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, the 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 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) 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. | but | b | option no. | but | 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 weak base cation and a weak acid anion (for example, ammonium fluoride NH 4 F), the weak base cations 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 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 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
for students of technological specialties of distance learning
Compiled by: Ogorodnikov Valery Anatolyevich
Editor T.L. Mateusz
Technical editor A.A. Shcherbakova
Signed for printing. Format 60´84 1/16
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"Mogilev State University of Food"
Water is a very weak electrolyte, dissociates to a small extent, forming hydrogen ions (H +) and hydroxide ions (OH -),
This process corresponds to the dissociation constant:
.
Since the degree of dissociation of water is very small, the equilibrium concentration of undissociated water molecules is equal to the total concentration of water with sufficient accuracy, i.e. 1000/18 = 5.5 mol / dm 3.
In dilute aqueous solutions, the concentration of water changes little and can be considered a constant value. Then the expression for the dissociation constant of water is transformed as follows:
.
The constant equal to the product of the concentration of H + and OH - ions is a constant value and is called ion product of water. In pure water at 25 ºС, the concentrations of hydrogen ions and hydroxide ions are equal and are
Solutions in which the concentrations of hydrogen ions and hydroxide ions are the same are called neutral solutions.
So, at 25 ºС
– neutral solution;
> - acidic solution;
< – щелочной раствор.
Instead of the concentrations of H + and OH ions it is more convenient to use their decimal logarithms, taken with the opposite sign; denoted by the symbols pH and pOH:
;
.
The decimal logarithm of the concentration of hydrogen ions, taken with the opposite sign, is called pH indicator(pH) .
Water ions in some cases can interact with the ions of the dissolved substance, which leads to a significant change in the composition of the solution and its pH.
table 2
Formulas for calculating the pH value (pH)
* Values of dissociation constants ( K) are listed in Appendix 3.
p K= -lg K;
HAN, acid; KtOH, base; KtAn - salt.
When calculating the pH of aqueous solutions, it is necessary:
1. Determine the nature of the substances that make up the solutions, and select a formula for calculating pH (table 2).
2. If a weak acid or base is present in the solution, look in the reference book or in Appendix 3 p K this connection.
3. Determine the composition and concentration of the solution ( FROM).
4. Substitute the numerical values of the molar concentration ( FROM) and p K
into the calculation formula and calculate the pH of the solution.
Table 2 shows the formulas for calculating pH in solutions of strong and weak acids and bases, buffer solutions and solutions of salts undergoing hydrolysis.
If only a strong acid (HAn) is present in the solution, which is a strong electrolyte and almost completely dissociates into ions 
, then the pH (pH)
will depend on the concentration of hydrogen ions (H +) in a given acid and is determined by formula (1).
If only a strong base is present in the solution, which is a strong electrolyte and almost completely dissociates into ions, then the pH (pH) will depend on the concentration of hydroxide ions (OH -) in the solution and is determined by formula (2).
If only a weak acid or only a weak base is present in the solution, then the pH of such solutions is determined by formulas (3), (4).
If a mixture of strong and weak acids is present in the solution, then the ionization of the weak acid is practically suppressed by the strong acid, so when calculating pH in such solutions, the presence of weak acids is neglected and the calculation formula used for strong acids, (1), is used. The same reasoning is also true for the case when a mixture of strong and weak bases is present in the solution. pH calculations lead according to the formula (2).
If a mixture of strong acids or strong bases is present in the solution, then pH calculations are carried out according to the formulas for calculating the pH for strong acids (1) or bases (2), having previously summed up the concentrations of the components.
If the solution contains a strong acid and its salt, or a strong base and its salt, then pH depends only on the concentration of a strong acid or strong base and is determined by formulas (1) or (2).
If a weak acid and its salt (for example, CH 3 COOH and CH 3 COONa; HCN and KCN) or a weak base and its salt (for example, NH 4 OH and NH 4 Cl) are present in the solution, then this mixture is buffer solution and pH is determined by formulas (5), (6).
If there is a salt in the solution formed by a strong acid and a weak base (hydrolyzed by a cation) or a weak acid and a strong base (hydrolyzed by an anion), a weak acid and a weak base (hydrolyzed by a cation and anion), then these salts, undergoing hydrolysis, change pH value, and the calculation is carried out according to formulas (7), (8), (9).
Example 1 Calculate the pH of an aqueous solution of NH 4 Br salt with concentration .
Solution. 1. In an aqueous solution, a salt formed by a weak base and a strong acid is hydrolyzed by the cation according to the equations:
In an aqueous solution, hydrogen ions (H +) remain in excess.
2. To calculate pH, we use the formula for calculating the pH value for a salt undergoing cation hydrolysis:
.
Dissociation constant of a weak base 
(R K = 4,74).
3. Substitute the numerical values into the formula and calculate the pH:
.
Example 2 Calculate the pH of an aqueous solution consisting of a mixture of sodium hydroxide, 
mol / dm 3 and potassium hydroxide, 
mol / dm 3.
Solution. 1. Sodium hydroxide (NaOH) and potassium hydroxide (KOH) are strong bases that almost completely dissociate in aqueous solutions into metal cations and hydroxide ions:
2. The pH will be determined by the amount of hydroxide ions. To do this, we summarize the concentrations of alkalis:
3. We substitute the calculated concentration into formula (2) to calculate the pH of strong bases:
Example 3 Calculate the pH of a buffer solution consisting of 0.10 M formic acid and 0.10 M sodium formate diluted 10 times.
Solution. 1. Formic acid HCOOH is a weak acid, in an aqueous solution it only partially dissociates into ions, in Appendix 3 we find formic acid 
:
2. Sodium formate HCOONa is a salt formed from a weak acid and a strong base; hydrolyzes by the anion, an excess of hydroxide ions appears in the solution:
3. To calculate pH, we use the formula for calculating the pH values of buffer solutions formed by a weak acid and its salt, according to formula (5)
Substitute the numerical values into the formula and get
4. The pH of buffer solutions does not change when diluted. If the solution is diluted 10 times, its pH will remain at 3.76.
Example 4 Calculate the pH value of a solution of acetic acid with a concentration of 0.01 M, the degree of dissociation of which is 4.2%.
Solution. Acetic acid is a weak electrolyte.
In a solution of a weak acid, the concentration of ions is less than the concentration of the acid itself and is defined as a∙C.
To calculate pH, we use formula (3):
Example 5 To 80 cm 3 0.1 n solution of CH 3 COOH was added 20 cm 3 0.2
n CH 3 COONa solution. Calculate the pH of the resulting solution if K(CH 3 COOH) \u003d 1.75 ∙ 10 -5.
Solution. 1. If the solution contains a weak acid (CH 3 COOH) and its salt (CH 3 COONa), then this is a buffer solution. We calculate the pH of the buffer solution of this composition according to the formula (5):
2. The volume of the solution obtained after draining the initial solutions is 80 + 20 = 100 cm 3, hence the concentrations of acid and salt will be equal:
3. We substitute the obtained values of the acid and salt concentrations
into the formula
.
Example 6 To 200 cm 3 0.1 N hydrochloric acid solution was added 200 cm 3 0.2 N potassium hydroxide solution, determine the pH of the resulting solution.
Solution. 1. A neutralization reaction occurs between hydrochloric acid (HCl) and potassium hydroxide (KOH), resulting in the formation of potassium chloride (KCl) and water:
HCl + KOH → KCl + H 2 O.
2. Determine the concentration of acid and base:
According to the reaction, HCl and KOH react as 1: 1, therefore, in such a solution, KOH remains in excess with a concentration of 0.10 - 0.05 = 0.05 mol / dm 3. Since the KCl salt does not undergo hydrolysis and does not change the pH of the water, the potassium hydroxide present in excess in this solution will affect the pH value. KOH is a strong electrolyte, we use formula (2) to calculate pH:
135. How many grams of potassium hydroxide are contained in 10 dm 3 of a solution whose pH is 11?
136. The hydrogen index (pH) of one solution is 2, and the other is 6. In 1 dm 3 of which solution is the concentration of hydrogen ions greater and how many times?
137. Indicate the reaction of the medium and find the concentration and ions in solutions for which the pH is: a) 1.6; b) 10.5.
138. Calculate the pH of solutions in which the concentration is (mol / dm 3): a) 2.0 ∙ 10 -7; b) 8.1∙10 -3; c) 2.7∙10 -10.
139. Calculate the pH of solutions in which the concentration of ions is (mol / dm 3): a) 4.6 ∙ 10 -4; b) 8.1∙10 -6; c) 9.3∙10 -9.
140. Calculate the molar concentration of a monobasic acid (NAn) in a solution if: a) pH = 4, α = 0.01; b) pH = 3, α = 1%; c) pH = 6,
α = 0.001.
141. Calculate the pH of a 0.01 N solution of acetic acid, in which the degree of dissociation of the acid is 0.042.
142. Calculate the pH of the following solutions of weak electrolytes:
a) 0.02 M NH 4 OH; b) 0.1 M HCN; c) 0.05 N HCOOH; d) 0.01 M CH 3 COOH.
143. What is the concentration of a solution of acetic acid, the pH of which is 5.2?
144. Determine the molar concentration of a solution of formic acid (HCOOH), whose pH is 3.2 ( K HCOOH = 1.76∙10 -4).
145. Find the degree of dissociation (%) and 0.1 M solution of CH 3 COOH, if the dissociation constant of acetic acid is 1.75∙10 -5.
146. Calculate the pH of 0.01 M and 0.05 N solutions of H 2 SO 4 .
147. Calculate the pH of a solution of H 2 SO 4 with a mass fraction of acid 0.5% ( ρ = 1.00 g/cm3).
148. Calculate the pH of a potassium hydroxide solution if 2 dm 3 of the solution contains 1.12 g of KOH.
149. Calculate and pH of 0.5 M ammonium hydroxide solution. \u003d 1.76 10 -5.
150. Calculate the pH of the solution obtained by mixing 500 cm 3 0.02 M CH 3 COOH with an equal volume of 0.2 M CH 3 COOK.
151. Determine the pH of the buffer mixture containing equal volumes of NH 4 OH and NH 4 Cl solutions with mass fractions of 5.0%.
152. Calculate the ratio in which sodium acetate and acetic acid should be mixed in order to obtain a buffer solution with pH = 5.
153. In what aqueous solution is the degree of dissociation the greatest: a) 0.1 M CH 3 COOH; b) 0.1 M HCOOH; c) 0.1 M HCN?
154. Derive a formula for calculating pH: a) acetate buffer mixture; b) ammonia buffer mixture.
155. Calculate the molar concentration of an HCOOH solution having pH = 3.
156. How will the pH change if it is diluted twice with water: a) 0.2 M HCl solution; b) 0.2 M solution of CH 3 COOH; c) a solution containing 0.1 M CH 3 COOH and 0.1 M CH 3 COOHa?
157*. A 0.1 N acetic acid solution was neutralized with a 0.1 N sodium hydroxide solution to 30% of its original concentration. Determine the pH of the resulting solution. 
158*. To 300 cm 3 0.2 M formic acid solution ( K\u003d 1.8 10 -4) added 50 cm 3 of 0.4 M NaOH solution. The pH was measured and then the solution was diluted 10 times. Calculate the pH of the dilute solution.
159*. To 500 cm 3 0.2 M solution of acetic acid ( K\u003d 1.8 ∙ 10 -5) added 100 cm 3 of 0.4 M NaOH solution. The pH was measured and then the solution was diluted 10 times. Calculate the pH of the dilute solution, write the chemical reaction equations.
160*. To maintain the required pH value, the chemist prepared a solution: to 200 cm 3 of a 0.4 M solution of formic acid, he added 10 cm 3 of a 0.2% KOH solution ( p\u003d 1 g / cm 3) and the resulting volume was diluted 10 times. What is the pH value of the solution? ( K HCOOH = 1.8∙10 -4).
Strong acids and bases(Table 2.1) half-
therefore, the concentration of hydrogen ions and hydroxyl ions is equal to
total concentration of a strong electrolyte.
For the strong grounds : [ OH - ] = C m;for the strong acids: [ H + ] = Cm.
Table 2.1
Strong electrolytes
Weak electrolyte It is customary to consider chemical compounds whose molecules, even in highly dilute solutions, do not completely dissociate into ions. The degree of dissociation of weak electrolytes for decimolar solutions (0.1 M) is less than 3%. Examples of weak electrolytes: all organic acids, some inorganic acids (eg H 2 S, HCN), most hydroxides (eg Zn(OH) 2 , Cu(OH) 2).
For solutions weak acids the concentration of hydrogen ions in solution is calculated by the formula:
where: Kc is the dissociation constant of a weak acid; Ck is the acid concentration, mol/dm 3 .
For solutions weak bases the concentration of hydroxyl ions is calculated by the formula:
where: Co is the dissociation constant of a weak base; Pine is the base concentration, mol/dm 3 .
Table 2.2
Dissociation constants of weak acids and bases at 25 °C
|
dissociation constant, cd |
|
2.2. Examples of solving an individual task
Example #1.
Job condition:Define concentration of hydrogen and hydroxide ions in solution, if pH = 5.5.
Decision
The concentration of hydrogen ions is calculated by the formula:
[H +] \u003d 10 -pH
[H +] \u003d 10 -5.5 \u003d 3.16 10 -6 mol / dm 3
The concentration of hydroxide ions is calculated by the formula:
10 -rOH
pOH \u003d 14 - pH \u003d 14 - 5.5 \u003d 8.5
10 -8.5 \u003d 3 10 -9 mol / dm 3
Example #2.
Job condition: Calculate the pH of a 0.001 M HCl solution.
Decision
Acid HC1 is a strong electrolyte (Table 2.1) and in dilute solutions almost completely dissociates into ions:
HC1⇄ H + + C1 -
Therefore, the concentration of ions [Н + ] is equal to the total concentration of the acid: [Н + ] \u003d Cm \u003d 0.001 M.
[H +] \u003d 0.001 \u003d 1 10 -3 mol / dm 3
pH \u003d - lg \u003d - lg 1 10 -3 \u003d 3
Example #3
Job condition: Calculate the pH of a 0.002 M NaOH solution.
Decision
The NaOH base is a strong electrolyte (Table 2.1) and in dilute solutions almost completely dissociates into ions:
NaOH ⇄Na + +OH -
Therefore, the concentration of hydroxide ions is equal to the total concentration of the base: [OH - ]= cm = 0.002 M.
pOH \u003d - lg [OH -] \u003d - lgSm \u003d - lg 2 10 -3 \u003d 2.7
pH = 14 - 2.7 = 11.3
Example number 4.
Job condition:Calculate the pH of a 0.04 M NH solution 4 Oh, if the dissociation constant Kd( NH 4 Oh) = 1.79 10 -5 (Table 2.2).
Decision
Founding NH 4 Oh is a weak electrolyte and in dilute solutions dissociates very slightly into ions.
The concentration of hydroxyl ions [OH - ] in a solution of a weak base is calculated by the formula:
pOH \u003d - lg [OH - ] \u003d - lg 8.5 10 -2 \u003d 1.1
Based on the formula: pH + pOH = 14, we find the pH of the solution:
pH = 14 - pOH = 14 - 1.1 = 12.9
Example number 5.
Job condition:Calculate pH 0.17 M solution of acetic acid (CH 3 COOH), if the dissociation constant Kd (CH 3 COOH) = 1.86 10 -5 (Table 2.2).
Decision
Acid CH 3 COOH is a weak electrolyte and in dilute solutions very slightly dissociates into ions.
The concentration of hydrogen ions in a weak acid solution is calculated by the formula:
Calculating pH solution according to the formula: pH = - lg
pH \u003d - lg 1.78 10 -3 \u003d 2.75
2.3. Individual tasks
Job conditions (Table 2.3):
Task number 1. Calculate the concentration of hydrogen and hydroxide ions in solution at a certain pH value (see example No. 1);
Task number 2. Calculate the pH of a strong electrolyte solution (acid, base) at a given concentration (see example No. 2, 3);
Task number 3. Calculate the pH of a weak electrolyte solution (acid, base) at a given concentration (see example No. 4, 5).
Table 2.3
The composition of the studied water
|
tasks |
Job conditions: |
||||
|
Task number 1 |
Task number 2 |
Task number 3 |
|||
|
Strong electrolyte |
Concentration, cm |
electrolyte |
Concentration, cm |
||
Continuation of the table. 2.3
