. Feldman. – M.: Education, OJSC “Moscow Textbooks”, 2009. (§§31,32)

6. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed. V.A. Volodin, Ved. scientific ed. I. Leenson. – M.: Avanta+, 2003.

Additional web resources

2. Indicators in neutralization reactions. titration().

Homework

1) p. 107-108 No. 4,5,7 from the Workbook in Chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.

2) p. 188 No. 1,4 from the textbook P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova “Chemistry: 8th grade,” 2013

The reaction between an acid and a base that produces salt and water is called a neutralization reaction.

We studied the reactions of acids with metals and metal oxides. These reactions produce a salt of the corresponding metal. Bases also contain metals. It can be assumed that acids will interact with bases to also form salts. Add a solution of hydrochloric acid HCl to a solution of sodium hydroxide NaOH.

The solution remains colorless and transparent, but by touch it can be determined that heat is released. The release of heat indicates that a chemical reaction has occurred between the alkali and acid.

To find out the essence of this reaction, let's do the following experiment. Place a piece of paper colored with purple litmus into the alkali solution. She will, of course, turn blue. Now, from the burette, we will begin to add the acid solution to the alkali solution in small portions until the color of the litmus again changes from blue to violet. If litmus turns from blue to purple, this means that there is no alkali in the solution. There was no more acid in the solution, since in its presence the litmus would have turned red. The solution became neutral. Having evaporated the solution, we obtained a salt - sodium chloride NaCl.

The formation of sodium chloride when sodium hydroxide reacts with hydrochloric acid is expressed by the equation:

NaOH + HCl = NaCl + H 2 O + Q

The essence of this reaction is that sodium and hydrogen atoms exchange places. As a result, the hydrogen atom of the acid combines with the hydroxyl group of the alkali to form a water molecule, and the sodium metal atom combines with the acid residue – Cl, forming a salt molecule. This reaction belongs to the familiar type of exchange reactions.

Do insoluble bases react with acids? Pour blue copper hydroxide into a glass. Let's add water. Copper hydroxide will not dissolve. Now let’s add a solution of nitric acid to it. The copper hydroxide will dissolve and a clear, blue-colored solution of copper nitrate will be obtained. The reaction is expressed by the equation:

Cu(OH) 2 + 2HNO 3 = Cu(NO 3) 2 + 2H 2 O

Bases that are insoluble in water, like alkalis, react with acids to form salts and water.

Using the neutralization reaction, insoluble acids and bases are determined experimentally. Oxide hydrates that react with neutralization with alkalis are classified as acids. Having verified from experience that this oxide hydrate is neutralized by alkalis, we write its formula as an acid formula, writing the chemical sign of hydrogen in first place: HNO3, H2SO4.

Acids do not interact with each other to form salts.

Oxide hydrates that undergo a neutralization reaction with m compounds are classified as bases. Having verified from experience that this oxide hydrate is neutralized by acids, we write its formula in the form Me(OH)n, i.e. we emphasize the presence of hydroxyl groups in it.

Bases do not interact with each other to form salts.

In the protolytic interactions considered so far (ionization of weak electrolytes and hydrolysis of salt ions), the obligatory component was water, the molecules of which, exhibiting the properties of an ampholyte, acted as either a donor or an acceptor of a proton, ensuring the occurrence of these interactions. Now let’s consider the direct interaction of acids and bases with each other, i.e. neutralization reactions.

A neutralization reaction is the protolytic interaction of an acid and a base, resulting in the formation of salt and water.

Depending on the strength of the acid and base involved, the neutralization reaction can be virtually irreversible or reversible to varying degrees.

When any strong acid interacts with any strong base (alkali), due to the fact that these reagents are completely dissociated into ions, the essence of such a reaction, regardless of the nature of the reagents, is expressed by the same molecular ionic equation:

In the process of neutralizing a strong acid with an alkali, the pH of the system changes, corresponding to the neutralization curve shown in Fig. 8.1. The neutralization curve in this case is characterized by a large and sharp jump in pH near the equivalence state (Veq) - The middle of this jump corresponds to the equivalence point, at which [H + ] = [OH-] = = 1 10 -7 mol/l, i.e. pH = 7.

Characteristic features of the reaction of neutralization of a strong acid with an alkali and vice versa are:

Irreversibility;

Exothermicity ( H 0= -57.6 kJ/mol);

Very high speed, since only mobile ions H + and OH- interact;

The pH jump during neutralization is large and sharp;

Equivalence point at pH = 7.

These features of the neutralization reaction between strong acids and bases have ensured its widespread use in analytical practice for the quantitative determination of acids and bases in the objects under study.

The most common case of a neutralization reaction is the interaction of acids and bases that differ in strength. Let's consider the neutralization of a weak acid HA with a strong base (alkali):

Since HA and H20 are weak electrolytes, protolytic equilibrium occurs due to competition for the proton between the strong bases OH- and A- and, therefore, this neutralization reaction will be characterized by the following features:

Reversibility;

The pH jump during neutralization is small and less sharp (Fig. 8.2), and with decreasing acid strength it decreases and smoothes out;

The equivalence point is located at pH > 7, since the hydrolysis reaction of the anion occurs in the system with the formation of OH- anions, the more of which the weaker the acid;

V E KB), when 50% alkali is added and [HA] = [A-], the pH value in the system is numerically equal to the value pK a of this weak acid.

The last position follows from the equation: pH = pK a+lg ([A-]/[NA]), according to which at [A - ] = [HA] pH = pK a(since lg ([A-]/[HA]) = 0). This circumstance allows not only to determine the value pK a weak acid, but also solve the inverse problem: by value pK a determine which weak acid is in the system.

Reactions of neutralization of bases of different strengths with a strong acid (Fig. 8.3) are characterized by features of equilibrium protolytic processes similar to those given above. However, you need to understand and remember that the following features are characteristic of neutralizing weak bases:

-
the equivalence point is at pH< 7 из-за проте­кающей параллельно реакции гидролиза по катиону с образо­ванием катионов Н + ;

In a state of semi-neutralization (1/2 V E KB), when 50% acid is added and [B] = [BH + ], the pH value in the system is numerically equal to the pK value (BH +) of the conjugate acid of a given weak base.

Thus, the study of the neutralization reaction makes it possible to determine not only the content of acids and bases in the system, but also the value pK a weak electrolytes, including proteins, as well as their isoelectric points.

Types of neutralization reactions. The reaction itself implies the extinguishing of foci (microbes, acids and toxins).

Neutralization reaction in medicine

The neutralization reaction is used in microbiology. This is based on the fact that some compounds are able to bind pathogens of various diseases, or their metabolisms. As a result, microorganisms are deprived of the opportunity to use their biological properties. This also includes viral inhibition reactions.

Neutralization of toxins occurs according to a similar principle. Various antitoxins are used as the main component, which block the action of toxins, preventing them from manifesting their properties.

Neutralization reaction in inorganic chemistry

Neutralization reactions are one of the foundations of inorganic. Neutralization is a type of exchange reaction. The output of the reaction is salt and water. Acids and bases are used for the reaction. Neutralization reactions are reversible and irreversible.

Irreversible reactions

The reversibility of the reaction depends on the degree of dissociation of the components. If two strong compounds are used, the neutralization reaction will not be able to return to the starting materials. This can be seen, for example, in the reaction of potassium hydroxide with nitric acid:
KOH + HNO3 – KNO3 + H2O;

The neutralization reaction in a particular case turns into a salt hydrolysis reaction.

In ionic form, the reaction looks like this:
H(+) + OH(-) > H2O;

From this we can conclude that when a strong acid reacts with a strong base, reversibility cannot occur.

Reversible reactions

If a reaction occurs between a weak base and a strong acid, or a weak acid and a strong base, or between a weak acid and a weak base, then the process is reversible.

Reversibility occurs as a result of a shift to the right in the equilibrium system. The reversibility of the reaction can be seen when using, for example, hydrocyanic acid and ammonia as starting materials.

Weak acid and strong base:
HCN+KOH=KCN+H2O;

In ionic form:
HCN+OH(-)=CN(-)+H2O.

Weak base and strong acid:
HCl+NH3-H2O=Nh4Cl+H2O;

In ionic form:
H(+)+NH3-H2O=NH4(+)+H2O.

Weak salt and weak base:
CH3COOH+NH3-H2O=CH3COONH4+H2O;

In ionic form:
CH3COOH+NH3-H2O=CH3COO(-)+NH4(+)+H2O.