The guy does not want to work full time, he always has excuses, they say, I will feel bad again and all that. He is studying for a point in the magistracy, once a week he is in pairs, he agreed with the teachers. I can’t officially get a job, so as not to miss my studies (we have points for attending) .. plus this is now not considered a good reason, there is no free correspondence in the specialty. Now there is no money, I ask him to stay at work in order to earn at least something. The company where he works does not take me yet. In response, he gave me 25-1000 excuses, then a university, then a job, then all of a sudden I feel bad like in winter, when I was lying with a layer of pressure. He always asks his mother for money for his trips, but from mine he shakes for rent. My parents are not yet able to give money, because. before that, the sisters needed money for competitions, and my mother and sister had heart problems and needed treatment and medicines, my brother doesn’t talk, my mother gave him about 8 thousand for injections and medicines (injections + vitamins). I don't think he cares about my parents. And in general, his mother allegedly “agreed” with my mother that they would give 3 thousand per month, but my mother said if possible. Before that, dad gave calmly, until the problems started. And his mother called out with a run-in to my mother saying that you don’t give money, we “supposedly” agreed, then she began to say they say prepare 10k (where did I get such an amount). In my family, only dad works, mom is employed in the tape, but they don’t call for work. In the city, the store does not fulfill half of the plan for sales in the city. In his family, they work in black, that his mother, that his stepfather. My parents are in white. There are 4 people in his family, including him, in mine there are 6 with me .. Today I asked about a part-time job, but there 600 rubles a day to work from 9-20:00 .. Xs when they call. Dad is on duty, we can’t collect documents for a social scholarship either ..
ViewDating, Love, Relationships I understand there is a young Johnny Depp (well, or whoever you like there most), handsome, gallant, with a velvety baritone. A man in a bed that women themselves dream of getting into. Then I understand my confidence. And then such a man will not offer anything directly, he will kindle passion and everything will happen smoothly and naturally. And how many cases are there when some Vasek sits and asks: why did you come to me?))) Does he consider himself irresistible like that? That any woman seeing him for the first time already dreams of continuing the banquet?
For electrolytes, the degree of dissociation of which is less than 5% at a concentration of more than 10 mol / dm3 or Kd less than 1 * 10 -4, the Ostwald dilution law is expressed:
For a weak base:
Examples of solving problems for calculating the pH of solutions of weak acids and bases.
Example 8 Calculate the pH of 0.001 N acetic acid if = 0.13
Solution:
Example 9 Define:
A) pH 0.01n CH 3 COOH, if 
C) pH 0.01n NH 4 OH, if 
Solution: Representing the equation 
in logarithmic form we get 
,
so for a monobasic acid 
Consequently 
, for a weak base 
;
, Consequently
Example 10 The concentration of H + ions in the solution is 2. 10 -4 mol/dm. Calculate the concentration of OH - , pH and pOH in this solution.
Solution:
;
Example 11. Calculate the pH of the solution, in 500 ml of which 2 g of NaOH is dissolved.
Solution:
PH=14-pC bases;
; pH = 14 - 1 = 13
Solution.
In an aqueous solution of ammonia, there is an equilibrium
NH 3 + H 2 O Û NH 4 + + OH -
Because ammonia solution is a weak base and K B< 1*10 -4 , то расчёт ведут следующим образом:
RON = - lg [OH -] \u003d - lg 4.2 10-3 \u003d 3-0.623 = 2,38
pH \u003d 14- pOH \u003d 14 - 2.38 = 11,62
Example 13 The degree of dissociation of CH 3 COOH in 0.1 mol / dm 3 solution is 1.32 * 10 -3. Calculate the concentrations of H + and CH3COO - ions, pH of the solution and K d of the acid.
Solution.
Write the equation for the dissociation of acetic acid
CH 3 COOH + H 2 O Û H 3 O + + CH 3 COO -
CH 3 COOH is a weak acid, therefore
A * CH 3 COOH \u003d 1.32 * 10 -2 * 0.1 \u003d 1.32 * 10 -3 mol / dm 3
pH = - lg = - lg 1.32 10 -3 \u003d 3 - 0.12 \u003d 2.88 [H +] \u003d [CH 3 COO -] \u003d 1.32 * 10 -3 mol / dm 3
From the Ostwald dilution law, K CH3COOH is found:
K CH3COOH \u003d a 2 * C CH3COOH \u003d (1.32 * 10 -2) 2 0.1 \u003d 1.74 * 10 -5
Tasks:
Calculation of ion concentration and ionic strength in solutions of strong electrolytes
1. Assuming complete dissociation, calculate the ion concentrations:
A) K + in a 0.5M solution of K 2 SO 4, K 3 PO 4;
B) Al 3+ in a 2M solution of Al 2 (SO 4) 3, AlCl 3.
2. Calculate the ionic strength in solutions:
0.3 M barium chloride, 0.06 M potassium orthophosphate, 0.02 M aluminum sulfate.
Answer: (0.82; 2.45 10 mol / dm 3)
5. Calculate the activity of Na +, H +, SO 4 2- ions in a solution with a concentration of 2 10 mol / dm 3 sodium sulfate and 5 10 mol / dm 3 sulfuric acid.
Answer: (3.16 * 10 mol / dm 3; 7.9x10 mol / dm 3; 8.2 * 10 mol / dm 3)
6. After dissolving potassium chloride, magnesium sulfate and iron (III) sulfate in water, the molar concentration of these salts is respectively: 0.05; 0.02 and 0.01 mol / dm 3. Calculate the ionic strength of the solution.
Calculation of ion concentration, pH and RON in solutions of weak electrolytes:
7. Calculate the pH of a 0.01N solution of ammonium hydroxide, the degree of dissociation of which is 0.1.
8. The active acidity of gastric juice is 0.047. Find the pH of gastric juice.
9. Find the pH of lactic acid, the dissociation constant of which is 1.44. 10 -4 , C=0.01.
10. Calculate the pH of a nitric acid solution if the mass fraction of acid in the solution is 4% ( 
).
11. Calculate the concentration and number of hydrogen ions in the blood with a volume of 100 ml, if blood pH = 7.36.
Pure water is a very weak electrolyte. The process of water dissociation can be expressed by the equation: HOH ⇆ H + + OH - . Due to the dissociation of water, any aqueous solution contains both H + ions and OH - ions. The concentrations of these ions can be calculated using ionic product equations for water
C (H +) × C (OH -) \u003d K w,
where Kw is ionic product constant of water ; at 25°C K w = 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 +) > C(OH -) and, as follows from the equation of the ionic product of water, C(H +) > 10 -7 mol / l, and C (OH -)< 10 –7 моль/л.
In an alkaline solution C (OH -) > C (H +); while in C(OH –) > 10 –7 mol/l, and C(H +)< 10 –7 моль/л.
pH is a value that characterizes the acidity or alkalinity of aqueous solutions; this value is called pH indicator and is calculated by the formula:
pH \u003d -lg C (H +)
In an acidic pH solution<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 = –lg C(OH –)
Hydrogen and hydroxyl indicators are related by the ratio
The hydroxyl index is used to calculate the pH in alkaline solutions.
Sulfuric acid is a strong electrolyte that dissociates in dilute solutions irreversibly and completely according to the scheme: H 2 SO 4 ® 2 H + + SO 4 2–. It can be seen from the dissociation process equation 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 = 14 - pOH = 14 - 1 = 13.
The dissociation of a weak electrolyte is an equilibrium process. The equilibrium constant written for the process of dissociation of a weak electrolyte is called dissociation constant . For example, for the process of dissociation of acetic acid
CH 3 COOH ⇆ CH 3 COO - + H +.
Each stage of the dissociation of a polybasic acid is characterized by its dissociation constant. Dissociation constant - reference value; cm. .
The calculation of ion concentrations (and pH) in solutions of weak electrolytes is reduced to solving the problem of chemical equilibrium for the case when the equilibrium constant is known and it is necessary to find the equilibrium concentrations of the substances involved in the reaction (see example 6.2 - type 2 problem).
In a 0.35% solution of NH 4 OH, the molar concentration of ammonium hydroxide is 0.1 mol / l (an example of converting a percentage concentration into a molar one - see example 5.1). This value is often referred to as C 0 . C 0 is the total electrolyte concentration in the solution (electrolyte concentration before dissociation).
NH 4 OH is considered to be a weak electrolyte that reversibly dissociates in an aqueous solution: NH 4 OH ⇆ NH 4 + + OH – (see also note 2 on page 5). Dissociation constant K = 1.8 10 -5 (reference value). Since a weak electrolyte dissociates incompletely, we will assume that x mol / l NH 4 OH has dissociated, then the equilibrium concentration of ammonium ions and hydroxide ions will also be equal to x mol / l: C (NH 4 +) \u003d C (OH -) \u003d x mol/l. The equilibrium concentration of undissociated NH 4 OH is: C (NH 4 OH) \u003d (C 0 -x) \u003d (0.1-x) mol / l.
We substitute the equilibrium concentrations of all particles expressed in terms of x into the dissociation constant equation:
.
Very weak electrolytes dissociate slightly (x ® 0) and the x in the denominator as a term can be neglected:
.
Usually, in problems of general chemistry, x in the denominator is neglected if (in this case, x - the concentration of the dissociated electrolyte - differs by 10 or less times from C 0 - the total concentration of the electrolyte in the solution).
C (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 = 14 - pOH = 14 - 2.87 = 11.13.
Degree of dissociation electrolyte can be calculated as the ratio of the concentration of the dissociated electrolyte (x) to the total electrolyte concentration (C 0):
(1,34%).
First, you should convert the percentage concentration to molar (see example 5.1). In this case, C 0 (H 3 PO 4) = 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 = C(H +) 1 stage each + C(H +) 2 stages each + C(H +) 3 stages each. However, the dissociation of weak electrolytes occurs mainly in the first stage, and in the second and subsequent stages - to a small 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 dissociate in 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 undissociated H 3 PO 4 will be equal to (3.6–x) mol/l. We substitute the concentrations of H + and H 2 PO 4 - ions and H 3 PO 4 molecules expressed in terms of x into the expression for the dissociation constant for the first stage (K 1 = 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) the pH of solutions of strong acids and bases; b) a weak electrolyte solution and the degree of electrolyte dissociation in this solution (table 8). Take the density of solutions equal to 1 g/ml.
Table 8 - Conditions of task No. 8
| option no. | a | b | option no. | a | b |
| 0.01M H 2 SO 4; 1% NaOH | 0.35% NH4OH | ||||
| 0.01MCa(OH) 2 ; 2%HNO3 | 1% CH3COOH | 0.04M H 2 SO 4 ; 4% NaOH | 1% NH4OH | ||
| 0.5M HClO 4 ; 1% Ba(OH)2 | 0.98% H3PO4 | 0.7M HClO 4 ; 4%Ba(OH)2 | 3% H3PO4 | ||
| 0.02M LiOH; 0.3% HNO3 | 0.34% H2S | 0.06M LiOH; 0.1% HNO3 | 1.36% H2S | ||
| 0.1M HMnO 4 ; 0.1% KOH | 0.031% H2CO3 | 0.2M HMnO 4 ; 0.2%KOH | 0.124% H 2 CO 3 | ||
| 0.4M HCl; 0.08%Ca(OH)2 | 0.47% HNO2 | 0.8 MHCl; 0.03%Ca(OH)2 | 1.4% HNO2 | ||
| 0.05M NaOH; 0.81% HBr | 0.4% H2SO3 | 0.07M NaOH; 3.24% HBr | 1.23% H2SO3 | ||
| 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% NH4OH | 0.06MH 2 SO 4; 0.8%NaOH | 5%CH3COOH | ||
| 0.7M HClO 4 ; 2%Ba(OH)2 | 1.96% H3PO4 | 0.08M H 2 SO 4 ; 3% NaOH | 4% H3PO4 | ||
| 0.04MLiOH; 0.63% HNO 3 | 0.68% H2S | 0.008MHI; 1.7%Ba(OH)2 | 3.4% H2S | ||
| 0.3MHMnO 4 ; 0.56%KOH | 0.062% H2CO3 | 0.08M LiOH; 1.3% HNO3 | 0.2% H2CO3 | ||
| 0.6M HCl; 0.05%Ca(OH)2 | 0.94% HNO2 | 0.01M HMnO 4 ; 1% KOH | 2.35% HNO2 | ||
| 0.03M NaOH; 1.62% HBr | 0.82% H2SO3 | 0.9MHCl; 0.01%Ca(OH)2 | 2% H2SO3 | ||
| 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%CH3COOH | 0.1M Ba(OH) 2 ; 6.4% HI | 6%CH3COOH | ||
| 0.002MHI; 3% Ba(OH)2 | 1%HF | 0.04MH 2 SO 4; 1.6%NaOH | 3.5% NH4OH | ||
| 0.005 MHBr; 0.24% LiOH | 1.64% H2SO3 | 0.001M HI; 0.4%Ba(OH)2 | 5% H3PO4 |
Example 7.5 200 ml of 0.2M H 2 SO 4 solution and 300 ml of 0.1M NaOH solution were mixed. Calculate the pH of the resulting solution and the concentrations of Na + and SO 4 2– ions in this solution.
Let's bring the reaction equation H 2 SO 4 + 2 NaOH → Na 2 SO 4 + 2 H 2 O to an abbreviated ion-molecular form: H + + OH - → H 2 O
It follows from the ion-molecular reaction equation that only H + and OH - ions enter the reaction and form a water molecule. 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 - - in short supply; they react completely. Together with them, the same amount (i.e. 0.03 mol) of H + ions will react.
Calculation of the number 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 +) = 0.03 mol; n(SO 4 2–) = 0.02 mol.
Because dilute solutions are mixed
V common. "Vsolution of H 2 SO 4 + V solution of NaOH" 200 ml + 300 ml \u003d 500 ml \u003d 0.5 l.
C(Na +) = n(Na +) / Vtot. \u003d 0.03 mol: 0.5 l \u003d 0.06 mol / l;
C(SO 4 2-) = n(SO 4 2-) / Vtot. \u003d 0.02 mol: 0.5 l \u003d 0.04 mol / l;
C(H +) = n(H +) / Vtot. \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 the strong acid solution with the 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 0.1M NaOH and 200 ml 0.2M 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 0.05 M Ba(OH) 2 and 200 ml 0.8 M HCl | ||
| 80 ml 0.15 M KOH and 20 ml 0.2 M H 2 SO 4 | 400ml 0.05M NaOH and 600ml 0.02M H 2 SO 4 | ||
| 100 ml 0.1 M Ba(OH) 2 and 20 ml 0.5 M 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 0.2 M Ba(OH) 2 and 150 ml 0.1 M 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 | 600 ml 0.01 M Ca(OH) 2 and 150 ml 0.12 M HCl | ||
| 250 ml 0.1M NaOH and 150 ml 0.1M H 2 SO 4 | 100 ml 0.2 M Ba(OH) 2 and 50 ml 1 M HCl | ||
| 1 l 0.05 M Ca (OH) 2 and 500 ml 0.1 M HNO 3 | 100 ml 0.5M NaOH and 100 ml 0.4M H 2 SO 4 | ||
| 100 ml 1M NaOH and 1900 ml 0.1M H 2 SO 4 | 25 ml 0.1M KOH and 75 ml 0.01M H 2 SO 4 | ||
| 300 ml 0.1 M Ba(OH) 2 and 200 ml 0.2 M HCl | 100 ml 0.02 M Ba(OH) 2 and 150 ml 0.04 M 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 | 250 ml 0.04 M Ba(OH) 2 and 500 ml 0.1 M HCl | ||
| 1 l 0.1M KOH and 2 l 0.05M H 2 SO 4 | 500 ml 1M NaOH and 1500 ml 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 0.25 M Ba(OH) 2 and 200 ml 0.3 M HCl | 1 l 0.03M Ca (OH) 2 and 500 ml 0.1M HNO 3 |
HYDROLYSIS OF SALT
When any salt is dissolved in water, this salt dissociates into cations and anions. If the salt is formed by a strong base cation and a weak acid anion (for example, potassium nitrite KNO 2), then nitrite ions will bind to H + ions, splitting them off from water molecules, resulting in the formation of 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 hydrolyzed by the anion (the reaction of the medium is alkaline; pH > 7).
If the salt is formed by a weak base cation and a strong acid anion (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 the 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), the cations of the weak base NH 4 + bind to OH - ions, splitting them off from water molecules, and the weak acid anions 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 a medium in a salt solution that is hydrolyzed by both the cation and the anion is determined by which of the weakly dissociating electrolytes formed as a result of hydrolysis is stronger (this can be found by comparing the dissociation constants). In the case of hydrolysis of NH 4 F, the environment 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, hydrolysis (i.e., decomposition by water) undergoes salts formed:
- a cation of a strong base and an anion of a weak acid (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 and/or anions of weak acids interact with water molecules; salts formed by cations of strong bases and anions of strong acids do not undergo hydrolysis.
The hydrolysis of salts formed by multiply charged cations and anions proceeds in steps; Below, specific examples show the sequence of reasoning that is recommended to follow when compiling the equations for the 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 acid 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 reversible 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 in molecular and ion-molecular form the equations of reactions of hydrolysis of sodium carbonate. Specify the pH of the solution (pH>7, pH<7 или pH=7).
1. Salt dissociation equation: Na 2 CO 3 ® 2Na + + CO 3 2–
2. Salt is formed by cations (Na +) of the strong base NaOH and anion (CO 3 2–) of a weak acid H2CO3. Therefore, the salt is hydrolyzed at the anion:
CO 3 2– + HOH ⇆ ... .
Hydrolysis in most cases proceeds reversibly (sign ⇄); for 1 ion participating in the hydrolysis process, 1 HOH molecule is recorded .
3. Negatively charged carbonate CO 3 2– ions bind to positively charged H + ions, splitting them off from HOH molecules, and form hydrocarbonate HCO 3 – ions; the solution is enriched with OH ions - (alkaline medium; pH> 7):
CO 3 2– + HOH ⇆ HCO 3 – + OH – .
This is the ion-molecular equation of the first stage of Na 2 CO 3 hydrolysis.
4. The equation of the first stage of hydrolysis in molecular form can be obtained by combining all CO 3 2– + HOH ⇆ HCO 3 – + OH – anions (CO 3 2–, HCO 3 – and OH –) present in the equation with Na + cations, forming salts Na 2 CO 3 , NaHCO 3 and 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 HCO 3 - ions bind to positively charged H + ions, splitting them off 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 -) present in the equation with Na + cations, forming a NaHCO 3 salt and a 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 in molecular and ion-molecular form the equations for the reactions of hydrolysis of aluminum sulfate. Specify the pH of the solution (pH>7, pH<7 или pH=7).
1. Salt dissociation equation: 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 recorded per 1 Al 3+ ion: Al 3+ + HOH ⇆ … .
3. Positively charged Al 3+ ions bind to negatively charged OH - ions, splitting them off from HOH molecules, and form hydroxoaluminum ions AlOH 2+; the solution is enriched with H + ions (acidic; pH<7):
Al 3+ + HOH ⇆ AlOH 2+ + H + .
This is the ion-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 Al 3+ + HOH ⇆ AlOH 2+ + H + cations (Al 3+ , AlOH 2+ and H +) present in the equation with SO 4 2– anions, forming salts of 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 in the first stage, hydroxoaluminum 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, splitting them off from HOH molecules).
6. The equation of the second stage of hydrolysis in molecular form can be obtained by linking all AlOH 2+ + HOH ⇆ Al(OH) 2 + + H + cations (AlOH 2+ , Al(OH) 2 + , and H +) present in the equation with anions SO 4 2–, 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, splitting them off from HOH molecules).
8. The equation of the third stage of hydrolysis in molecular form can be obtained by linking the Al(OH) 2 + + HOH ⇆ Al(OH) 3 + H + cations (Al(OH) 2 + and H +) present in the equation with SO 4 anions 2–, forming a salt (Al (OH) 2) 2 SO 4 and 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 in molecular and ion-molecular form the equations of reactions of hydrolysis of ammonium orthophosphate. Specify the pH of the solution (pH>7, pH<7 или pH=7).
1. Salt dissociation equation: (NH 4) 3 PO 4 ® 3NH 4 + + PO 4 3–
2. Salt is formed cations (NH 4 +) of a weak base NH4OH and anions
(PO 4 3–) weak acid H3PO4. Consequently, salt hydrolyzes both cation and anion : NH 4 + + PO 4 3– +HOH ⇆ … ; ( per 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, splitting them off from HOH molecules, forming a weak base NH 4 OH, and negatively charged PO 4 3– ions bind to H + ions, forming hydrogen phosphate ions HPO 4 2–:
NH 4 + + PO 4 3– + HOH ⇆ NH 4 OH + HPO 4 2– .
This is the ion-molecular equation of the first stage of hydrolysis (NH 4) 3 PO 4 .
4. The equation of the first stage of hydrolysis in molecular form can be obtained by linking the anions (PO 4 3–, HPO 4 2–) present in the equation with cations NH 4 +, forming 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 to OH - ions, HPO 4 2– ions - to H + ions, splitting them off 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 linking the NH 4 + + HPO 4 2– + HOH ⇆ NH 4 OH + H 2 PO 4 – anions present in the equation (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, dihydrophosphate 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 to OH - ions, H 2 PO 4 - ions to H + ions, splitting them off 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 NH 4 + + H 2 PO 4 - + HOH ⇆ NH 4 OH + H 3 PO 4 anions present in the equation H 2 PO 4 - and NH 4 + cations 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 hydrolysis process proceeds predominantly in the first stage, so the reaction of the medium in the salt solution, which is hydrolyzed by both the cation and the anion, is determined by which of the weakly dissociating electrolytes formed in 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> 7), since the HPO 4 2– ion is a weaker electrolyte than NH 4 OH: KNH 4 OH = 1.8 10 –5 > KHPO 4 2– = K III H 3 PO 4 = 1.3 × 10 -12 (the dissociation of the HPO 4 2– ion is the dissociation of H 3 PO 4 in the third stage, therefore KHPO 4 2– \u003d K III H 3 PO 4).
Task number 10
Write down in molecular and ion-molecular form the equations for the reactions of hydrolysis of salts (table 10). Specify the pH of the solution (pH>7, pH<7 или pH=7).
Table 10 - Conditions of task No. 10
| option number | List of salts | option number | 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 |
Table 10 continued
| option number | List of salts | option number | 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. Handbook of analytical chemistry / Yu.Yu. Lurie. - M.: Chemistry, 1989. - 448 p.
2. Rabinovich, V.A. Brief 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: a 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: in 2 hours / Mogilev State University of Food; auth.-stat. V.A. Ogorodnikov. - Mogilev, 2002. - Part 1: General questions of chemistry. – 96 p.
Educational edition
GENERAL CHEMISTRY
Methodical instructions and control tasks
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1. Calculation of pH in solutions of strong acids and bases.
Calculation of pH in solutions of strong monobasic acids and bases is carried out according to the formulas:
pH \u003d - lg C to and pH \u003d 14 + lg C o
Where C to, C o is the molar concentration of an 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 - lg C O)
3. Calculation of pH in solutions of hydrolyzable salts
There are 3 cases of hydrolysis of salts:
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 = 7 + 1/2 pK to + 1/2 lg C s
b) salt hydrolysis by cation (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 = 7 - 1/2 pK o - 1/2 lg C s
c) hydrolysis of the salt by 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 calculation of pH is carried out according to the formula:
pH \u003d 7 + 1/2 pK to - 1/2 pK o
If the salt is formed by a weak polybasic acid or a weak multiprotonic base, then in the formulas (7-9) listed above for calculating pH, the values of pK k and pK o according to the last stage of dissociation are substituted
4. Calculation of pH in solutions of various mixtures of acids and bases
When acid and base are poured, 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 of:
a) a weak acid and its salt, 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 acid salts of different acidity, for example NaH 2 PO 4 + Na 2 HPO 4
d) a mixture of acid and medium salts, for example NaНCO 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.
Calculation of pH in buffer systems is carried out according to the formulas: pH = pK to - lg C to / C s and pH = 14 - pK o + lg C o / C s
Acid-base buffer solutions, 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 a parameter (pH, system potential, etc.) when the composition of the system changes.
Acid-base called buffer solution , which maintains a roughly 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 their conjugate bases. A strong acid, when added to a buffer solution, "turns" into a weak acid, and a strong base into a weak base. Formula for calculating the pH of a buffer solution: pH = pK about + lg C about /FROM With This equation Henderson-Hasselbach . It follows from this equation that the pH of a buffer solution depends on the ratio of the concentrations of a weak acid and its conjugate base. Since this ratio does not change when diluted, the pH of the solution remains constant. Dilution cannot be unlimited. With a very significant dilution, the pH of the solution will change, because, firstly, the concentrations of the components will become so small 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 when 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 its conjugate base, 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 a solution (without a change in volume) to the change in pH caused by this increase (p. 239, 7.79)
In a strongly acidic and strongly alkaline environment, the buffer capacity increases significantly. Solutions in which a sufficiently high concentration of a strong acid or strong base also have buffering properties.
Buffer capacity is maximum at pH=pKa. To maintain a certain pH value, a buffer solution should be used, in which the pKa value of the weak acid included in its composition is as close as possible to this pH. It makes sense to use a buffer solution to maintain the pH in the pKa + _ 1 range. This interval is called the working force of the buffer.
19. Basic concepts related to 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 the 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 exist independently in the environment where the complexation occurs.
A complex compound consists of inner and outer spheres. K3(Fe(CN)6)- K3-outer sphere, Fe-complexing agent, CN- ligand, complexing agent + ligand=inner sphere.
Dentality is the number of ligand donor centers 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-coordination number. (Ag (NH3) 2) + - coordination number 2, since ammonia is a monodentate ligand, and in (Ag (S2O3) 2) 3- - 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, while a polynuclear complex contains two or more. Polynuclear complex particles containing identical 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 complex compounds 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 reference data on stepwise instability constants are not available, the general instability constant of the complex ion is used:
The general instability constant is equal to the product of stepwise instability constants.
In analytical chemistry, instead of the instability constants, the stability constants of the complex ion have recently been used: 
The stability constant refers to the process of formation of a complex ion and is equal to the reciprocal of the instability constant: Kst = 1/Knest.
The stability constant characterizes the equilibrium of complex formation.
See page 313 for thermodynamic and concentration constants.
20. Influence of various factors on the process of complex formation and 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 pattern of changes in the stability of many metal complexes with various ligands can be explained with help. Theories of hard and soft acids and bases (HMCA): soft acids form more stable compounds with soft bases, and hard acids with hard ones (for example, Al3 +, B3 + (l. to-you) form complexes with O- and N-soda Ligands (l. bases), and Ag + or Hg2 + (m. to-you) with S-sod. Ligands (m. basic.) Complexes of metal cations with polydentate ligands are more stable than complexes with similar monodentate ligands.
2) ionic strength. With an increase in ionic strength and a decrease in the activity coefficients of ions, the stability of the complex decreases.
3) temperature. If, during the formation of the complex, delta H is greater than 0, then with increasing temperature, the stability of the complex increases; 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 involved in the formation of the complex occur. The effect 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 ligand concentration 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 bound into complexes
Molar fraction of complex particles
