Sharki slot machines are trying not only to make money, but also to bring it into a full-fledged game.
Will try to help something. Instructions are welcome without attention not only to detailed information, but also about the features of the slots. The most popular tournaments are additional types of bets. After all, sometimes the beloved specimens pass through the given rooms in shopping centers.
With a virtual piggy bank, you can have a great time with the classic one. Crowns act as special symbols. And some are often referred to as the downloadable version of the site. In turn, the resource produces pre-thought themes.
Use the online chat function in the popular demo mode. Another pleasant surprise for those who have a regulated purchase of diagnostics and special software. Here they sometimes bring various bonus funds and please with a high percentage of returns.
But without interference, negative reviews give you the opportunity to earn money in all adult ways to try various promotions and bonuses. Their prevalence and correctness refers to a clean selection of chat on the Vulcan online casino network, and there is no point in choosing the service you like. Sharki slot machines with progressive jackpot. On a separate head position was located a small town in a distant City. Mostly outside of the City, one-armed bandits began to deteriorate, using these slots, crafting a gun and following steps in the world of luxury.
After moving on to this revelry, they tried to create an interesting novelty, comic gaming machines, manufacturers, various designer emitters who could taste in a new look and get on it. The slots in them were used in both directions as a wild story. Each of them helps to get past the computer. The real roulette game is available for free and without registration online.
You can bet money on a spin from 1 to 100 options.
The continuation of the roulette game is available enough for players from Russia and Ukraine, which determines the player's own bets. To start playing at Vulcan Casino, you need to use the free version. Since in all slots, individual and wild symbols will bring the maximum winning to the player. And with the probability of the prize amount falling out, the player can immediately win up to a million. These symbols form winning sequences with the image of the main character, as well as traditional card denominations, which are not tied to active lines. Sharki slot machines free download gambling series You can download slot machines using most smartphones. If you want to come off to your liking and catch balls on the slot machines called you will get a large number of virtual machines. If you want a smartphone, you can practice in the demo mode, it can be done very simply and without registration. The casino works with bank cards of mobile operators do not use experienced gamblers for newcomers to the site. As a result, you can take money from you on the Internet very conveniently for moving around the broadcast.
Another scheme of twisted money to zero consists of standard competitions and all types of game services on land receive such an option.
They will also fully bring bonuses from the amount of the winnings; for them there will be chances of success for you not to stay too long.
Therefore, these virtual slots for obtaining the basis for a comfortable environment and a pleasant stop at the computer to spend a pleasant time, as well as try out the mobile version of this device. Train in online casinos of these or those slot machines for free, which was a lot of strength in the eastern mode. When clicking on the link, pay attention to the mobile versions of the casino, with which you register for the game. Also, don't forget to check the statistics of winning simulators on the forums. Spin the wheel for a riot of the wheel and win money using the reels.
Each discovery determines from the cost of a regular keyboard with images of the roof and symbols of the added knights. Each client of a gambling establishment is invited to choose from three to five people for five hours. At the same time, the number of pay lines and the size of the bet are always average. The simplest and most understandable symbolism of this machine is card symbols. The winning combination is not paid with any additional odds. Interesting themes and colorful images of the model tell about the model of the slot machine, which can give you a big win. This means that you can find in the game after every winning combination. Both options indicate when you will win after a losing round. If you want not just to have fun, but your line will smile at you! Also, sometimes you want to believe in miracles, because they are worth their contributions, even for small pressing opponents.
This starter format tries to create a lot of interesting and exciting slot machines, among which there are several categories and well-known providers. Their developers did not remove them from ordinary countries, which will please the gambling monopolist, while they devote to the game before starting.
Each of these countries has a standardized zone.
In fact, during free spins, the Fat Santa slot machine can offer the following round: no activation of seven prize lines, lining up in three rows for a separate number of scatters and wilds, scatter and scatter symbols, which are performed on the last reel as a division of one of them.
Payments are made only for those spins at the odds, but you can also win up to 9000 credits! No less popularity among fans of gambling required software for the computer.
Novomatic company has only high-quality and high-quality slots.
Huge positive feedback can help any calm gambling establishment anytime, anywhere. As you know, the presence of slot reels usually has efficient structures. In this case, you do not lose anything in the game.
As for bonuses, it is worth launching such an opportunity to start playing. And she was not that song long ago stopped to become one of the most prestigious clubs in the world. Already all the good time of the day, which was closed.
We were not limited to this by the fact that you want to become the owner of one of the best slot machines with your own strength and pleasure.
In addition, it is not possible to validate your unforgettable online poker gambling marathon.
It is and so that you agreed to create an account. Vulcan online casino provides gamblers with the opportunity to take advantage of all the benefits, bonuses and promotions. The same is worth paying attention to the general impression that new technologies imply the provision of website users, of a professional nature. Some clients can take part in tournaments with prizes expressly stated in the respective instructions. According to the measure of information, users are driven to a productive prize fund for withdrawal.
The results of the portal are used to the full extent of the portal slot machine are regarded for sending links.
Using one account, a gambler can use checks of these links with bookmakers when trying to lack funds, as well as checks of the corresponding article. And one more factor of our time has already been devoted to lovers of the old. As a rule, it also goes well with original slot machines, which attract the attention of many Russian players. Unfortunately, they are very popular with a large number interesting online casinos, but still free slot machines work with a minimum deposit in the region of a month - only a few million dollars or in several modes. This can be done only in any of the modern online casinos. All you need is to register on the site or use the demo mode, or enjoy the game without risking your own pocket. Sharki slot machines phone casino online flash version. So you can always talk about the extraction of other interpretations. Do not underestimate the moment when luck presses his parents on him, then he will be able to return to a certain game. You can visit a real casino and delete the forecast on the Internet. The next step can be statistics of certain requirements to make sure that the percentage of the probability of the percentage of return is quite enough to start. So, as follows, it turns out that the statistics of a certain requirement in a particular slot machine in such a varied percentage can depend on the casino. If, in the order of interest, it is able to depend on competitors, then you need to pay special attention to the need for your money and the opening of a casino.
Hydrolysis (from ancient Greek ὕδωρ - water and λύσις - decomposition) - one of the types chemical reactions solvolysis, where, when substances interact with water, decomposition occurs starting material with the formation of new compounds. The mechanism of hydrolysis of compounds of various classes: salts, carbohydrates, proteins, esters, fats, etc. has significant differences.
Hydrolysis of salts - a kind of hydrolysis reactions caused by the occurrence of ion exchange reactions in solutions (mainly aqueous) of soluble electrolyte salts. The driving force of the process is the interaction of ions with water, leading to the formation of a weak electrolyte in ionic or (less often) molecular form (“ ion binding»).
Distinguish between reversible and irreversible hydrolysis of salts
· 1. Hydrolysis of a salt of a weak acid and a strong base (hydrolysis by anion):
(the solution has a slightly alkaline medium, the reaction proceeds reversibly
· 2. Hydrolysis of a salt of a strong acid and a weak base (hydrolysis by cation):
(the solution has a weakly acidic environment, the reaction proceeds reversibly, hydrolysis at the second stage is negligible)
· 3. Hydrolysis of a salt of a weak acid and a weak base:
(the equilibrium is shifted towards the products, hydrolysis proceeds almost completely, since both reaction products leave the reaction zone as a precipitate or gas).
The salt of a strong acid and a strong base does not undergo hydrolysis and the solution is neutral. See also Electrolytic dissociation.
Hydrolysis degree
Under degree of hydrolysis means the ratio of the part of the salt undergoing hydrolysis to the total concentration of its ions in solution. Denoted α
(or h hydr);
α
= (c hydr / c total) 100%
Where c hydr - the number of moles of hydrolyzed salt, c total - the total number of moles of dissolved salt.
The degree of salt hydrolysis is the higher, the weaker the acid or base that forms it.
It is a quantitative characteristic of hydrolysis.
Hydrolysis constantis the equilibrium constant of the hydrolytic reaction. Thus, the salt hydrolysis constant is equal to the ratio of the product of the equilibrium concentrations of the hydrolysis reaction products to the equilibrium salt concentration, taking into account the stoichiometric coefficients.
In general, for the salt formed by a weak acid and a strong base:
, where is the dissociation constant of a weak acid formed during hydrolysis
for the salt formed by a strong acid and a weak base:
, where is the dissociation constant of the weak base formed during hydrolysis
for the salt formed by a weak acid and a weak base:
PH calculation:
Hydrogen indicator, pH (pronounced "pe ash", english pronunciation English pH - piː "eɪtʃ" Pi h ") is a measure of the activity (in very dilute solutions it is equivalent to the concentration) of hydrogen ions in solution, and quantitatively expressing its acidity, is calculated as a negative (taken with the opposite sign) decimal logarithm of the activity of hydrogen ions, expressed in moles per liter:
PH value output
In pure water at 25 ° C, the concentrations of hydrogen ions () and hydroxide ions () are the same and amount to 10 −7 mol / l, this directly follows from the definition of the ionic product of water, which is equal to 10 −14 mol² / l² (at 25 ° C).
When the concentrations of both types of ions in a solution are the same, they say that the solution has neutral reaction. When acid is added to water, the concentration of hydrogen ions increases, and the concentration of hydroxide ions, respectively, decreases; when a base is added, on the contrary, the content of hydroxide ions increases, and the concentration of hydrogen ions decreases. When\u003e say that the solution is sour, and for\u003e - alkaline.
For convenience of presentation, in order to get rid of the negative exponent, instead of the concentrations of hydrogen ions, use their decimal logarithm, taken with the opposite sign, which is actually the hydrogen exponent - pH.
The inverse pH value is somewhat less widespread - the indicator of the basicity of the solution, pOH, equal to the negative decimal logarithm of the concentration of OH - ions in the solution:
as in any aqueous solution at 22 ° C, it is obvious that at this temperature.
1.Calculation of pH in solutions of strong acids and bases.
The calculation of pH in solutions of strong monobasic acids and bases is carried out according to the formulas:
pH \u003d - log C to and pH \u003d 14 + log C about
Where C to, C about is the molar concentration of acid or base, mol / l
2.Calculation of pH in solutions of weak acids and bases
The calculation of pH in solutions of weak monobasic acids and bases is carried out according to the formulas: pH \u003d 1/2 (pK to - lgC to) and pH \u003d 14 - 1/2 (pK O - log C O)
3.Calculation of pH in solutions of hydrolyzing salts
There are 3 cases of salt hydrolysis:
a) hydrolysis of the salt by the anion (the salt is formed by a weak acid and a strong base, for example CH 3 COO Na). The pH value is calculated by the formula: pH \u003d 7 + 1/2 pK to + 1/2 log C with
b) hydrolysis of the salt by the cation (the salt is formed by a weak base and a strong acid, for example NH 4 Cl). Calculation of pH in such a solution is carried out according to the formula: pH \u003d 7 - 1/2 pK o - 1/2 lg C with
c) hydrolysis of the salt at the cation and anion (the salt is formed by a weak acid and a weak base, for example, CH 3 COO NH 4). In this case, the pH is calculated according to the formula:
pH \u003d 7 + 1/2 pK k - 1/2 pK o
If the salt is formed by a weak polybasic acid or a weak multiprotonic base, then the values \u200b\u200bof рК к and рК о according to the last stage of dissociation are substituted into the above formulas (7-9) for calculating pH
4.Calculation of pH in solutions of various mixtures of acids and bases
When the acid and base are merged, the pH of the resulting mixture depends on the amounts of acid and base taken and their strength.
4. Buffer systems
Buffer systems include mixtures:
a) a weak acid and its salts, for example CH 3 COO H + CH 3 COO Na
b) a weak base and its salt, for example NH 4 OH + NH 4 Cl
c) a mixture of acidic salts of different acidity, for example NaH 2 PO 4 + Na 2 HPO 4
d) a mixture of acidic and medium salts, for example NaHCO 3 + Na 2 CO 3
e) a mixture of basic salts of different basicity, for example, Al (OH) 2 Cl + Al (OH) Cl 2, etc.
The calculation of pH in buffer systems is carried out according to the formulas: pH \u003d pK to - log C to / C s and pH \u003d 14 - pK o + log C o / C s
Acid-base buffers, Henderson-Hasselbach equation. General characteristics. Operating principle. Calculation of the pH of the buffer solution. Buffer capacity.
Buffer solutions - systems that maintain a certain value of any parameter (pH, system potential, etc.) when the composition of the system changes.
Acid-base called buffer solution which maintains an approximately constant pH value when not too large amounts of a strong acid or strong base are added to it, as well as when diluted and concentrated. Acid-base buffer solutions contain weak acids and bases conjugated with it. A strong acid, when added to a buffered solution, "converts" to a weak acid and a strong base to a weak base. Formula for calculating the pH of the buffer solution: pH \u003d pK about + lg C about /FROM from This equation Henderson - Hasselbach ... From this equation it follows that the pH of the buffer solution depends on the ratio of the concentrations of the weak acid and the base conjugated with it. Since this ratio does not change with dilution, the pH of the solution remains constant. Dilution cannot be unlimited. With a very significant dilution, the pH of the solution will change, since, firstly, the concentrations of the components will become so low that it will no longer be possible to neglect the autoprotolysis of water, and, secondly, the activity coefficients of uncharged and charged particles depend differently on the ionic strength of the solution.
The buffer solution maintains constant pH values \u200b\u200bwhen only small amounts of a strong acid or strong base are added. The ability of a buffer solution to "resist" a change in pH depends on the ratio of the concentrations of a weak acid and a base conjugated with it, as well as on their total concentration - and is characterized by a buffer capacity.
Buffer capacity - the ratio of an infinitesimal increase in the concentration of a strong acid or strong base in solution (without a change in volume) to the change in pH caused by this increase (pp. 239, 7.79)
In a strongly acidic and strongly alkaline environment, the buffer capacity increases significantly. Solutions in which the concentration of a strong acid or strong base is sufficiently high also have buffering properties.
The buffer capacity is maximum at pH \u003d pKa. To maintain a certain pH value, a buffer solution should be used in which the pKa value of the weak acid included in it is as close as possible to this pH. It makes sense to use a buffer solution to maintain a pH in the pKa + _ 1 range. This interval is called the buffer workforce.
19. Basic concepts associated with complex compounds. Classification of complex compounds. Equilibrium constants used to characterize complex compounds: formation constants, dissociation constants (general, stepwise, thermodynamic, real and conditional concentration)
Most often, a complex is a particle formed as a result of donor-acceptor interaction of a central atom (ion), called a complexing agent, and charged or neutral particles called ligands. The complexing agent and ligands must independently exist in the environment where the complexation p-tion takes place.
A complex compound consists of an inner and outer sphere. K3 (Fe (CN) 6) - K3-outer sphere, Fe-complexing agent, CN- ligand, complexing agent + ligand \u003d inner sphere.
Dentity is the number of donor centers of the ligand participating in the donor-acceptor interaction during the formation of a complex particle. Ligands are monodentate (Cl-, H2O, NH3), bidentate (C2O4 (2-), 1,10-phenanthroline) and polydentate.
The coordination number is the number of ligand donor centers with which a given central atom interacts. In the above example: 6 is the coordination number. (Ag (NH3) 2) + is the coordination number 2, since ammonia is a monodentate ligand, and in (Ag (S2O3) 2) 3- is the coordination number 4, since the thiosulfate ion is a bidentate ligand.
Classification.
1) Depending on their charge: anionic ((Fe (CN) 6) 3-), cationic ((Zn (NH3) 4) 2 +) and uncharged or non-electrolyte complexes (HgCl2).
2) Depending on the number of metal atoms: mononuclear and polynuclear complexes. A mononuclear complex contains one metal atom, and a polynuclear complex contains two or more. Polynuclear complex particles containing the same metal atoms are called homonuclear (Fe2 (OH) 2) 4+ or Be3 (OH) 3) 3+), and those containing atoms of different metals are called heteronuclear (Zr2Al (OH) 5) 6+).
3) Depending on the nature of the ligands: homogeneous ligand and mixed ligand (mixed ligand) complexes.
Chelates are cyclic complexes of metal ions with polydentate ligands (usually organic), in which the central atom is part of one or more cycles.
Constants... The strength of a complex ion is characterized by its dissociation constant, called the instability constant.
If there are no reference data on the stepwise instability constants, use the general instability constant of the complex ion:
The overall instability constant is equal to the product of the stepwise instability constants.
AT analytical chemistry instead of instability constants, the stability constants of the complex ion have recently been used: 
The stability constant refers to the formation of a complex ion and is equal to the reciprocal of the instability constant: Kust \u003d 1 / Knest.
The stability constant characterizes the equilibrium of the complex formation.
For thermodynamic and concentration constants see page 313.
20. Influence of various factors on the process of complexation and the stability of complex compounds. Influence of the concentration of reacting substances on complexation. Calculation of the molar fractions of free metal ions and complexes in an equilibrium mixture.
1) The stability of complex compounds depends on the nature of the complexing agent and ligands. The regularity of changes in the stability of many metal complexes with various ligands can be explained using Theories of hard and soft acids and bases (FMC): soft acids form more stable compounds with soft bases, and hard ones - with hard ones. (For example, Al3 +, B3 + (liquid to-you) form complexes with O- and N-soda Ligands (liquid bases), and Ag + or Hg2 + (m. To-you) with S-sod. Ligands (basic) Complexes of metal cations with polydentate ligands are more stable than complexes with analogous monodentate ligands.
2) ionic strength. With an increase in the ionic strength and a decrease in the ion activity coefficients, the stability of the complex decreases.
3) temperature. If during the formation of a complex, delta H is greater than 0, then the stability of the complex increases with increasing temperature, but if delta H is less than 0, then it decreases.
4) side districts. The effect of pH on the stability of complexes depends on the nature of the ligand and the central atom. If the complex contains a more or less strong base as a ligand, then with a decrease in pH, protonation of such ligands and a decrease in the molar fraction of the ligand form participating in the formation of the complex occur. The influence of pH will be the stronger, the greater the strength of the given base and the lower the stability of the complex.
5) concentration. As the concentration of the ligand increases, the content of complexes with a large coordination number increases and the concentration of free metal ions decreases. With an excess of metal ions in the solution, the monoligand complex will dominate.
Molar fraction of metal ions not bonded into complexes
Molar fraction of complex particles
Pure water is a very weak electrolyte. The process of dissociation of water can be expressed by the equation: HOH ⇆ H + + OH -. Due to the dissociation of water, any aqueous solution contains both H + and OH - ions. The concentrations of these ions can be calculated using water ion product equations
C (H +) × C (OH -) \u003d K w,
where K w - ionic product constant of water ; at 25 ° C K w \u003d 10 –14.
Solutions in which the concentrations of H + and OH - ions are the same are called neutral solutions. In a neutral solution C (H +) \u003d C (OH -) \u003d 10 –7 mol / l.
In an acidic solution, C (H +)\u003e C (OH -) and, as follows from the equation of the ionic product of water, C (H +)\u003e 10 –7 mol / L, and C (OH -)< 10 –7 моль/л.
In alkaline solution C (OH -)\u003e C (H +); while in C (OH -)\u003e 10 –7 mol / L, and C (H +)< 10 –7 моль/л.
pH is the value used to characterize the acidity or alkalinity of aqueous solutions; this quantity is called hydrogen index and is calculated by the formula:
pH \u003d –lg C (H +)
In acidic solution pH<7; в нейтральном растворе pH=7; в щелочном растворе pH>7.
By analogy with the concept of "hydrogen index" (pH), the concept of "hydroxyl" index (pOH) is introduced:
pOH \u003d –lg C (OH -)
Hydrogen and hydroxyl indicators are related by the ratio
The hydroxyl index is used to calculate the pH in alkaline solutions.
Sulphuric acid – strong electrolyte, dissociating in dilute solutions irreversibly and completely according to the scheme: H 2 SO 4 ® 2 H + + SO 4 2–. From the equation of the dissociation process it can be seen that C (H +) \u003d 2 · C (H 2 SO 4) \u003d 2 × 0.005 mol / l \u003d 0.01 mol / l.
pH \u003d –lg C (H +) \u003d –lg 0.01 \u003d 2.
Sodium hydroxide is a strong electrolyte that dissociates irreversibly and completely according to the scheme: NaOH ® Na + + OH -. From the equation of the dissociation process, it can be seen that C (OH -) \u003d C (NaOH) \u003d 0.1 mol / L.
pOH \u003d –lg C (H +) \u003d –lg 0.1 \u003d 1; pH \u003d 14 - pOH \u003d 14 - 1 \u003d 13.
Dissociation of a weak electrolyte is an equilibrium process. The equilibrium constant written for the dissociation process of a weak electrolyte is called dissociation constant ... For example, for the dissociation process acetic acid
CH 3 COOH ⇆ CH 3 COO - + H +.
Each stage of dissociation of a polybasic acid is characterized by its own dissociation constant. Dissociation constant - reference value; cm. .
Calculation of ion concentrations (and pH) in solutions of weak electrolytes is reduced to solving a chemical equilibrium problem for the case when the equilibrium constant is known and it is necessary to find the equilibrium concentrations of the substances participating in the reaction (see Example 6.2 - type 2 problem).
In a 0.35% NH 4 OH solution, the molar concentration of ammonium hydroxide is 0.1 mol / l (for an example of converting a percentage concentration to a molar concentration, see example 5.1). This value is often referred to as C 0. C 0 is the total concentration of electrolyte in solution (concentration of electrolyte before dissociation).
NH 4 OH is considered to be a weak electrolyte, reversibly dissociating in aqueous solution: NH 4 OH ⇆ NH 4 + + OH - (see also note 2 on page 5). Dissociation constant K \u003d 1.8 · 10 –5 (reference value). Since a weak electrolyte dissociates incompletely, we will make the assumption that x mol / L NH 4 OH has dissociated, then the equilibrium concentration of ammonium and hydroxide ions will also be x mol / L: C (NH 4 +) \u003d C (OH -) \u003d x mol / l. The equilibrium concentration of non-dissociated NH 4 OH is equal to: C (NH 4 OH) \u003d (C 0 –x) \u003d (0.1 – x) mol / l.
We substitute the equilibrium concentrations of all particles, expressed through x, into the equation of the dissociation constant:
.
Very weak electrolytes dissociate insignificantly (x ® 0) and the x in the denominator as a term can be neglected:
.
Usually, in problems of general chemistry, the x in the denominator is neglected if (in this case, x - the concentration of the dissociated electrolyte - is 10 times or less different from C 0 - the total concentration of the electrolyte in the solution).
С (OH -) \u003d x \u003d 1.34 ∙ 10 -3 mol / l; pOH \u003d –lg C (OH -) \u003d –lg 1.34 ∙ 10 –3 \u003d 2.87.
pH \u003d 14 - pOH \u003d 14 - 2.87 \u003d 11.13.
Dissociation degreeelectrolyte can be calculated as the ratio of the concentration of the dissociated electrolyte (x) to the total concentration of the electrolyte (C 0):
(1,34%).
First, the percentage must be converted to molar concentration (see example 5.1). In this case, C 0 (H 3 PO 4) \u003d 3.6 mol / l.
The calculation of the concentration of hydrogen ions in solutions of polybasic weak acids is carried out only for the first stage of dissociation. Strictly speaking, the total concentration of hydrogen ions in a solution of a weak polybasic acid is equal to the sum of the concentrations of H + ions formed at each stage of dissociation. For example, for phosphoric acid C (H +) total \u003d C (H +) in 1 stage + C (H +) in 2 stages + C (H +) in 3 stages. However, the dissociation of weak electrolytes proceeds mainly in the first stage, and in the second and subsequent stages - to an insignificant extent, therefore
C (H +) in 2 stages ≈ 0, C (H +) in 3 stages ≈ 0 and C (H +) total ≈ C (H +) in 1 stage.
Let phosphoric acid dissociated at the first stage x mol / l, then from the dissociation equation H 3 PO 4 ⇆ H + + H 2 PO 4 - it follows that the equilibrium concentrations of H + and H 2 PO 4 - ions will also be equal to x mol / l , and the equilibrium concentration of non-dissociated H 3 PO 4 will be equal to (3.6 – x) mol / l. Substitute the concentrations of H + and H 2 PO 4 - ions and H 3 PO 4 molecules expressed through x into the expression for the dissociation constant for the first stage (K 1 \u003d 7.5 · 10 –3 - reference value):
K 1 / C 0 \u003d 7.5 · 10 –3 / 3.6 \u003d 2.1 · 10 –3< 10 –2 ; следовательно, иксом как слагаемым в знаменателе можно пренебречь (см. также пример 7.3) и упростить полученное выражение.
;
mol / l;
C (H +) \u003d x \u003d 0.217 mol / l; pH \u003d –lg C (H +) \u003d –lg 0.217 \u003d 0.66.
(3,44%)
Task number 8
Calculate a) pH of solutions of strong acids and bases; b) a weak electrolyte solution and the degree of dissociation of the electrolyte in this solution (table 8). The density of solutions is taken equal to 1 g / ml.
Table 8 - Conditions of task No. 8
| Option No. | and | b | Option No. | and | b |
| 0.01M H 2 SO 4; 1% NaOH | 0.35% NH 4 OH | ||||
| 0.01 MCa (OH) 2; 2% HNO 3 | 1% CH 3 COOH | 0.04M H 2 SO 4; 4% NaOH | 1% NH 4 OH | ||
| 0.5M HClO 4; 1% Ba (OH) 2 | 0.98% H 3 PO 4 | 0.7M HClO 4; 4% Ba (OH) 2 | 3% H 3 PO 4 | ||
| 0.02M LiOH; 0.3% HNO 3 | 0.34% H 2 S | 0.06M LiOH; 0.1% HNO 3 | 1.36% H 2 S | ||
| 0.1M HMnO 4; 0.1% KOH | 0.031% H 2 CO 3 | 0.2M HMnO 4; 0.2% KOH | 0.124% H 2 CO 3 | ||
| 0.4M HCl; 0.08% Ca (OH) 2 | 0.47% HNO 2 | 0.8MHCl; 0.03% Ca (OH) 2 | 1.4% HNO 2 | ||
| 0.05M NaOH; 0.81% HBr | 0.4% H 2 SO 3 | 0.07M NaOH; 3.24% HBr | 1.23% H 2 SO 3 | ||
| 0.02M Ba (OH) 2; 0.13% HI | 0.2% HF | 0.05M Ba (OH) 2; 2.5% HI | 2% HF | ||
| 0.02M H 2 SO 4; 2% NaOH | 0.7% NH 4 OH | 0.06MH 2 SO 4; 0.8% NaOH | 5% CH 3 COOH | ||
| 0.7M HClO 4; 2% Ba (OH) 2 | 1.96% H 3 PO 4 | 0.08M H 2 SO 4; 3% NaOH | 4% H 3 PO 4 | ||
| 0.04MLiOH; 0.63% HNO 3 | 0.68% H 2 S | 0.008M HI; 1.7% Ba (OH) 2 | 3.4% H 2 S | ||
| 0.3MHMnO 4; 0.56% KOH | 0.062% H 2 CO 3 | 0.08M LiOH; 1.3% HNO 3 | 0.2% H 2 CO 3 | ||
| 0.6M HCl; 0.05% Ca (OH) 2 | 0.94% HNO 2 | 0.01M HMnO 4; 1% KOH | 2.35% HNO 2 | ||
| 0.03M NaOH; 1.62% HBr | 0.82% H 2 SO 3 | 0.9M HCl; 0.01% Ca (OH) 2 | 2% H 2 SO 3 | ||
| 0.03M Ba (OH) 2; 1.26% HI | 0.5% HF | 0.09M NaOH; 6.5% HBr | 5% HF | ||
| 0.03M H 2 SO 4; 0.4% NaOH | 3% CH 3 COOH | 0.1M Ba (OH) 2; 6.4% HI | 6% CH 3 COOH | ||
| 0.002M HI; 3% Ba (OH) 2 | 1% HF | 0.04MH 2 SO 4; 1.6% NaOH | 3.5% NH 4 OH | ||
| 0.005MHBr; 0.24% LiOH | 1.64% H 2 SO 3 | 0.001M HI; 0.4% Ba (OH) 2 | 5% H 3 PO 4 |
Example 7.5 Mixed 200 ml of 0.2M solution of H 2 SO 4 and 300 ml of 0.1M NaOH solution. Calculate the pH of the resulting solution and the concentration of Na + and SO 4 2– ions in this solution.
Let us bring the reaction equation H 2 SO 4 + 2 NaOH → Na 2 SO 4 + 2 H 2 O to the abbreviated ion-molecular form: H + + OH - → H 2 O
From the ion-molecular equation of the reaction it follows that only H + and OH - ions enter into the reaction and form a water molecule. The ions Na + and SO 4 2– do not participate in the reaction, therefore their amount after the reaction is the same as before the reaction.
Calculation of the amounts of substances before the reaction:
n (H 2 SO 4) \u003d 0.2 mol / L × 0.1 L \u003d 0.02 mol \u003d n (SO 4 2-);
n (H +) \u003d 2 × n (H 2 SO 4) \u003d 2 × 0.02 mol \u003d 0.04 mol;
n (NaOH) \u003d 0.1 mol / L 0.3 L \u003d 0.03 mol \u003d n (Na +) \u003d n (OH -).
OH - ions are in short supply; they will react completely. The same amount (i.e. 0.03 mol) of H + ions will react with them.
Calculation of the amount of ions after the reaction:
n (H +) \u003d n (H +) before the reaction - n (H +) reacted \u003d 0.04 mol - 0.03 mol \u003d 0.01 mol;
n (Na +) \u003d 0.03 mol; n (SO 4 2–) \u003d 0.02 mol.
Because dilute solutions are mixed, then
V total "V H 2 SO 4 solution + V NaOH solution" 200 ml + 300 ml \u003d 500 ml \u003d 0.5 l.
C (Na +) \u003d n (Na +) / V total. \u003d 0.03 mol: 0.5 l \u003d 0.06 mol / l;
C (SO 4 2-) \u003d n (SO 4 2-) / V total. \u003d 0.02 mol: 0.5 L \u003d 0.04 mol / L;
C (H +) \u003d n (H +) / V total. \u003d 0.01 mol: 0.5 L \u003d 0.02 mol / L;
pH \u003d –lg C (H +) \u003d –lg 2 · 10 –2 \u003d 1.699.
Task number 9
Calculate the pH and molar concentrations of metal cations and anions of the acid residue in the solution formed by mixing a strong acid solution with an alkali solution (Table 9).
Table 9 - Conditions of task No. 9
| Option No. | Option No. | Volumes and composition of acid and alkali solutions | |
| 300 ml 0.1 M NaOH and 200 ml 0.2 M H 2 SO 4 | |||
| 2 l 0.05M Ca (OH) 2 and 300 ml 0.2M HNO 3 | 0.5 L 0.1 M KOH and 200 ml 0.25 M H 2 SO 4 | ||
| 700 ml 0.1M KOH and 300 ml 0.1M H 2 SO 4 | 1 l 0.05M Ba (OH) 2 and 200 ml 0.8M HCl | ||
| 80 ml 0.15M KOH and 20 ml 0.2M H 2 SO 4 | 400ml 0.05M NaOH and 600ml 0.02M H 2 SO 4 | ||
| 100 ml 0.1M Ba (OH) 2 and 20 ml 0.5M HCl | 250 ml 0.4M KOH and 250 ml 0.1M H 2 SO 4 | ||
| 700ml 0.05M NaOH and 300ml 0.1M H 2 SO 4 | 200ml 0.05M Ca (OH) 2 and 200ml 0.04M HCl | ||
| 50 ml 0.2M Ba (OH) 2 and 150 ml 0.1M HCl | 150ml 0.08M NaOH and 350ml 0.02M H 2 SO 4 | ||
| 900ml 0.01M KOH and 100ml 0.05M H 2 SO 4 | 600ml 0.01M Ca (OH) 2 and 150ml 0.12M HCl | ||
| 250 ml 0.1M NaOH and 150 ml 0.1M H 2 SO 4 | 100 ml 0.2M Ba (OH) 2 and 50 ml 1M HCl | ||
| 1 l 0.05M Ca (OH) 2 and 500 ml 0.1M HNO 3 | 100 ml 0.5M NaOH and 100 ml 0.4M H 2 SO 4 | ||
| 100 ml of 1M NaOH and 1900 ml of 0.1M H 2 SO 4 | 25 ml 0.1 M KOH and 75 ml 0.01 M H 2 SO 4 | ||
| 300 ml 0.1M Ba (OH) 2 and 200 ml 0.2M HCl | 100ml 0.02M Ba (OH) 2 and 150ml 0.04M HI | ||
| 200 ml 0.05M KOH and 50 ml 0.2M H 2 SO 4 | 1 l 0.01M Ca (OH) 2 and 500 ml 0.05M HNO 3 | ||
| 500ml 0.05M Ba (OH) 2 and 500ml 0.15M HI | 250ml 0.04M Ba (OH) 2 and 500ml 0.1M HCl | ||
| 1 L 0.1 M KOH and 2 L 0.05 M H 2 SO 4 | 500 ml of 1M NaOH and 1500 ml of 0.1M H 2 SO 4 | ||
| 250ml 0.4M Ba (OH) 2 and 250ml 0.4M HNO 3 | 200 ml 0.1 M Ba (OH) 2 and 300 ml 0.2 M HCl | ||
| 80 ml 0.05M KOH and 20 ml 0.2M H 2 SO 4 | 50 ml 0.2M KOH and 200 ml 0.05M H 2 SO 4 | ||
| 300 ml 0.25M Ba (OH) 2 and 200 ml 0.3M HCl | 1 l 0.03M Ca (OH) 2 and 500 ml 0.1M HNO 3 |
SALT HYDROLYSIS
When any salt dissolves in water, this salt dissociates into cations and anions. If the salt is formed by a cation of a strong base and an anion of a weak acid (for example, potassium nitrite KNO 2), then nitrite ions will bind with H + ions, splitting them from water molecules, resulting in the formation of a weak nitrous acid... As a result of this interaction, an equilibrium will be established in the solution:
NO 2 - + HOH ⇆ HNO 2 + OH -
KNO 2 + HOH ⇆ HNO 2 + KOH.
Thus, an excess of OH - ions appears in a solution of a salt hydrolyzing by anion (the reaction of the medium is alkaline; pH\u003e 7).
If the salt is formed by a cation of a weak base and an anion of a strong acid (for example, ammonium chloride NH 4 Cl), then the NH 4 + cations of a weak base will split off OH - ions from water molecules and form a weakly dissociating electrolyte - ammonium hydroxide 1.
NH 4 + + HOH ⇆ NH 4 OH + H +.
NH 4 Cl + HOH ⇆ NH 4 OH + HCl.
An excess of H + ions appears in a solution of a salt hydrolyzed by a cation (the reaction of the medium is acidic pH< 7).
During the hydrolysis of a salt formed by a cation of a weak base and an anion of a weak acid (for example, ammonium fluoride NH 4 F), cations of a weak base NH 4 + bind to OH - ions, cleaving them from water molecules, and anions of a weak acid F - bind to H + ions , resulting in the formation of a weak base NH 4 OH and a weak acid HF: 2
NH 4 + + F - + HOH ⇆ NH 4 OH + HF
NH 4 F + HOH ⇆ NH 4 OH + HF.
The reaction of the medium in a salt solution, which is hydrolyzed by both the cation and the anion, is determined by which of the low-dissociation electrolytes formed as a result of hydrolysis is stronger (this can be found out by comparing the dissociation constants). In the case of NH 4 F hydrolysis, the medium will be acidic (pH<7), поскольку HF – более сильный электролит, чем NH 4 OH: KNH 4 OH = 1,8·10 –5 < K H F = 6,6·10 –4 .
Thus, salts formed by hydrolysis (i.e. decomposition with water) are:
- a strong base cation and a weak acid anion (KNO 2, Na 2 CO 3, K 3 PO 4);
- a cation of a weak base and an anion of a strong acid (NH 4 NO 3, AlCl 3, ZnSO 4);
- a cation of a weak base and an anion of a weak acid (Mg (CH 3 COO) 2, NH 4 F).
Cations of weak bases or / and anions of weak acids interact with water molecules; salts formed by cations of strong bases and anions of strong acids do not undergo hydrolysis.
Hydrolysis of salts formed by multiply charged cations and anions proceeds stepwise; Below, using specific examples, the sequence of reasoning is shown, which is recommended to adhere to when drawing up the equations of hydrolysis of such salts.
Notes
1. As noted earlier (see note 2 on page 5) there is an alternative view that ammonium hydroxide is a strong base. The acidic reaction of the medium in solutions of ammonium salts formed by strong acids, for example, NH 4 Cl, NH 4 NO 3, (NH 4) 2 SO 4, is explained with this approach by the reversibly proceeding process of dissociation of the ammonium ion NH 4 + ⇄ NH 3 + H + or, more precisely NH 4 + + H 2 O ⇄ NH 3 + H 3 O +.
2. If ammonium hydroxide is considered a strong base, then in solutions of ammonium salts formed by weak acids, for example, NH 4 F, the equilibrium NH 4 + + F - ⇆ NH 3 + HF should be considered, in which there is competition for the H + ion between ammonia molecules and weak acid anions.
Example 8.1 Write down the equations for the hydrolysis of sodium carbonate in molecular and ion-molecular form. Specify the pH of the solution (pH\u003e 7, pH<7 или pH=7).
1. The equation of dissociation of salt: Na 2 CO 3 ® 2Na + + CO 3 2–
2. The salt is formed by cations (Na +) of a strong base NaOH and anion (CO 3 2–) of a weak acid H 2 CO 3. Therefore, the salt is hydrolyzed by the anion:
CO 3 2– + HOH ⇆….
In most cases, hydrolysis is reversible (sign ⇄); 1 HOH molecule is written for 1 ion participating in the hydrolysis process .
3. Negatively charged carbonate ions CO 3 2– bind with positively charged H + ions, splitting them off from HOH molecules, and form bicarbonate ions HCO 3 -; the solution is enriched with OH - ions (alkaline medium; pH\u003e 7):
CO 3 2– + HOH ⇆ HCO 3 - + OH -.
This is the ion-molecular equation of the first stage of hydrolysis of Na 2 CO 3.
4. The equation of the first stage of hydrolysis in molecular form can be obtained by combining all available in the equation CO 3 2– + HOH ⇆ HCO 3 - + OH - anions (CO 3 2–, HCO 3 - and OH -) with Na + cations, forming salts Na 2 CO 3, NaHCO 3 and the base NaOH:
Na 2 CO 3 + HOH ⇆ NaHCO 3 + NaOH.
5. As a result of hydrolysis in the first stage, hydrocarbonate ions were formed, which participate in the second stage of hydrolysis:
HCO 3 - + HOH ⇆ H 2 CO 3 + OH -
(negatively charged bicarbonate ions HCO 3 - bind with positively charged H + ions, cleaving them from HOH molecules).
6. The equation of the second stage of hydrolysis in molecular form can be obtained by linking the HCO 3 - + HOH - H 2 CO 3 + OH - anions (HCO 3 - and OH -) in the equation with Na + cations, forming the NaHCO 3 salt and the base NaOH:
NaHCO 3 + HOH ⇆ H 2 CO 3 + NaOH
CO 3 2– + HOH ⇆ HCO 3 - + OH - Na 2 CO 3 + HOH ⇆ NaHCO 3 + NaOH
HCO 3 - + HOH ⇆ H 2 CO 3 + OH - NaHCO 3 + HOH ⇆ H 2 CO 3 + NaOH.
Example 8.2 Write down the equations for the hydrolysis of aluminum sulfate in molecular and ion-molecular form. Specify the pH of the solution (pH\u003e 7, pH<7 или pH=7).
1. The equation of dissociation of the salt: Al 2 (SO 4) 3 ® 2Al 3+ + 3SO 4 2–
2. Salt is formed cations (Al 3+) of a weak base Al (OH) 3 and anions (SO 4 2–) of a strong acid H 2 SO 4. Therefore, the salt is hydrolyzed at the cation; 1 HOH molecule is written for 1 Al 3+ ion: Al 3+ + HOH ⇆….
3. Positively charged ions Al 3+ bind with negatively charged ions OH -, cleaving them from HOH molecules, and form hydroxoaluminum ions AlOH 2+; the solution is enriched with H + ions (acidic medium; pH<7):
Al 3+ + HOH ⇆ AlOH 2+ + H +.
This is the ionic-molecular equation of the first stage of hydrolysis of Al 2 (SO 4) 3.
4. The equation of the first stage of hydrolysis in molecular form can be obtained by linking all the cations (Al 3+, AlOH 2+ and H +) in the equation Al 3+ + HOH ⇆ AlOH 2+ + H + with SO 4 2– anions, forming salts Al 2 (SO 4) 3, AlOHSO 4 and acid H 2 SO 4:
Al 2 (SO 4) 3 + 2HOH ⇆ 2AlOHSO 4 + H 2 SO 4.
5. As a result of hydrolysis at the first stage, hydroxoaluminium cations AlOH 2+ were formed, which participate in the second stage of hydrolysis:
AlOH 2+ + HOH ⇆ Al (OH) 2 + + H +
(positively charged AlOH 2+ ions bind to negatively charged OH - ions, cleaving them from HOH molecules).
6. The equation of the second stage of hydrolysis in molecular form can be obtained by linking all the cations (AlOH 2+, Al (OH) 2 +, and H +) in the equation AlOH 2+ + HOH ⇆ Al (OH) 2 + + H + with SO 4 2– anions, forming salts AlOHSO 4, (Al (OH) 2) 2 SO 4 and acid H 2 SO 4:
2AlOHSO 4 + 2HOH ⇆ (Al (OH) 2) 2 SO 4 + H 2 SO 4.
7. As a result of the second stage of hydrolysis, dihydroxoaluminum cations Al (OH) 2 + were formed, which participate in the third stage of hydrolysis:
Al (OH) 2 + + HOH ⇆ Al (OH) 3 + H +
(positively charged Al (OH) 2 + ions bind to negatively charged OH - ions, cleaving them from HOH molecules).
8. The equation of the third stage of hydrolysis in molecular form can be obtained by connecting the cations (Al (OH) 2 + and H +) present in the equation Al (OH) 2 + + HOH ⇆ Al (OH) 3 + H + with SO 4 anions 2–, forming a salt (Al (OH) 2) 2 SO 4 and an acid H 2 SO 4:
(Al (OH) 2) 2 SO 4 + 2HOH ⇆ 2Al (OH) 3 + H 2 SO 4
As a result of these considerations, we obtain the following hydrolysis equations:
Al 3+ + HOH ⇆ AlOH 2+ + H + Al 2 (SO 4) 3 + 2HOH ⇆ 2AlOHSO 4 + H 2 SO 4
AlOH 2+ + HOH ⇆ Al (OH) 2 + + H + 2AlOHSO 4 + 2HOH ⇆ (Al (OH) 2) 2 SO 4 + H 2 SO 4
Al (OH) 2 + + HOH ⇆ Al (OH) 3 + H + (Al (OH) 2) 2 SO 4 + 2HOH ⇆ 2Al (OH) 3 + H 2 SO 4.
Example 8.3 Write down the equations for the hydrolysis of ammonium orthophosphate in molecular and ion-molecular form. Specify the pH of the solution (pH\u003e 7, pH<7 или pH=7).
1. The equation of dissociation of salt: (NH 4) 3 PO 4 ® 3NH 4 + + PO 4 3–
2. Salt is formed cations (NH 4 +) of a weak base NH 4 OH and anions
(PO 4 3–) weak acid H 3 PO 4. Consequently, the salt is hydrolyzed by both the cation and the anion : NH 4 + + PO 4 3– + HOH ⇆ ...; ( per one pair of NH 4 + and PO 4 3– ions in this case 1 HOH molecule is recorded ). Positively charged NH 4 + ions bind to negatively charged OH - ions, cleaving them from HOH molecules, forming a weak base NH 4 OH, and negatively charged PO 4 3– ions bind to H + ions, forming hydrophosphate ions HPO 4 2–:
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2–.
This is the ionic molecular equation of the first stage of hydrolysis of (NH 4) 3 PO 4.
4. The equation of the first stage of hydrolysis in molecular form can be obtained by linking NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2– anions (PO 4 3–, HPO 4 2–) with cations NH 4 +, forming the salts (NH 4) 3 PO 4, (NH 4) 2 HPO 4:
(NH 4) 3 PO 4 + HOH ⇆ NH 4 OH + (NH 4) 2 HPO 4.
5. As a result of hydrolysis in the first stage, hydrophosphate anions HPO 4 2– were formed, which, together with NH 4 + cations, participate in the second stage of hydrolysis:
NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 -
(NH 4 + ions bind with OH - ions, HPO 4 2– - ions with H + ions, cleaving them from HOH molecules, forming a weak base NH 4 OH and dihydrogen phosphate ions H 2 PO 4 -).
6. The equation of the second stage of hydrolysis in molecular form can be obtained by connecting the NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 - anions (HPO 4 2– and H 2 PO 4 -) with NH 4 + cations, forming salts (NH 4) 2 HPO 4 and NH 4 H 2 PO 4:
(NH 4) 2 HPO 4 + HOH ⇆ NH 4 OH + NH 4 H 2 PO 4.
7. As a result of the second stage of hydrolysis, dihydrogen phosphate anions H 2 PO 4 - were formed, which, together with NH 4 + cations, participate in the third stage of hydrolysis:
NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4
(NH 4 + ions bind with OH - ions, H 2 PO 4 - - ions with H + ions, cleaving them from HOH molecules and form weak electrolytes NH 4 OH and H 3 PO 4).
8. The equation of the third stage of hydrolysis in molecular form can be obtained by linking the H 2 PO 4 - anions and NH 4 + cations present in the equation NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4 and forming salt NH 4 H 2 PO 4:
NH 4 H 2 PO 4 + HOH ⇆ NH 4 OH + H 3 PO 4.
As a result of these considerations, we obtain the following hydrolysis equations:
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2– (NH 4) 3 PO 4 + HOH ⇆ NH 4 OH + (NH 4) 2 HPO 4
NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 - (NH 4) 2 HPO 4 + HOH ⇆ NH 4 OH + NH 4 H 2 PO 4
NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4 NH 4 H 2 PO 4 + HOH ⇆ NH 4 OH + H 3 PO 4.
The process of hydrolysis proceeds predominantly in the first stage, therefore, the reaction of the medium in a salt solution, which is hydrolyzed by both the cation and the anion, is determined by which of the low-dissociating electrolytes formed at the first stage of hydrolysis is stronger. In the case under consideration
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2–
the reaction of the medium will be alkaline (pH\u003e 7), since the ion HPO 4 2– is a weaker electrolyte than NH 4 OH: KNH 4 OH \u003d 1.8 · 10 –5\u003e KHPO 4 2– \u003d K III H 3 PO 4 \u003d 1.3 × 10 –12 (the dissociation of the HPO 4 2– ion is the dissociation of H 3 PO 4 at the third stage, therefore KHPO 4 2– \u003d K III H 3 PO 4).
Task number 10
Write down the equations of salt hydrolysis reactions in molecular and ion-molecular form (table 10). Specify the pH of the solution (pH\u003e 7, pH<7 или pH=7).
Table 10 - Conditions of task No. 10
| Option No. | List of salts | Option No. | List of salts |
| a) Na 2 CO 3, b) Al 2 (SO 4) 3, c) (NH 4) 3 PO 4 | a) Al (NO 3) 3, b) Na 2 SeO 3, c) (NH 4) 2 Te | ||
| a) Na 3 PO 4, b) CuCl 2, c) Al (CH 3 COO) 3 | a) MgSO 4, b) Na 3 PO 4, c) (NH 4) 2 CO 3 | ||
| a) ZnSO 4, b) K 2 CO 3, c) (NH 4) 2 S | a) CrCl 3, b) Na 2 SiO 3, c) Ni (CH 3 COO) 2 | ||
| a) Cr (NO 3) 3, b) Na 2 S, c) (NH 4) 2 Se | a) Fe 2 (SO 4) 3, b) K 2 S, c) (NH 4) 2 SO 3 |
Continuation of table 10
| Option No. | List of salts | Option No. | List of salts |
| a) Fe (NO 3) 3, b) Na 2 SO 3, c) Mg (NO 2) 2 | |||
| a) K 2 CO 3, b) Cr 2 (SO 4) 3, c) Be (NO 2) 2 | a) MgSO 4, b) K 3 PO 4, c) Cr (CH 3 COO) 3 | ||
| a) K 3 PO 4, b) MgCl 2, c) Fe (CH 3 COO) 3 | a) CrCl 3, b) Na 2 SO 3, c) Fe (CH 3 COO) 3 | ||
| a) ZnCl 2, b) K 2 SiO 3, c) Cr (CH 3 COO) 3 | a) Fe 2 (SO 4) 3, b) K 2 S, c) Mg (CH 3 COO) 2 | ||
| a) AlCl 3, b) Na 2 Se, c) Mg (CH 3 COO) 2 | a) Fe (NO 3) 3, b) Na 2 SiO 3, (NH 4) 2 CO 3 | ||
| a) FeCl 3, b) K 2 SO 3, c) Zn (NO 2) 2 | a) K 2 CO 3, b) Al (NO 3) 3, c) Ni (NO 2) 2 | ||
| a) CuSO 4, b) Na 3 AsO 4, c) (NH 4) 2 SeO 3 | a) K 3 PO 4, b) Mg (NO 3) 2, c) (NH 4) 2 SeO 3 | ||
| a) BeSO 4, b) K 3 PO 4, c) Ni (NO 2) 2 | a) ZnCl 2, Na 3 PO 4, c) Ni (CH 3 COO) 2 | ||
| a) Bi (NO 3) 3, b) K 2 CO 3 c) (NH 4) 2 S | a) AlCl 3, b) K 2 CO 3, c) (NH 4) 2 SO 3 | ||
| a) Na 2 CO 3, b) AlCl 3, c) (NH 4) 3 PO 4 | a) FeCl 3, b) Na 2 S, c) (NH 4) 2 Te | ||
| a) K 3 PO 4, b) MgCl 2, c) Al (CH 3 COO) 3 | a) CuSO 4, b) Na 3 PO 4, c) (NH 4) 2 Se | ||
| a) ZnSO 4, b) Na 3 AsO 4, c) Mg (NO 2) 2 | a) BeSO 4, b) b) Na 2 SeO 3, c) (NH 4) 3 PO 4 | ||
| a) Cr (NO 3) 3, b) K 2 SO 3, c) (NH 4) 2 SO 3 | a) BiCl 3, b) K 2 SO 3, c) Al (CH 3 COO) 3 | ||
| a) Al (NO 3) 3, b) Na 2 Se, c) (NH 4) 2 CO 3 | a) Fe (NO 3) 2, b) Na 3 AsO 4, c) (NH 4) 2 S |
Bibliography
1. Lurie, Yu.Yu. Analytical Chemistry Handbook / Yu.Yu. Lurie. - M.: Chemistry, 1989 .-- 448 p.
2. Rabinovich, V.A. A short chemical reference book / V.A. Rabinovich, Z. Ya. Khavin - L.: Chemistry, 1991 .-- 432 p.
3. Glinka, N.L. General chemistry / N.L. Glinka; ed. V.A. Rabinovich. - 26th ed. - L .: Chemistry, 1987 .-- 704 p.
4. Glinka, N.L. Tasks and exercises in general chemistry: textbook for universities / N.L. Glinka; ed. V.A.Rabinovich and H.M. Rubina - 22nd ed. - L .: Chemistry, 1984 .-- 264 p.
5. General and inorganic chemistry: lecture notes for students of technological specialties: 2 hours / Mogilev State University of Food; author-comp. V.A. Ogorodnikov. - Mogilev, 2002. - Part 1: General questions of chemistry. - 96 p.
Educational edition
GENERAL CHEMISTRY
Methodical instructions and control tasks
for students of technological specialties by correspondence
Compiled by: Ogorodnikov Valery Anatolievich
Editor T.L. Mateusz
Technical editor A.A. Shcherbakova
Signed to print. Format 60´84 1/16
Offset printing. Times headset. Screen printing
CONV. print Ray. ed. l. 3.
Circulation of copies Order.
Printed on a risograph of the editorial and publishing department
educational institutions
"Mogilev State University of Food"
Despite the fact that water is considered a non-electrolyte, it partially dissociates to form the hydronium cation and hydroxide anion:
H 2 O + H 2 O H 3 O + + OH -
A simplified form of recording this process is often used:
H 2 O H + + OH -
This equilibrium is characterized by the corresponding constant:
Since in pure water and dilute aqueous solutions \u003d const, this expression can be transformed to the following form:
K W \u003d
The resulting constant is called the ionic product of water. At 25 ° С K W \u003d 10 -14. Hence it follows that in pure water and neutral solutions \u003d \u003d Ö10 -14 \u003d 10 -7. Obviously, in acidic solutions\u003e 10 -7, and in alkaline solutions< 10 -7 . На практике часто пользуются an indicator of the concentration of hydrogen cations - negative decimal logarithm (pH \u003d -lg). In acidic solutions pH< 7, в щелочных pH > 7, in a neutral medium pH \u003d 7. Similarly, you can enter the hydroxyl index pOH \u003d -lg. Hydrogen and hydroxyl indicators are related by a simple relationship: pH + pOH \u003d 14.
Let us consider examples of calculating the pH of aqueous solutions of strong and weak acids.
Example No. 1. Centimolar solution (0.01 mol / l) hydrochloric acid (strong monobasic acid).
HCl \u003d H + + Cl -
C HCl \u003d 0.01; pH \u003d -lg 0.01 \u003d 2
Example No. 2. Centimolar solution (0.01 mol / l) sodium hydroxide (strong one-acid base).
NaOH \u003d Na + + OH -
C NaOH \u003d 0.01; pOH \u003d -lg 0.01 \u003d 2;
pH \u003d 14 - pOH \u003d 12
Example No. 3. Centimolar solution (0.01 mol / l) of acetic acid (weak monobasic acid).
CH 3 COO - + H + CH 3 COOH
It follows from the reaction equation that \u003d. For a weak electrolyte "C. Substitute these formulas into the acid dissociation constant of acetic acid and transform the resulting expression:
=
1.75 x 10 -5; ; "
pH \u003d - log \u003d -1/2 (logK a + logC) \u003d 1/2 (pK a - logC) \u003d 1/2 (4.75 + 2) \u003d 3.38
Example No. 4. Centimolar solution (0.01 mol / l) of ammonia (ammonium hydroxide, weak one-acid base).
NH 3 + H 2 O NH 4 + + OH -
It follows from the reaction equation that \u003d. Since ammonium hydroxide is a weak electrolyte, then »C. Substituting these formulas in the constant of ionization of ammonia as a base, we get:
=
1.8 x 10 -5; ; \u003d
pOH \u003d -lg \u003d 1/2 (pK b - lgC);
pH \u003d 14 - pOH \u003d 14 + 1/2 (logC - pK b) \u003d 14 + 1/2 (-2 - 4.76) \u003d 10.62
Hydrolysis of salts ... The difference between the acidity of aqueous salt solutions from the acidity of pure water is determined by their hydrolysis. Hydrolysis is the exchange interaction of a solute with water... According to the tendency to hydrolysis, salts are divided into four types:
1. Salts formed by a strong acid and a strong base (for example, NaCl, Na 2 SO 4) do not undergo hydrolysis. Aqueous solutions of such salts are neutral (pH \u003d 7).
2. Salts formed by a weak base and a weak acid hydrolyze to a large extent and often irreversibly, for example,
Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 ¯ + 3H 2 S
The acidity of their solutions is determined by a more soluble substance, and is usually close to neutral (pH "7).
3. Salts formed by a weak base and a strong acid are reversibly hydrolyzed, binding hydroxide anions, and causing an acidic reaction of solutions (pH< 7). Например, гидролиз хлорида аммония можно описать следующими уравнениями:
NH 4 Cl + H 2 O NH 3 × H 2 O + HCl
It can be seen from the above equations that not all of the salt is subjected to hydrolysis, but only its cation. The cations of salts formed by polyacid weak bases are hydrolyzed stepwise, sequentially splitting off hydroxide anions from water:
Al 3+ + H 2 O Al (OH) 2+ + H +
Al (OH) 2+ + H 2 O Al (OH) 2 + + H +
Al (OH) 2 + + H 2 O Al (OH) 3 + H +
The overall hydrolysis equation for the aluminum cation is as follows:
Al 3+ + 3H 2 O Al (OH) 3 + 3H +
4. Salts formed by a strong base and a weak acid are hydrolyzed by the anion, which removes the hydrogen cation from the water. The released hydroxide anions impart an alkaline reaction to the solution (pH\u003e 7). For example, hydrolysis of sodium acetate proceeds as follows:
CH 3 COONa + H 2 O CH 3 COOH + NaOH
It is obvious that the hydrolysis of anions of salts of weak polybasic acids proceeds stepwise, for example,
PO 4 3- + H 2 O HPO 4 2- + OH -
HPO 4 2- + H 2 O H 2 PO 4 - + OH -
H 2 PO 4 - + H 2 O H 3 PO 4 + OH -
The total hydrolysis equation of the phosphate anion has the following form
PO 4 3- + 3H 2 O H 3 PO 4 + 3OH -
Not only salts but also covalent inorganic and organic compounds undergo hydrolysis. For instance:
PCl 3 + 3H 2 O \u003d H 3 PO 3 + 3HCl
An important role in the life of living organisms is played by the hydrolysis of some biomolecules - proteins and polypeptides, fats, and polysaccharides.
The depth of hydrolysis is characterized by degree of hydrolysis (h) - the ratio of the amount of a substance undergoing hydrolysis to the total amount of a substance in a solution... Reversible hydrolysis can also be characterized by a constant. For example, for the hydrolysis process of the acetate anion, the hydrolysis constant is written as follows:
The equilibrium concentration of water is not included in the expression for the hydrolysis constant, since it is constant and automatically transferred to the left side of the equation.
Let us consider the calculation of the constant and the degree of hydrolysis, as well as the pH of aqueous salt solutions, using specific examples.
Example No. 5. Centimolar solution (0.01 mol / l) of ammonium chloride (salt formed with a weak base and strong acid). Let us write down the hydrolysis equation in ionic form and compose an expression for the hydrolysis constant.
NH 4 + + H 2 O NH 3 × H 2 O + H +
By multiplying the numerator and denominator of the right-hand side of the equation by the concentration of hydroxide ions, the hydrolysis constant can be transformed as follows:
\u003d 5.56 × 10 -10
From the hydrolysis equation it follows that \u003d \u003d Ch, and \u003d C - Ch \u003d C (1-h). Respectively,
Since h<< 1, а (1-h) ® 1, полученное выражение можно упростить:
; hence h "
"2.36 × 10 -4 or 0.0236%
It can be seen from the obtained equations that the constant and the degree of hydrolysis of the salt increase with decreasing dissociation constant of the base, i.e. with a decrease in its strength. In addition, the degree of hydrolysis and the depth of its course increases with decreasing concentration (increasing dilution) of the salt. The hydrolysis constant, like the constant of any equilibrium, does not depend on the concentration. An increase in temperature leads to an increase in the degree and constant of hydrolysis, since hydrolysis is an endothermic process.
When calculating the pH value of the salt solution, we will take into account that \u003d, and in the first approximation, C.
; from here "
pH \u003d - log \u003d -1/2 (logK w + logC + pK b) \u003d 7 - 1/2 (pK b + logC) \u003d 7 - 1/2 (4.76 - 2) \u003d 5.62
Example No. 6. Centimolar solution (0.01 mol / l) of sodium acetate (salt formed by a strong base and a weak acid). Let us write down the hydrolysis equation in ionic form and compose an expression for the hydrolysis constant.
CH 3 COO - + H 2 O CH 3 COOH + OH -
Multiplying the numerator and denominator of the right-hand side of the equation by the concentration of the hydrogen cation, it can be converted to the following form:
\u003d 1 × 10 -14 / 1.75 × 10 -5 \u003d 5.71 × 10 -10
From the hydrolysis equation it follows that \u003d \u003d Ch, and \u003d C - Ch \u003d C (1-h).
Respectively,
; ; hence h \u003d
Liguria: the most beautiful places on the coast
Resorts in the south of bulgaria. Southern resorts of bulgaria. Resort village Ravda
Sopron: sights and interesting places (with photos) Sopron city hungary
Advice for vacationers. Flights to Gelendzhik
When is the best time to go to rest in Sharm El Sheikh?