Examples of organic substances that undergo hydrolysis. Hydrolysis reaction: what is it
Chemistry, like most exact sciences, requiring a lot of attention and solid knowledge, has never been the favorite discipline of schoolchildren. And in vain, because with its help it is possible to understand the many processes taking place around and inside a person. Take, for example, the hydrolysis reaction: at first glance, it seems that it matters only for chemical scientists, but in fact, without it, no organism could function fully. Let's find out about the features of this process, as well as about its practical significance for humanity.
Hydrolysis reaction: what is it?
This phrase is a specific reaction of exchange decomposition between water and a substance dissolved in it with the formation of new compounds. Hydrolysis can also be called solvolysis in water.
This chemical term is derived from 2 Greek words: "water" and "decomposition".
Hydrolysis products
The reaction under consideration can occur when H 2 O interacts with both organic and inorganic substances. Its result directly depends on what the water was in contact with, as well as whether additional catalyst substances were used, whether the temperature and pressure were changed.
For example, the salt hydrolysis reaction promotes the formation of acids and alkalis. And when it comes to organic substances, other products are obtained. Water solvolysis of fats promotes the formation of glycerol and higher fatty acids. If the process takes place with proteins, various amino acids are formed as a result. Carbohydrates (polysaccharides) are decomposed into monosaccharides.
In the human body, which is unable to fully assimilate proteins and carbohydrates, the hydrolysis reaction "simplifies" them to substances that the body is able to digest. So solvolysis in water plays an important role in the normal functioning of every biological individual.
Hydrolysis of salts
Having learned hydrolysis, it is worth familiarizing yourself with its course in substances of inorganic origin, namely, salts.
The peculiarities of this process is that when these compounds interact with water, the ions of a weak electrolyte in the composition of the salt are detached from it and form new substances with H 2 O. It can be either acid or both. As a result of all this, there is a shift in the equilibrium of dissociation of water.
Reversible and irreversible hydrolysis
In the above example, in the latter, instead of one arrow, you can see two, and both are directed in different directions. What does it mean? This sign signals that the hydrolysis reaction is reversible. In practice, this means that, interacting with water, the taken substance simultaneously not only decomposes into components (which allow new compounds to arise), but also forms again.
However, not every hydrolysis is reversible, otherwise it would not make sense, since new substances would be unstable.
There are a number of factors that can contribute to making this reaction irreversible:
- Temperature. Whether it rises or falls depends on the direction in which the equilibrium shifts in the ongoing reaction. If it gets higher, there is a shift towards an endothermic reaction. If, on the contrary, the temperature decreases, the advantage is on the side of the exothermic reaction.
- Pressure. This is another thermodynamic quantity that actively influences ionic hydrolysis. If it rises, the chemical equilibrium is shifted towards the reaction, which is accompanied by a decrease in the total amount of gases. If it goes down, vice versa.
- High or low concentration of substances participating in the reaction, as well as the presence of additional catalysts.
Types of hydrolysis reactions in saline solutions
- By anion (ion with a negative charge). Solvolysis in water of salts of acids of weak and strong bases. This reaction is reversible due to the properties of the interacting substances.
Hydrolysis degree
Studying the features of hydrolysis in salts, it is worth paying attention to such a phenomenon as its degree. This word means the ratio of salts (which have already entered into a decomposition reaction with H 2 O) to the total amount of this substance contained in the solution.
The weaker the acid or base involved in hydrolysis, the higher its degree. It is measured in the range of 0-100% and is determined by the formula below.
N is the number of molecules of the substance that have undergone hydrolysis, and N 0 is their total number in solution.
In most cases, the degree of aqueous solvolysis in salts is low. For example, in a 1% sodium acetate solution, it is only 0.01% (at a temperature of 20 degrees).
Hydrolysis in organic substances
The process under study can also occur in organic chemical compounds.
In almost all living organisms, hydrolysis occurs as part of energy metabolism (catabolism). With its help, proteins, fats and carbohydrates are broken down into easily digestible substances. At the same time, water itself is often rarely able to start the process of solvolysis, so organisms have to use various enzymes as catalysts.
If we are talking about a chemical reaction with organic substances aimed at obtaining new substances in laboratory or production conditions, then strong acids or alkalis are added to the solution to accelerate and improve it.
Hydrolysis in triglycerides (triacylglycerols)
This difficult to pronounce term refers to fatty acids, which most of us know as fats.
They are of both animal and plant origin. However, everyone knows that water is not able to dissolve such substances, how does the hydrolysis of fats occur?
The reaction in question is called fat saponification. It is an aqueous solvolysis of triacylglycerols under the influence of enzymes in an alkaline or acidic environment. Depending on it, alkaline hydrolysis and acidic hydrolysis are released.
In the first case, as a result of the reaction, salts of higher fatty acids (better known to everyone as soaps) are formed. Thus, from NaOH, ordinary solid soap is obtained, and from KOH, liquid soap. So alkaline hydrolysis in triglycerides is the process of forming detergents. It should be noted that it can be freely carried out in fats of both vegetable and animal origin.
The reaction in question is the reason that soap is rather difficult to wash in hard water and does not wash at all in salt water. The fact is that H 2 O is called hard, which contains an excess of calcium and magnesium ions. Soap, once it gets into water, undergoes hydrolysis again, breaking down into sodium ions and a hydrocarbon residue. As a result of the interaction of these substances in water, insoluble salts are formed, which look like white flakes. To prevent this from happening, sodium bicarbonate NaHCO 3, more commonly known as baking soda, is added to the water. This substance increases the alkalinity of the solution and thereby helps the soap to perform its functions. By the way, in order to avoid such troubles, in modern industry, synthetic detergents are made from other substances, for example, from salts of esters of higher alcohols and sulfuric acid. Their molecules contain from twelve to fourteen carbon atoms, so that they do not lose their properties in salt or hard water.
If the environment in which the reaction takes place is acidic, this process is called acid hydrolysis of triacylglycerols. In this case, under the action of a certain acid, substances evolve to glycerol and carboxylic acids.
Hydrolysis of fats has another option - the hydrogenation of triacylglycerols. This process is used in some types of purification, for example, when removing traces of acetylene from ethylene or oxygen impurities from various systems.
Hydrolysis of carbohydrates
The substances under consideration are among the most important constituents of human and animal food. However, the body is not able to assimilate sucrose, lactose, maltose, starch and glycogen in their pure form. Therefore, as in the case of fats, these carbohydrates are broken down into digestible elements using a hydrolysis reaction.
Also, water solvolysis of carbon is actively used in industry. From starch, as a result of the considered reaction with H 2 O, glucose and molasses are extracted, which are part of almost all sweets.
Another polysaccharide that is actively used in industry for the manufacture of many useful substances and products is cellulose. Technical glycerin, ethylene glycol, sorbitol and ethyl alcohol, well-known to all, are extracted from it.
Cellulose hydrolysis occurs with prolonged exposure to high temperatures and the presence of mineral acids. The end product of this reaction is, as is the case with starch, glucose. It should be borne in mind that the hydrolysis of cellulose is more difficult than that of starch, since this polysaccharide is more resistant to the effects of mineral acids. However, since cellulose is the main component of the cell walls of all higher plants, the raw materials containing it are cheaper than for starch. At the same time, cellulose glucose is more used for technical needs, while the starch hydrolysis product is considered better suitable for nutrition.
Protein hydrolysis
Proteins are the main building blocks for the cells of all living organisms. They are composed of numerous amino acids and are very important for the normal functioning of the body. However, being high molecular weight compounds, they can be poorly absorbed. To simplify this task, they are hydrolyzed.
As with other organic substances, this reaction breaks down proteins into low molecular weight products that are easily absorbed by the body.
Hydrolysis is the exchange reaction of salt with water ( salt derivative In this case, the original substance is destroyed by water, with the formation of new substances.
Since hydrolysis is an ion exchange reaction, its driving force is the formation of a weak electrolyte (precipitation and / or gas evolution). It is important to remember that the hydrolysis reaction is a reversible reaction (in most cases), but there is also irreversible hydrolysis (proceeds to the end, there will be no starting substance in the solution). Hydrolysis is an endothermic process (with an increase in temperature, both the rate of hydrolysis and the yield of hydrolysis products increase).
As can be seen from the definition that hydrolysis is an exchange reaction, it can be assumed that the metal goes to the OH - group (+ a possible acid residue if a basic salt is formed (during the hydrolysis of a salt formed by a strong acid and a weak polyacid base)), and to the acid residue there is a hydrogen proton H + (+ a possible metal ion and a hydrogen ion, with the formation of an acid salt if the salt formed by a weak polybasic acid is hydrolyzed)).
There are 4 types of hydrolysis:
1. Salt formed by a strong base and strong acid. Since it was already mentioned above, hydrolysis is an ion exchange reaction, and it occurs only in the case of the formation of a weak electrolyte. As described above, the OH - group goes to the metal, and the hydrogen proton H + goes to the acid residue, but neither the strong base nor the strong acid are weak electrolytes, therefore, hydrolysis in this case does not go:
NaCl + HOH ≠ NaOH + HCl
The reaction of the medium is close to neutral: pH≈7
2. The salt is formed by a weak base and a strong acid. As indicated above: the OH - group goes to the metal, and the hydrogen proton H + goes to the acid residue. For example:
NH 4 Cl + HOH↔NH 4 OH + HCl
NH 4 + + Cl - + HOH↔NH 4 OH + H + + Cl -
NH 4 + + HOH↔NH 4 OH + H +
As can be seen from the example, hydrolysis proceeds by cation, the reaction of the medium is acidic pH < 7.При написании уравнений гидролиза для солей, образованных сильной кислотой и слабым многокислотным основанием, то в правой части следует писать основную соль, так как гидролиз идёт только по первой ступени:
FeCl 2 + HOH ↔ FeOHCl + HCl
Fe 2+ + 2Cl - + HOH↔FeO + + H + + 2Cl -
Fe 2+ + HOH ↔ FeOH + + H +
3. The salt is formed by a weak acid and a strong base. As indicated above: the OH - group goes to the metal, and the hydrogen proton H + goes to the acid residue. For example:
CH 3 COONa + HOH↔NaOH + CH 3 COOH
CH 3 COO - + Na + + HOH↔Na + + CH 3 COOH + OH -
CH 3 COO - + HOH↔ + CH 3 COOH + OH -
Hydrolysis proceeds by anion, the reaction of the medium is alkaline, pH > 7.When writing the equations for the hydrolysis of a salt formed by a weak polybasic acid and a strong base, the formation of an acidic salt should be written on the right side, hydrolysis proceeds in 1 step. For example:
Na 2 CO 3 + HOH↔NaOH + NaHCO 3
2Na + + CO 3 2- + HOH↔HCO 3 - + 2Na + + OH -
CO 3 2- + HOH↔HCO 3 - + OH -
4. The salt is formed by a weak base and a weak acid. This is the only case when the hydrolysis goes to the end, is irreversible (until the initial salt is completely consumed). For example:
CH 3 COONH 4 + HOH↔NH 4 OH + CH 3 COOH
This is the only case when the hydrolysis goes to the end. Hydrolysis proceeds both by the anion and by the cation; it is difficult to predict the reaction of the medium, but it is close to neutral: pH≈7.
There is also a constant of hydrolysis, consider it using the example of the acetate ion, denoting it Ac - ... As can be seen from the examples above, acetic (ethanoic) acid is a weak acid, and, therefore, its salts are hydrolyzed according to the scheme:
Ac - + HOH↔HAc + OH -
Let's find the equilibrium constant for this system:
Knowing ionic product of water, we can express the concentration through it [ OH] -,
Substituting this expression into the equation for the hydrolysis constant, we get:
Substituting the constant of water ionization into the equation, we get:
But the constant dissociation of acid (for example, hydrochloric acid) is equal to:
Where is the hydrated hydrogen proton: . The same is for acetic acid, as in the example. Substituting the value for the acid dissociation constant into the hydrolysis constant equation, we get:
As follows from the example, if the salt is formed with a weak base, then the denominator will be the dissociation constant of the base, calculated by the same criterion as the dissociation constant of the acid. If the salt is formed by a weak base and a weak acid, then the denominator will be the product of the dissociation constants of the acid and base.
The degree of hydrolysis.
There is also another value that characterizes hydrolysis - the degree of hydrolysis -α which is equal to the ratio of the amount (concentration) of the salt undergoing hydrolysis to the total amount (concentration) of the dissolved saltThe degree of hydrolysis depends on the salt concentration and the temperature of the solution. It increases with dilution of the salt solution and with an increase in the temperature of the solution. Recall that the more diluted the solution, the lower the molar concentration of the initial salt; and the degree of hydrolysis increases with increasing temperature, since hydrolysis is an endothermic process, as mentioned above.
The degree of salt hydrolysis is the higher, the weaker the acid or base that forms it. As follows from the equation of the degree of hydrolysis and types of hydrolysis: with irreversible hydrolysisα≈1.
The degree of hydrolysis and the hydrolysis constant are interrelated through the Ostwald equation (Wilhelm Friedrich Ostwald-sOstwald dilution act, displayed in 1888yearThe dilution law shows that the degree of dissociation of the electrolyte depends on its concentration and the dissociation constant. Let us take the initial concentration of the substance asC 0, and the dissociated part of the substance - forγ, we recall the scheme for the dissociation of a substance in a solution:
AB↔A + + B -
Then Ostwald's law can be expressed as follows:
Recall that the equation contains concentrations at the moment of equilibrium. But if the substance is poorly dissociated, then (1-γ) → 1, which brings the Ostwald equation into the form: K d = γ 2 C 0.
The degree of hydrolysis is similarly related to its constant:
In the overwhelming majority of cases, this formula is used. But if necessary, you can express the degree of hydrolysis through the following formula:
Special cases of hydrolysis:
1) Hydrolysis of hydrides (compounds of hydrogen with elements (here we will consider only metals of groups 1 and 2 and metam), where hydrogen exhibits an oxidation degree of -1):
NaH + HOH → NaOH + H 2
CaH 2 + 2HOH → Ca (OH) 2 + 2H 2
CH 4 + HOH → CO + 3H 2
The reaction with methane is one of the industrial methods for producing hydrogen.
2) Hydrolysis of peroxides.Peroxides of alkali and alkaline earth metals decompose with water, with the formation of the corresponding hydroxide and hydrogen (or oxygen) peroxide:
Na 2 O 2 +2 H 2 O → 2 NaOH + H 2 O 2
Na 2 O 2 + 2H 2 O → 2NaOH + O 2
3) Hydrolysis of nitrides.
Ca 3 N 2 + 6HOH → 3Ca (OH) 2 + 2NH 3
4) Hydrolysis of phosphides.
K 3 P + 3HOH → 3KOH + PH 3
Evolved gas PH 3 -phosphine, very toxic, affects the nervous system. It is also capable of spontaneous combustion when in contact with oxygen. Have you ever walked in a swamp at night or walked past cemeteries? We saw rare bursts of lights - "wandering lights" appear, as phosphine is burning.
5) Hydrolysis of carbides. Here will be given two reactions that have practical application, since with their help 1 members of the homologous series of alkanes (reaction 1) and alkynes (reaction 2) are obtained:
Al 4 C 3 +12 HOH → 4 Al (OH) 3 + 3CH 4 (reaction 1)
СaC 2 +2 HOH → Ca (OH) 2 + 2C 2 H 2 (reaction 2, the product is acitalene, according to UPA S etin)
6) Hydrolysis of silicides. As a result of this reaction, 1 representative of the homologous series of silanes (there are 8 of them in total) is formed SiH 4 - a monomeric covalent hydride.
Mg 2 Si + 4HOH → 2Mg (OH) 2 + SiH 4
7) Hydrolysis of phosphorus halides. Here we will consider phosphorus chlorides 3 and 5, which are acid chlorides of phosphorous and phosphoric acids, respectively:
PCl 3 + 3H 2 O = H 3 PO 3 + 3HCl
PCl 5 + 4H 2 O = H 3 PO 4 + 5HCl
8) Hydrolysis of organic substances. Fats are hydrolyzed to form glycerol (C 3 H 5 (OH) 3) and carboxylic acid (an example of a saturated carboxylic acid) (C n H (2n + 1) COOH)
Esters:
CH 3 COOCH 3 + H 2 O↔CH 3 COOH + CH 3 OH
Alcoholates:
C 2 H 5 ONa + H 2 O↔C 2 H 5 OH + NaOH
Living organisms hydrolyze various organic substances in the course of reactions catabolism with the participation enzymes. For example, during hydrolysis with the participation of digestive enzymes proteins are broken down into amino acids, fats - into glycerol and fatty acids, polysaccharides - into monosaccharides (for example, glucose).
Hydrolysis of fats in the presence of alkalis produces soap; hydrolysis of fats in the presence catalysts used to obtain glycine and fatty acids.
Tasks
1) The degree of dissociation of a acetic acid in a 0.1 M solution at 18 ° C is equal to 1.4 · 10 –2. Calculate the dissociation constant of acid K d. (Hint - use the Ostwald equation.)
2) What mass of calcium hydride needs to be dissolved in water in order to reduce the released gas to iron 6.96 g of iron oxide ( II, III)?
3) Write the reaction equation Fe 2 (SO 4) 3 + Na 2 CO 3 + H 2 O
4) Calculate the degree, the constant of hydrolysis of the salt Na 2 SO 3 for the concentration Cm = 0.03 M, taking into account only the 1st stage of hydrolysis. (The dissociation constant of sulfurous acid should be taken equal to 6.3 ∙ 10 -8)
Solutions:
a) Substitute these problems into the Ostwald dilution law:
b) K d = · [C] = (1.4 · 10 –2) · 0.1 / (1 - 0.014) = 1.99 · 10 –5
Answer. K d = 1.99 · 10 –5.
c) Fe 3 O 4 + 4H 2 → 4H 2 O + 3Fe
CaH 2 + HOH → Ca (OH) 2 + 2H 2
We find the number of moles of iron oxide (II, III), it is equal to the ratio of the mass of a given substance to its molar mass, we get 0.03 (mol). By UHR we find that the moles of calcium hydride are 0.06 (mol). So the mass of calcium hydride is equal to 2.52 (grams).
Answer: 2.52 (grams).
d) Fe 2 (SO 4) 3 + 3Na 2 CO 3 + 3H 2 O → 3CO2 + 2Fe (OH) 3 ↓ + 3Na 2 SO 4
e) Sodium sulfite undergoes hydrolysis by anion, the reaction of the salt solution medium is alkaline (pH> 7):
SO 3 2- + H 2 O<-->OH - + HSO 3 -
The hydrolysis constant (see the equation above) is: 10 -14 / 6.3 * 10 -8 = 1.58 * 10 -7
The degree of hydrolysis is calculated by the formula α 2 / (1 - α) = K h / C 0.
So, α = (K h / C 0) 1/2 = (1.58 * 10 -7 / 0.03) 1/2 = 2.3 * 10 -3
Answer: K h = 1.58 * 10 -7; α = 2.3 * 10 -3
Editor: Kharlamova Galina Nikolaevna
Investigating the effect of a universal indicator on solutions of some salts 
As we can see, the medium of the first solution is neutral (pH = 7), the second is acidic (pH< 7), третьего щелочная (рН >7). How can we explain such an interesting fact? 🙂
First, let's remember what pH is and what it depends on.
pH is a hydrogen index, a measure of the concentration of hydrogen ions in a solution (according to the first letters of the Latin words potentia hydrogeni - the strength of hydrogen).
pH is calculated as the negative decimal logarithm of the concentration of hydrogen ions, expressed in moles per liter:
In pure water at 25 ° C, the concentrations of hydrogen ions and hydroxide ions are the same and amount to 10 -7 mol / l (pH = 7).
When the concentrations of both types of ions in a solution are the same, the solution has a neutral reaction. When> the solution is acidic, and when> it is alkaline.
Due to what, in some aqueous solutions of salts, does the violation of the equality of the concentrations of hydrogen ions and hydroxide ions occur?
The fact is that there is a shift in the equilibrium of dissociation of water due to the binding of one of its ions (or) with salt ions with the formation of a poorly dissociated, hardly soluble or volatile product. This is the essence of hydrolysis.
- This is the chemical interaction of salt ions with water ions, leading to the formation of a weak electrolyte-acid (or acidic salt), or base (or basic salt).
The word "hydrolysis" means decomposition with water ("hydro" - water, "lysis" - decomposition).
Depending on which salt ion interacts with water, there are three types of hydrolysis:
- ž hydrolysis by cation (only cation reacts with water);
- žhydrolysis by anion (only anion reacts with water);
- ž joint hydrolysis - hydrolysis by cation and anion (both cation and anion reacts with water).
Any salt can be considered as a product formed by the interaction of a base and an acid:
Salt hydrolysis is the interaction of its ions with water, leading to the appearance of an acidic or alkaline medium, but not accompanied by the formation of a precipitate or gas.
The hydrolysis process takes place only with the participation soluble salts and consists of two stages:
1)dissociation salt in solution - irreversible reaction (degree of dissociation, or 100%);
2) actually , i.e. interaction of salt ions with water, - reversible reaction (degree of hydrolysis ˂ 1, or 100%)
The equations of the 1st and 2nd stages - the first of them is irreversible, the second is reversible - you cannot add!
Note that the salts formed by cations alkalis and anions strong acids, do not undergo hydrolysis, they only dissociate when dissolved in water. In solutions of KCl, NaNO 3, NaSO 4 and BaI salts, the medium neutral.
Anion hydrolysis
In case of interaction anions dissolved salt with water, the process is called by hydrolysis of the salt by the anion.
1) KNO 2 = K + + NO 2 - (dissociation)
2) NO 2 - + H 2 O ↔ HNO 2 + OH - (hydrolysis)
The dissociation of the KNO 2 salt proceeds completely, the hydrolysis of the NO 2 anion - to a very small extent (for a 0.1 M solution - by 0.0014%), but this turns out to be enough for the solution to become alkaline(among the products of hydrolysis there is an OH - ion), it contains p H = 8.14.
Only anions undergo hydrolysis weak acids (in this example, the nitrite ion NO 2 corresponding to the weak nitrous acid HNO 2). The anion of a weak acid attracts the hydrogen cation present in water and forms a molecule of this acid, while the hydroxide ion remains free:
NO 2 - + H 2 O (H +, OH -) ↔ HNO 2 + OH -
Examples:
a) NaClO = Na + + ClO -
ClO - + H 2 O ↔ HClO + OH -
b) LiCN = Li + + CN -
CN - + H 2 O ↔ HCN + OH -
c) Na 2 CO 3 = 2Na + + CO 3 2-
CO 3 2- + H 2 O ↔ HCO 3 - + OH -
d) K 3 PO 4 = 3K + + PO 4 3-
PO 4 3- + H 2 O ↔ HPO 4 2- + OH -
e) BaS = Ba 2+ + S 2-
S 2- + H 2 O ↔ HS - + OH -
Please note that in the examples (c - e) it is impossible to increase the number of water molecules and instead of hydroanions (HCO 3, HPO 4, HS) write the formulas of the corresponding acids (H 2 CO 3, H 3 PO 4, H 2 S). Hydrolysis is a reversible reaction, and it cannot proceed "to the end" (before the formation of acid).
If such an unstable acid as H 2 CO 3 was formed in a solution of its NaCO 3 salt, then CO 2 gas (H 2 CO 3 = CO 2 + H 2 O) would be released from the solution. However, when soda dissolves in water, a transparent solution is formed without gas evolution, which is evidence of incomplete hydrolysis of the anion with the appearance of only hydranions of carbonic acid HCO 3 - in the solution.
The degree of hydrolysis of the salt with respect to the anion depends on the degree of dissociation of the hydrolysis product, the acid. The weaker the acid, the higher the degree of hydrolysis. For example, the ions CO 3 2-, PO 4 3- and S 2- undergo hydrolysis to a greater extent than the NO 2 ion, since the dissociation of H 2 CO 3 and H 2 S is at the 2nd stage, and H 3 PO 4 is 3rd stage is much less than the dissociation of the acid HNO 2. Therefore, solutions, for example, Na 2 CO 3, K 3 PO 4 and BaS will be strongly alkaline(which is easy to verify by the soapyness of soda to the touch) .
An excess of OH ions in a solution can be easily detected with an indicator or measured with special devices (pH meters).
If in a concentrated solution of a salt that is strongly hydrolyzed by the anion,
for example Na 2 CO 3, add aluminum, then the latter (due to amphotericity) will react with alkali and hydrogen evolution will be observed. This is an additional proof of the hydrolysis, because we did not add NaOH alkali to the soda solution!
Pay special attention to the salts of acids of medium strength - phosphoric and sulfurous. In the first stage, these acids dissociate quite well, therefore, their acidic salts are not subjected to hydrolysis, and the solution medium of such salts is acidic (due to the presence of a hydrogen cation in the composition of the salt). And the average salts are hydrolyzed by the anion - the medium is alkaline. So, hydrosulfites, hydrogen phosphates and dihydrogen phosphates are not hydrolyzed by the anion, the medium is acidic. Sulfites and phosphates - hydrolyzed by anion, the medium is alkaline.
Hydrolysis by cation
In the case of interaction of a dissolved salt cation with water, the process is called
by hydrolysis of salt by cation
1) Ni (NO 3) 2 = Ni 2+ + 2NO 3 - (dissociation)
2) Ni 2+ + H 2 O ↔ NiOH + + H + (hydrolysis)
The dissociation of the Ni (NO 3) 2 salt proceeds entirely, the hydrolysis of the Ni 2+ cation - to a very small extent (for a 0.1 M solution - by 0.001%), but this turns out to be enough for the medium to become acidic (the H + ion is present among the hydrolysis products ).
Only cations of poorly soluble basic and amphoteric hydroxides and ammonium cation undergo hydrolysis NH 4 +. The metal cation cleaves the hydroxide ion from the water molecule and releases the hydrogen cation H +.
As a result of hydrolysis, the ammonium cation forms a weak base - ammonia hydrate and a hydrogen cation:
NH 4 + + H 2 O ↔ NH 3 · H 2 O + H +
Please note that you cannot increase the number of water molecules and write hydroxide formulas (for example, Ni (OH) 2) instead of hydroxocations (for example, NiOH +). If hydroxides were formed, then precipitation would fall out of the salt solutions, which is not observed (these salts form transparent solutions).
An excess of hydrogen cations can be easily detected with an indicator or measured with special instruments. Magnesium or zinc is added to a concentrated solution of a salt that is highly hydrolyzed by the cation, then the latter react with an acid with the release of hydrogen.
If the salt is insoluble, then there is no hydrolysis, because the ions do not interact with water.
1). Hydrolysis is an endothermic reaction, so an increase in temperature increases hydrolysis.
2). An increase in the concentration of hydrogen ions weakens hydrolysis, in the case of cation hydrolysis. Similarly, increasing the concentration of hydroxide ions weakens the hydrolysis, in the case of anionic hydrolysis.
3). When diluted with water, the equilibrium shifts towards the reaction, i.e. to the right, the degree of hydrolysis increases.
4). The addition of foreign substances can affect the equilibrium position when these substances react with one of the participants in the reaction. So, when adding copper sulfate to the solution
2CuSO4 + 2H2O<=>(CuOH) 2SO4 + H2SO4
sodium hydroxide solution, the hydroxide ions contained in it will interact with hydrogen ions. As a result, their concentration will decrease, and, according to Le Chatelier's principle, the equilibrium in the system will shift to the right, and the degree of hydrolysis will increase. And if a sodium sulfide solution is added to the same solution, then the equilibrium will not shift to the right, as one would expect (mutual enhancement of hydrolysis), but, on the contrary, to the left, due to the binding of copper ions to practically insoluble copper sulfide.
5). Salt concentration. Consideration of this factor leads to a paradoxical conclusion: the equilibrium in the system shifts to the right, in accordance with Le Chatelier's principle, but the degree of hydrolysis decreases.
Example,
Al (NO 3 ) 3
The salt is cationically hydrolyzed. The hydrolysis of this salt can be enhanced if:
- heat or dilute the solution with water;
- add alkali solution (NaOH);
- add a solution of salt hydrolyzed by the anion of Na 2 CO 3;
- dissolution lead in the cold;
- prepare the most concentrated solution of Al (NO 3) 3 as possible;
- add acid to the solution, for example HCl
Hydrolysis of salts of polyacid bases and polybasic acids takes place in steps
For example, the hydrolysis of iron (II) chloride includes two stages:
1st stage
FeCl 2 + HOH<=>Fe (OH) Cl + HCl
Fe 2+ + 2Cl - + H + + OH -<=>Fe (OH) + + 2Cl - + H +
2nd stage
Fe (OH) Cl + HOH<=>Fe (OH) 2 + HCl
Fe (OH) + + Cl - + H + + OH -<=>Fe ( OH) 2 + H + + Cl -
Hydrolysis of sodium carbonate includes two stages:
1st stage
Na 2 CO 3 + HOH<=>NaNSO 3 + NaOH
CO 3 2- + 2Na + + H + + OH - => HCO 3 - + OH - + 2Na +
2nd stage
NaHCO 3 + H 2 O<=>NaOH + H 2 CO 3
HCO 3 - + Na + + H + + OH -<=>H 2 CO 3 + OH - + Na +
Hydrolysis is a reversible process. An increase in the concentration of hydrogen ions and hydroxide ions prevents the reaction from proceeding to the end. In parallel with hydrolysis, a neutralization reaction takes place, when the formed weak base (Fe (OH) 2) interacts with a strong acid, and the resulting weak acid (H 2 CO 3) - with an alkali.
Hydrolysis is irreversible if, as a result of the reaction, an insoluble base and (or) a volatile acid is formed:
Al 2 S 3 + 6H 2 O => 2Al (OH) 3 ↓ + 3H 2 S
Salts completely biodegradable - Al 2 S 3 , cannot be obtained by the exchange reaction in aqueous solutions, since instead of exchange, the reaction of joint hydrolysis proceeds:
2AlCl 3 + 3Na 2 S ≠ Al 2 S 3 + 6NaCl
2AlCl 3 + 3Na 2 S + 6H 2 O = 2Al (OH) 3 ↓ + 6NaCl + 3H 2 S(mutual enhancement of hydrolysis)
Therefore, they are obtained in anhydrous media by sintering or other methods, for example:
2Al + 3S = t ° C= Al 2 S 3
Examples of hydrolysis reactions
(NH 4) 2 CO 3 ammonium carbonate - salt, weak acid and weak base. Soluble. It is hydrolyzed by cation and anion at the same time. The number of steps is 2.
Stage 1: (NH 4) 2 CO 3 + H 2 O ↔ NH 4 OH + NH 4 HCO 3
Stage 2: NH 4 HCO 3 + H 2 O ↔NH 4 OH + H 2 CO 3
The reaction of the solution is slightly alkaline, pH> 7, because ammonium hydroxide is a stronger electrolyte than carbonic acid. K d (NH 4 OH)> K d (H 2 CO 3)
CH 3 COONH 4 ammonium acetate - salt, weak acid and weak base. Soluble. It is hydrolyzed by cation and anion at the same time. The number of steps is 1.
CH 3 COONH 4 + H 2 O ↔NH 4 OH + CH 3 COOH
The reaction of the solution is neutral pH = 7, because K d (CH 3 COO H) = K d (NH 4 OH)
K 2 HPO 4- potassium hydrogen phosphate- salt, weak acid and strong base. Soluble. Anion hydrolyzed. The number of steps is 2.
Stage 1: K 2 HPO 4 + H 2 O ↔KH 2 PO 4 + KOH
Stage 2: KH 2 PO 4 + H 2 O ↔H 3 PO 4 + KOH
Solution reaction 1 step slightly alkalinepH=8,9 , since as a result of hydrolysis, OH - ions accumulate in the solution and the hydrolysis process prevails over the dissociation of HPO 4 2- ions, giving H + ions (HPO 4 2- ↔H + + PO 4 3-)
Solution reaction 2 steps slightly acidicpH=6,4 , since the process of dissociation of dihydroorthophosphate ions prevails over the process of hydrolysis, while hydrogen ions not only neutralize hydroxide ions, but also remain in excess, which causes a weakly acidic reaction of the medium.
Task: Determine the medium of sodium bicarbonate and sodium hydrogen sulfite solutions.
Solution:
1) Consider the processes in a sodium bicarbonate solution. The dissociation of this salt occurs in two stages, hydrogen cations are formed in the second stage:
NaHCO 3 = Na + + HCO 3 - (I)
HCO 3 - ↔ H + + CO 3 2- (II)
The dissociation constant for the second stage is K 2 of carbonic acid, equal to 4.8 ∙ 10 -11.
The hydrolysis of sodium bicarbonate is described by the equation:
NaHCO 3 + H 2 O ↔ H 2 CO 3 + NaOH
HCO 3 - + H 2 O ↔H 2 CO 3 + OH -, the constant of which is
K g = K w / K 1 (H 2 CO 3) = 1 ∙ 10 -14 / 4.5 ∙ 10 -7 = 2.2 ∙ 10 -8.
The hydrolysis constant is noticeably larger than the dissociation constant, therefore solutionNaHCO 3 has an alkaline environment.
2) Consider the processes in a solution of sodium hydrosulfite. The dissociation of this salt occurs in two stages, hydrogen cations are formed in the second stage:
NaHSO 3 = Na + + HSO 3 - (I)
HSO 3 - ↔ H + + SO 3 2- (II)
The dissociation constant for the second stage is K 2 of sulfurous acid, equal to 6.2 ∙ 10 -8.
Hydrolysis of sodium hydrosulfite is described by the equation:
NaHSO 3 + H 2 O ↔H 2 SO 3 + NaOH
HSO 3 - + H 2 O ↔H 2 SO 3 + OH -, the constant of which is
K g = K w / K 1 (H 2 SO 3) = 1 ∙ 10 -14 / 1.7 ∙ 10 -2 = 5.9 ∙ 10 -13.
In this case, the dissociation constant is greater than the hydrolysis constant, therefore solution
NaHSO 3 has an acidic environment.
Task: Determine the environment of the ammonium cyanide salt solution.
Solution:
NH 4 CN ↔NH 4 + + CN -
NH 4 + + 2H 2 O ↔NH 3. H 2 O + H 3 O +
CN - + H 2 O ↔HCN + OH -
NH 4 CN + H 2 O↔ NH 4 OH + HCN
K d (HCN) = 7.2 ∙ 10 -10; K d (NH 4 OH) = 1.8 ∙ 10 -5
Answer: Hydrolysis by cation and anion, because K o> K k, slightly alkaline medium, pH> 7
The process of formation of weakly dissociated compounds with a change in the pH of the medium during the interaction of water and salt is called hydrolysis.
Salt hydrolysis occurs in the case of the binding of one water ion with the formation of poorly soluble or weakly dissociated compounds due to a shift in the dissociation equilibrium. For the most part, this process is reversible and intensifies with dilution or increasing temperature.
To find out which salts undergo hydrolysis, you need to know which bases and acids were used in strength in its formation. There are several types of their interactions.
Preparation of a salt from a base and a weak acid
Examples include sulphide of aluminum and chromium, as well as ammonium acid and ammonium carbonate. When dissolved in water, these salts form bases and weakly dissociating acids. To trace the reversibility of the process, it is necessary to draw up an equation for the salt hydrolysis reaction:
Ammonium acetic acid + water ↔ ammonia + acetic acid
In ionic form, the process looks like:
CH 3 COO- + NH 4 + + H 2 O ↔ CH 3 COOH + NH 4 OH.
In the above hydrolysis reaction, ammonia and acetic acid are formed, that is, weakly dissociating substances.
The pH of aqueous solutions (pH) directly depends on the relative strength, that is, the dissociation constants of the reaction products. The above reaction will be slightly alkaline, since the decay constant of acetic acid is less than the constant of ammonium hydroxide, that is, 1.75 ∙ 10 - 5 is less than 6.3 ∙ 10 -5. If bases and acids are removed from the solution, then the process continues to the end.
Consider an example of irreversible hydrolysis:
Aluminum sulfate + water = aluminum hydroxide + hydrogen sulfide
In this case, the process is irreversible, because one of the reaction products is removed, that is, it precipitates.
Hydrolysis of compounds obtained by the reaction of a weak base with a strong acid
This type of hydrolysis describes the decomposition reactions of aluminum sulfate, copper chloride or bromide, as well as iron or ammonium chloride. Consider the reaction of ferric chloride, which takes place in two stages:
Stage one:
Iron chloride + water ↔ iron hydroxychloride + hydrochloric acid
The ionic equation for the hydrolysis of ferric chloride salts takes the form:
Fe 2+ + H 2 O + 2Cl - ↔ Fe (OH) + + H + + 2Cl -
Second stage of hydrolysis:
Fe (OH) + + H 2 O + Cl - ↔ Fe (OH) 2 + H + + Cl -
Due to the deficiency of hydroxyl ions and the accumulation of hydrogen ions, the hydrolysis of FeCl 2 proceeds in the first stage. Strong hydrochloric acid and a weak base, iron hydroxide, are formed. In the case of such reactions, the medium turns out to be acidic.
Non-hydrolysable salts obtained by the interaction of strong bases and acids
Examples of such salts are calcium or sodium chloride, potassium sulfate and rubidium bromide. However, these substances do not hydrolyze, since they have a neutral medium when dissolved in water. The only low-dissociating substance in this case is water. To confirm this statement, one can draw up an equation for the hydrolysis of sodium chloride salts with the formation of hydrochloric acid and sodium hydroxide:
NaCl + H 2 O ↔ NaOH + HCl
Reaction in ionic form:
Na + + Cl - + Н 2 О↔ Na + + ОН - + Н + + Cl -
H 2 O ↔ H + + OH -
Salts as a reaction product of strong alkali and weak acid
In this case, the hydrolysis of salts proceeds along the anion, which corresponds to the alkaline medium of the pH value. Mention may be made, as examples, of sodium acetate, sulfate and carbonate, potassium silicate and sulfate, and sodium hydrocyanic acid. For example, let's compose the ion-molecular equations for the hydrolysis of sodium sulfide and acetate salts:
Dissociation of sodium sulfide:
Na 2 S ↔ 2Na + + S 2-
The first stage of hydrolysis of a polybasic salt occurs at the cation:
Na 2 S + H 2 O ↔ NaH S + NaOH
Ionic recording:
S 2- + H 2 O ↔ HS - + OH -
The second stage is feasible in case of an increase in the reaction temperature:
HS - + H 2 O ↔ H 2 S + OH -
Consider another hydrolysis reaction using sodium acetate as an example:
Sodium acetic acid + water ↔ acetic acid + caustic soda
In ionic form:
CH 3 COO - + H 2 O ↔ CH 3 COOH + OH -
The reaction produces weak acetic acid. In both cases, the reactions will be alkaline.
Equilibrium reaction according to Le Chatelier's principle
Hydrolysis, like other chemical reactions, is reversible and irreversible. In the case of reversible reactions, one of the reagents is not completely consumed, while irreversible processes proceed with the full consumption of the substance. This is due to a shift in the equilibrium of reactions, which is based on changes in physical characteristics such as pressure, temperature, and mass fraction of reagents.
According to the concept of Le Chatelier's principle, the system will be considered equilibrium until one or more external conditions of the process are changed to it. For example, with a decrease in the concentration of one of the substances, the equilibrium of the system will gradually begin to shift towards the formation of the same reagent. Salt hydrolysis also has the ability to obey the Le Chatelier principle, which can be used to weaken or enhance the process.
Enhanced hydrolysis
Hydrolysis can be enhanced to complete irreversibility in several ways:
- Increase the rate of formation of OH - and H + ions. For this, the solution is heated, and by increasing the absorption of heat by water, that is, endothermic dissociation, this indicator rises.
- Add water.
- Convert one of the products to a gaseous state or bind into a heavily soluble substance.
Suppression of hydrolysis
It is possible to suppress the hydrolization process, as well as to enhance it, in several ways.
Introduce one of the substances formed in the process into the solution. For example, alkalinize the solution, if the pH is 7, or, on the contrary, acidify, where the reaction medium is less than 7 in terms of pH.
Mutual enhancement of hydrolysis
Mutual enhancement of hydrolysis is applied when the system has become equilibrium. Let's analyze a specific example, where the systems in different vessels have become equilibrium:
Al 3+ + H 2 O ↔ AlOH 2+ + H +
CO 3 2- + H 2 O ↔ HCO 3 - + OH -
Both systems are poorly hydrolyzed, therefore, if mixed with each other, the binding of hydroxoins and hydrogen ions will occur. As a result, we obtain the molecular equation of salt hydrolysis:
Aluminum chloride + sodium carbonate + water = sodium chloride + aluminum hydroxide + carbon dioxide.
According to Le Chatelier's principle, the equilibrium of the system will move towards the reaction products, and the hydrolysis will proceed to the end with the formation of aluminum hydroxide precipitated. Such an intensification of the process is possible only if one of the reactions proceeds through the anion, and the other through the cation.
Anion hydrolysis
Hydrolysis of aqueous solutions of salts is carried out by combining their ions with water molecules. One of the methods of hydrolysis is carried out by anion, that is, the addition of an aqueous H + ion.
For the most part, salts are susceptible to this hydrolysis method, which are formed through the interaction of a strong hydroxide and a weak acid. Examples of anionically decomposing salts are sodium sulfate or sulfite, and potassium carbonate or phosphate. At the same time, the hydrogen index is more than seven. As an example, let us analyze the dissociation of sodium acetate:
In solution, this compound is divided into a cation - Na +, and an anion - CH 3 COO -.
The dissociated sodium acetate cation formed by a strong base cannot react with water.
In this case, acid anions easily react with H 2 O molecules:
CH 3 COO - + HOH = CH 3 COOH + OH -
Consequently, hydrolysis is carried out by the anion, and the equation takes the form:
CH3COONa + HOH = CH 3 COOH + NaOH
If polybasic acids undergo hydrolysis, the process takes place in several stages. Under normal conditions, these substances are hydrolyzed in the first stage.
Hydrolysis by cation
Salts formed by the interaction of a strong acid and a low strength base are mainly susceptible to cationic hydrolysis. Examples are ammonium bromide, copper nitrate, and zinc chloride. In this case, the medium in solution during hydrolysis corresponds to less than seven. Let us consider the process of hydrolysis by cation using the example of aluminum chloride:
In an aqueous solution, it dissociates into an anion - 3Cl - and a cation - Al 3+.
Strong hydrochloric acid ions do not interact with water.
Ions (cations) of the base, on the other hand, are susceptible to hydrolysis:
Al 3+ + HOH = AlOH 2+ + H +
In molecular form, hydrolysis of aluminum chloride is as follows:
AlCl3 + H 2 O = AlOHCl + HCl
Under normal conditions, it is preferable to neglect the hydrolysis in the second and third stages.
Dissociation degree
Any salt hydrolysis reaction is characterized by the degree of dissociation, which shows the ratio between the total number of molecules and molecules capable of passing into the ionic state. The degree of dissociation is characterized by several indicators:
- The temperature at which the hydrolysis takes place.
- Concentration of the dissociated solution.
- The origin of the salt being dissolved.
- The nature of the solvent itself.
According to the degree of dissociation, all solutions are divided into strong and weak electrolytes, which, in turn, when dissolved in various solvents, exhibit a different degree.
Dissociation constant
A quantitative indicator of the ability of a substance to decay into ions is the dissociation constant, also called the equilibrium constant. In simple terms, the equilibrium constant is the ratio of electrolytes decomposed into ions to non-dissociated molecules.
Unlike the degree of dissociation, this parameter does not depend on the external conditions and the concentration of the saline solution during hydrolysis. During the dissociation of polybasic acids, the degree of dissociation at each stage becomes an order of magnitude lower.
Indicator of acid-base properties of solutions
The pH or pH is a measure for determining the acid-base properties of a solution. Water in a limited amount dissociates into ions and is a weak electrolyte. When calculating the pH, a formula is used, which is the negative decimal logarithm of the accumulation of hydrogen ions in solutions:
pH = -lg [H +]
- For an alkaline environment, this figure will be more than seven. For example, [H +] = 10 -8 mol / l, then pH = -lg = 8, that is, pH ˃ 7.
- For an acidic environment, on the other hand, the pH should be less than seven. For example, [H +] = 10 -4 mol / l, then pH = -lg = 4, that is, pH ˂ 7.
- For a neutral medium, pH = 7.
Very often, to determine pH solutions, an express method is used for indicators that, depending on pH, change their color. For a more accurate determination, use ionomers and pH meters.
Quantitative characteristics of hydrolysis
Salt hydrolysis, like any other chemical process, has a number of characteristics, in accordance with which the process becomes possible. The most significant quantitative characteristics are the constant and the degree of hydrolysis. Let's dwell on each of them in more detail.
Hydrolysis degree
To find out which salts undergo hydrolysis and in what quantity, a quantitative indicator is used - the degree of hydrolysis, which characterizes the completeness of the course of hydrolysis. The degree of hydrolysis is called the part of the substance from the total number of molecules capable of hydrolysis, it is written as a percentage:
h = n / N ∙ 100%,
where the degree of hydrolysis is h;
the number of salt particles subjected to hydrolysis - n;
the total sum of the salt molecules participating in the reaction is N.
Factors affecting the degree of hydrolysis include:
- constant hydrolysis;
- the temperature, with an increase in which the degree increases due to the enhancement of the interaction of ions;
- salt concentration in solution.
Hydrolysis constant
It is the second most important quantitative characteristic. In general, the salt hydrolysis equations can be written as:
MA + NON ↔ MON + NA
Hence it follows that the equilibrium constant and the concentration of water in the same solution are constant values. Accordingly, the product of these two indicators will also be a constant value, which means the hydrolysis constant. In general, Kg can be written as:
Kg = ([HA] ∙ [MON]) / [MA],
where HA is an acid,
MON is the foundation.
In a physical sense, the hydrolysis constant describes the ability of a particular salt to undergo a hydrolysis process. This parameter depends on the nature of the substance and its concentration.
