PHformula
pHformula (pashformula) is the first system of pharmaceutical-cosmeceutical products and procedures created as a result of the union of cosmeceuticals and medicine. This system allows you to cope with a number of skin conditions: acne, excessive pigmentation, rosacea, severe sensitivity and premature aging. At the same time, pHformula products not only solve existing problems, but also act as a preventive measure, preventing the situation from recurring in the future.
History
The laboratories in which pHformula was created were founded at the end of the 19th century in Barcelona. They are now run by the fourth generation of a family of pharmacists specializing in dermatology. The brand actively invests in research activities to scientifically substantiate and prove the effectiveness of its products, actively collaborating with the best medical institutions. All active ingredients in the formulas are pharma-cosmeceutical ingredients, and studies demonstrating their effectiveness have been published in the public domain.
Brand strengths
- pharma-cosmeceutical products
- clinical effectiveness of formulas in aesthetic cosmetology
- use of the most modern scientific developments
- system dermatologically tested
- simple system of prescribing and using home care products
- a unique opportunity to create multifunctional combinations of skin renewal procedures
- high efficiency of procedures
- pharmaceutical activity level of ingredients
- products do not contain lanolin and artificial colors
- pHformula are non-comedogenic products (do not clog pores)
- system of preservatives does not contain parabens
- unique transport complex PH-DVC ™ for delivery of active substances *
- reliable UV protection designed to preserve and restore skin cell DNA
* The unique PH-DVC ™ transport complex helps the active ingredients evenly penetrate into the deep layers of the skin, thereby increasing their bioavailability and lengthening their period of action. The use of the PH-DVC ™ complex allows you to use the maximum concentration of ingredients without the risk of negative reactions and complications typical for most traditional peels.
PHformula controlled skin renewal system. Professional care
The pHformula controlled skin renewal system consists of 3 sequential stages: preparation of the skin for renewal procedures, a course of professional renewing procedures, and post-cycle recovery. Home care products for skin preparation and repair have the most active formulations and their use is necessary to obtain optimal results and reduce the risk of complications.
PHformula treatments are personalized with selectable products to address a particular skin problem, but each treatment focuses on exfoliating (exfoliating) and actively stimulating cell regeneration and repair.
pHformula is the first line of products to use a combination of alpha keto, alpha hydroxy, alpha beta and poly hydroxy acids. This complex of acids is less traumatic than products based on a single acid in a high concentration.
In addition to acid combinations, all pHformula formulations contain components for skin regeneration: vitamins, antioxidants, trace elements, oxygen carriers, metabolizers. These substances help the skin recover faster after renewal procedures and reduce the likelihood of complications.
Phformula laboratory has developed a wide range of skin rejuvenation treatments that can correct various skin conditions such as acne, rosacea, signs of aging, hyperpigmentation. Also in the arsenal of possibilities of pHformula there is a procedure for the effect, similar to microdermabrasion and methods that combine the action of renewing products and mesoscooter therapy. In the spring-summer season, rejuvenating treatments for the skin of the hands, neck and décolleté and around the eyes can also be performed.
The pHformula specialist will select the procedure that suits you, taking into account the characteristics of your skin and the desired results during the consultation stage.
Indications for use of the pHformula system
1. Aging
- Photoaging (damage caused by UV rays)
- Uneven pigmentation
- Lentigo
- Telangiectasia
- Dull skin color
- Hyperkeratosis
- Uneven skin texture
- Superficial and moderate wrinkles
2. Hyperpigmentation
- Melasma
- Chloasma
- Photopigmentation
- Superficial hyperpigmentation (epidermal)
- Post-inflammatory hyperpigmentation
- Solar lentigo
- Freckles
3 degrees of acne:
- Grade 1: open and closed comedones, excess sebum production, enlarged pores
- Grade 2: open and closed comedones, single papules and pustules, minor inflammation
- Grade 3: inflamed papulopustular acne, the appearance of single nodular elements
Post-acne
4. Chronic redness (rosacea)
- Redness, sensitivity
- Telangiectasia
5. Home care
- Pharma-cosmeceutical products for skin renewal
PHformula's pre- and post-skincare recommendations have been specially formulated to accelerate recovery and get the best results without damaging the skin. PHformula home products supply the skin with all the essential active ingredients (vitamins, antioxidants, amino acids, etc.) that have been clinically proven to be useful in preparing skin for renewal and recovery procedures: active prep concentrates and revitalizing concentrates for problem solving aging, hyperpigmentation, acne and chronic redness of the skin, as well as additional products for all conditions and types of skin (cleansing, UV protection, creams for face, body, hands, toning agents).
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 an 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.
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–. 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 of 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 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 does not fully dissociate, 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 slightly (x ® 0) and the x in the denominator can be neglected as a term:
.
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 or less times different from C 0 - the total electrolyte concentration 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 through the first stage, and through 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 - 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 in terms of 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 | 0.8M HCl; 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 1M NaOH and 1900 ml 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 0.1M Ba (OH) 2 and 300 ml 0.2M 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 the nitrite ions will bind with H + ions, cleaving them from water molecules, as a result of which a weak nitrous acid is formed. 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 with OH - ions, cleaving them from water molecules, and anions of a weak acid F - bind with ions H + , 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 both by the cation and by 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:
- cation of a strong base and 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 / and anions of weak acids interact with water molecules; salts formed by strong base cations and strong acid anions 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 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 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 знак); 1 HOH molecule is recorded for 1 ion participating in the hydrolysis process .
3. Negatively charged carbonate ions CO 3 2– bind with positively charged H + ions, cleaving them 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 the anions present in the equation CO 3 2– + HOH ⇆ HCO 3 - + OH - (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, bicarbonate 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 connecting 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 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 cationically hydrolyzed; 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, dihydroxoaluminium 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 +) weak base NH 4 OH and anions
(PO 4 3–) weak acid H 3 PO 4. Hence, 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 linking 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 connecting 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 |
List of references
1. Lurie, Yu.Yu. Handbook of analytical chemistry / 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: at 2 pm / 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
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Tasks for the section Ionic product of water:
Problem 1. What is called the ionic product of water? What is it equal to? Give a derivation of the expression of the ionic product of water. How does temperature affect the ionic product of water?
Decision.
Water is a weak electrolyte, its molecules are slightly decomposed into ions:
H 2 O ↔ H + + OH -
Equilibrium constant the dissociation reaction of water is as follows:
K \u003d /
at 22 ° K \u003d 1.8 × 10 -16.
Neglecting the concentration of dissociated water molecules and taking the mass of 1 liter of water per 1000 g, we obtain:
1000/18 \u003d 55.56 g
K \u003d · / 55.56 \u003d 1.8 × 10 -16
· \u003d 1.8 × 10 -16 · 55.56 \u003d 1 · 10 -14
Determines the acidity of the solution, - determines the alkalinity of the solution.
In clean water \u003d \u003d 1 × 10 -7.
The work is called
K H 2 O \u003d · \u003d 1 · 10 -14
Ionic product of water increases with increasing temperature, since in this case the dissociation of water also increases.
The acidity of a solution is usually expressed through:
Lg \u003d pOH
pH< 7 in an acidic environment
pH\u003e 7 in an alkaline environment
pH \u003d 7 in a neutral environment.
The acidity of the medium can be determined using.
Problem 2. How many grams of sodium hydroxide is in a state of complete dissociation in 100 ml of a solution with a pH of 13?
Decision.
pH \u003d -lg
10 -13 M
Decision.
For determining pH solution must be converted into:
Suppose the density of the solution is 1, then V (solution) \u003d 1000 ml, m (solution) \u003d 1000 g.
Let's find how many grams of ammonium hydroxide is contained in 1000 g of solution:
100 g of solution contains 2 g of NH 4 OH
1000 g - x g NH 4 OH
M (NH 4 OH) \u003d 14 + 1 4 + 16 + 1 \u003d 35 g / mol
1 mol of solution contains 35 g of NH 4 OH
y mol - 20 g NH 4 OH
For weak grounds, which is NH 4 OH, the ratio
\u003d K H 2 O / (K d. Main C main) 1/2
According to the reference data, we find K d (NH 4 OH) \u003d 1.77 · 10 -5, then
10 -14 / (1.77 · 10 -5 · 0.57) 1/2 \u003d 3.12 · 10 -12
pH \u003d -lg \u003d - lg 3.12 10 -12 \u003d 11.5
Decision.
pH \u003d -lg
10 - pH
10 -12.5 \u003d 3.16 10 -13 M
pOH \u003d 14 -12.5 \u003d 1.5
pOH \u003d -lg
10 - pOH
10 -1.5 \u003d 3.16 10 -2 M
Problem 5. Find the pH of a concentrated solution of a strong electrolyte - 0.205 MHCl.
Decision.With significant concentration strong electrolyte, his active concentration is different from the true one. Correction for electrolyte activity should be made. We define ionic strength solution:
I \u003d 1 / 2ΣC i z i 2 where
C i and z i - respectively, the concentration and charges of individual ions
I \u003d ½ (0.205 · 1 2 + 0.205 · 1 2) \u003d 0.205
f H + \u003d 0.83, then
a H + \u003d f H + \u003d 0.205 0.83 \u003d 0.17
pH \u003d -lg [ a H +] \u003d -lg 0.17 \u003d 0.77
Categories ,Water is a very weak electrolyte, it dissociates to a small extent, forming hydrogen ions (H +) and hydroxide ions (OH -),
The dissociation constant corresponds to this process:
.
Since the degree of dissociation of water is very small, the equilibrium concentration of undissociated water molecules is, with sufficient accuracy, equal to the total concentration of water, that is, 1000/18 \u003d 5.5 mol / dm 3.
In dilute aqueous solutions, the water concentration changes little and can be considered constant. Then the expression for the dissociation constant of water is transformed as follows:
.
A constant equal to the product of the concentration of H + and OH - ions is a constant and is called ionic product of water... In pure water at 25 ºС, the concentrations of hydrogen ions and hydroxide ions are equal and amount to
Solutions in which the concentrations of hydrogen and hydroxide ions are the same are called neutral solutions.
So, at 25 ºС
- neutral solution;
\u003e - acidic solution;
< – щелочной раствор.
Instead of the concentrations of H + and OH - it is more convenient to use their decimal logarithms taken with the opposite sign; indicated by the symbols pH and pOH:
;
.
The decimal logarithm of the concentration of hydrogen ions, taken with the opposite sign, is called hydrogen index(pH) .
In some cases, water ions can interact with ions of the solute, which leads to a significant change in the composition of the solution and its pH.
table 2
Formulas for calculating pH (pH)
* Values \u200b\u200bof dissociation constants ( K) are specified in Appendix 3.
p K \u003d - lg K;
HAn is acid; KtOH - base; KtAn is 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, find by reference or in Appendix 3 p K this connection.
3. Determine the composition and concentration of the solution ( FROM).
4. Substitute the numerical values \u200b\u200bof 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 that undergo hydrolysis.
If the solution contains only a strong acid (HАn), which is a strong electrolyte and almost completely dissociates into ions 
, then the pH
will depend on the concentration of hydrogen ions (H +) in a given acid and is determined by formula (1).
If the solution contains only a strong base, which is a strong electrolyte and almost completely dissociates into ions, then the pH value (pH) will depend on the concentration of hydroxide ions (OH -) in the solution and be determined by formula (2).
If only a weak acid or only a weak base is present in a 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, therefore, when calculating the pH in such solutions, the presence of weak acids is neglected and the calculation formula used for strong acids is used (1). The same reasoning is true for the case when a mixture of strong and weak bases is present in the solution. Calculating pH 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 pH for strong acids (1) or bases (2), having previously summed the concentrations of the components.
If the solution contains a strong acid and its salt or a strong base and its salt, then the pH depends only on the concentration of a strong acid or strong base and is determined by formulas (1) or (2).
If the solution contains 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), then this mixture is buffer solution and pH is determined by formulas (5), (6).
If the solution contains a salt 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.
Decision. 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:
Hydrogen ions (H +) remain in excess in an aqueous solution.
2. To calculate the pH, we will use the formula for calculating the pH for salt undergoing hydrolysis by cation:
.
Dissociation constant of a weak base 
(R K = 4,74).
3. Substitute the numerical values \u200b\u200binto 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.
Decision.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 sum of the hydroxide ions. To do this, we summarize the concentration of alkalis:
3. 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.
Decision.1. Formic acid HCOOH is a weak acid, in an aqueous solution only partially dissociates into ions, in Appendix 3 we find formic acid 
:
2. Sodium formate HCOONa is a salt formed by a weak acid and a strong base; hydrolyzed by anion, an excess of hydroxide ions appears in the solution:
3. To calculate the pH, we will use the formula for calculating the pH of buffer solutions formed by a weak acid and its salt, according to the formula (5)
Substitute the numerical values \u200b\u200binto the formula and get
4. The pH of buffer solutions does not change upon dilution. If the solution is diluted 10 times, its pH remains equal to 3.76.
Example 4. Calculate the pH of an acetic acid solution with a concentration of 0.01 M, the degree of dissociation of which is 4.2%.
Decision. 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 the pH, we use the formula (3):
Example 5. To 80 cm 3 0.1 N CH 3 COOH solution was added 20 cm 3 0.2
n solution of CH 3 COONa. Calculate the pH of the resulting solution if K(CH 3 COOH) \u003d 1.75 ∙ 10 –5.
Decision.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 a buffer solution of a given composition using the formula (5):
2. The volume of the solution obtained after merging the original solutions is 80 + 20 \u003d 100 cm 3, hence the concentration of acid and salt will be equal:
3. Substitute the obtained values \u200b\u200bof 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.
Decision.1. Between hydrochloric acid (HCl) and potassium hydroxide (KOH), a neutralization reaction takes place, 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 \u003d 0.05 mol / dm 3. Since the KCl salt does not undergo hydrolysis and does not change the pH of water, the pH value will be affected by the excess potassium hydroxide in this solution. KOH is a strong electrolyte, to calculate pH we use the formula (2):
135. How many grams of potassium hydroxide is contained in 10 dm 3 of a solution with a pH value of 11?
136. The hydrogen index (pH) of one solution is equal to 2, and of the other - 6. In 1 dm 3 of which solution the concentration of hydrogen ions is 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 equal to (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 monobasic acid (HAn) in solution, if: a) pH \u003d 4, α \u003d 0.01; b) pH \u003d 3, α \u003d 1%; c) pH \u003d 6,
α \u003d 0.001.
141. Calculate the pH of 0.01 N acetic acid solution, 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 the acetic acid solution, the pH of which is 5.2?
144. Determine the molar concentration of the formic acid solution (HCOOH), the pH of which is 3.2 ( K НСООН \u003d 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 and pH 0.01 M and 0.05 N solutions of H 2 SO 4.
147. Calculate and the pH of the H 2 SO 4 solution with a mass fraction of acid 0.5% ( ρ \u003d 1.00 g / cm 3).
148. Calculate the pH of the potassium hydroxide solution if 2 dm 3 of the solution contains 1.12 g of KOH.
149. Calculate and the 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 solutions of NH 4 OH and NH 4 Cl with a mass fraction of 5.0%.
152. Calculate the ratio of sodium acetate and acetic acid to be mixed to obtain a buffer solution with pH \u003d 5.
153. In which aqueous solution the degree of dissociation is the greatest: a) 0.1 M CH 3 COOH; b) 0.1 M HCOOH; c) 0.1 M HCN?
154. Derive the formula for calculating pH: a) acetate buffer mixture; b) ammonia buffer mixture.
155. Calculate the molar concentration of the HCOOH solution having pH \u003d 3.
156. How will the pH change if diluted by half 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 COONa?
157 *. 0.1 N acetic acid solution was neutralized with 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) 50 cm 3 of a 0.4 M NaOH solution was added. The pH was measured and then the solution was diluted 10 times. Calculate the pH of the diluted solution.
159 *. To 500 cm 3 0.2 M acetic acid solution ( K \u003d 1.8 ∙ 10 –5) added 100 cm 3 of a 0.4 M NaOH solution. The pH was measured and then the solution was diluted 10 times. Calculate the pH of the diluted solution, write the equations of the chemical reaction.
160 *. To maintain the required pH, the chemist prepared a solution: to 200 cm 3 of a 0.4 M formic acid solution 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 \u003d 1.8 ∙ 10 –4).
| Electrolyte solution | Formula for calculating pH |
| pure water | pH \u003d -lg |
| strong acid | pH \u003d -lgC (1 / z strong acid) |
| strong base | pH \u003d 14 - pОH \u003d 14 + logC (1 / z strong base) |
| weak acid | pH \u003d ½рК а - ½lgС acids |
| weak base | pH \u003d 14 - ½ pK in + ½ logС base |
| ampholyte solutions | |
| salt, anionic hydrolysis | pH \u003d 7 + ½ pK a (strong acid) + ½ lgC salt |
| salt, cation hydrolysis | pH \u003d 7 - ½ pK in (strong base) - ½ lgC salt |
| salt, mixed hydrolysis | pH \u003d 7 - ½ pK in (strong base) + + ½ pK a (strong acid) |
| buffer system type 1 | pH \u003d pK a + lg (C salt / C acid) |
| buffer system type 2 | pH \u003d 14 - рК в + log (С base / С salt) or рН \u003d рК а (ВН +) + log (С base / С salt) |
Ostwald's dilution law: α \u003d √K a / C acid;
\u003d 10 - pH;
В а \u003d n (1 / z acid) / (V buffer solution ∙ ∆рН) \u003d
\u003d (С (1 / z acid) ∙ V acid) / (V buffer solution ∙ (pH 2 - pH 1));
В в \u003d n (1 / z base) / (V buffer solution ∙ ∆рН) \u003d
\u003d (С (1 / z base) ∙ V base) / (V buffer solution ∙ (pH 2 - pH 1)).
Situational tasks:
1. Calculate the pH of 0.01 mol / L solutions of sulfuric acid, potassium hydroxide, phosphoric acid, ammonia, zinc hydroxide, sodium bicarbonate, magnesium sulfate, ammonium acetate.
2. Calculate the degree of dissociation and pH of acetic acid, if K a (CH 3 COOH) \u003d 1.8 · 10 -5, C (x) \u003d 0.18 mol / l.
3. Determine the concentration of hydrogen ions in the blood plasma if pH \u003d 7.4.
4. Determine the pH of the buffer solution, which is obtained by mixing 0.1 mol / L NH 4 Cl solution and 0.1 mol / L NH 4 OH solution in the ratio: a) 1: 1; b) 1: 4; c) 4: 1. K B (NH 4 OH) \u003d 1.79 · 10 -5.
5. Calculate the C salt / C acid ratio for the sodium hydrogen phosphate / sodium dihydrogen phosphate buffer system if pH \u003d 7.4.
6. Calculate the volume of 0.2 mol / L sodium hydroxide solution to be added to 50 ml 0.2 mol / L sodium dihydrogen phosphate solution to obtain a buffer solution with pH \u003d 7.4.
7. Calculate the mass of sodium acetate that should be added to a solution of acetic acid C (CH 3 COOH) \u003d 0.316 mol / l and a volume of 2 liters to obtain a buffer solution with pH \u003d 4.87.
8. How many moles of the equivalent of ascorbic acid must be administered to the patient to normalize the acid-base state, if the pH of his blood is 7.5 (the norm is 7.4), the total amount of blood is 5 liters, the acid capacity is 0.05 mol / l?
Laboratory work No. 4. "Methods for determining the pH of the medium, properties of buffer solutions."
Topic number 5.
Complexation. Properties of complex compounds. Heterogeneous equilibrium. Oxidation-reduction equilibrium.
Topic meaning:
The study of the topic will contribute to the formation of the following competencies OK-1; OK-5; OPK-1; OPK-7; OPK-9; PC-5
The purpose of the lesson: after studying the topic, the student should
Know:
ü safety and work regulations in physical, chemical, biological laboratories with reagents, instruments, animals;
ü physicochemical essence of the processes occurring in a living organism on a molecular, cellular, tissue, organ levels;
ü physicochemical methods of analysis in medicine (potentiometric);
ü main types of chemical equilibria (protolithic , heterogeneous, ligand-exchange, redox) in the processes of life;
ü fundamentals of hemoglobin chemistry, its participation in gas exchange and maintenance of the acid-base state.
Be able to:
ü use educational, scientific, popular science literature, the Internet for professional activities;
ü use physical, chemical and biological equipment;
ü predict the direction and result of physical and chemical processesand chemical transformations of biologically important substances;
ü make calculations based on the results of the experiment.
Form of organization of the educational process:laboratory lesson.
Location of the lesson:educational and scientific chemical and biochemical laboratory.
Lesson equipment:chemical glassware and reagents, interactive whiteboard, projection equipment, labor protection instructions, reference books, personal protective equipment.
Lesson plan:
Questions to study the topic:
1. Complex compounds. Werner's coordination theory. The structure of complex compounds.
2. Classification and nomenclature. Obtaining complex compounds.
3. Intracomplex compounds and their role in biological processes. Polydentate ligands. The structure of the active center of biological complexes: chlorophyll, hemoglobin, cyanocobalamin, catalase. Toxicity of salts of heavy metals, their interaction with complexes of biogenic metals.
4. Antidotes: Unithiol (2,3-dimercaptopropanesulfonate sodium), Trilon A (ethylenediaminetetraacetate), Trilon B (ethylenediaminetetraacetic acid disodium salt), British anti-Lewisite (BAL) (2,3-dimercaptopropanol), tetacinacetic acid (ethylene diamine) Penicillamine (2-amino-3-mercapto-3-methylbutanoic acid), Acyzole (Zinc bisvinylimidazole diacetate).
5. Stability of complex compounds in solutions. Primary and secondary dissociation of complex compounds. Stability constant and instability constant of a complex ion and their relationship with the stability of the complex.
6. Complexometric titration. Determination of water hardness by complexometric method. Disodium salt of ethylenediaminetetraacetic acid (EDTA) - Trilon B. Metallicators - acid chromium black (eriochrome black T).
7. Heterogeneous equilibria and processes. Solubility constant. Conditions for the formation and dissolution of sediments. Reactions underlying the formation of inorganic bone tissue calcium hydroxide phosphate. The phenomenon of isomorphism: substitution of hydroxide ions in calcium hydroxide phosphate for fluorine ions, calcium ions for strontium ions. Osteotropy of metals.
8. Mechanism of calcium buffer functioning.
9. Reactions underlying the formation of calculi: urates, oxalates, carbonates. The use of calcium chloride and magnesium sulfate as antidotes.
10. Classification and essence of precipitation titration methods. Argentometry.
11. Electronic theory of redox reactions (ORR) (LV Pisarzhevsky).
12. Redox properties of elements and their compounds, depending on the position of the element in the Periodic Table of the elements and the oxidation state of the elements in the compounds.
13. Conjugated pairs of oxidizing agent-reducing agent. Redox duality.
14. Types of redox reactions: intermolecular, intramolecular, disproportionation. Compilation of redox reactions by the method of electronic and ion-electronic balance.
15. Mechanism of origin of electrode and redox potentials. Standard, real, formal electrode and redox potentials (redox potentials). Nernst-Peters equation. Comparative strength of oxidizing and reducing agents.
16. Standard variation of the Gibbs and Helmholtz energies of the redox reaction. Predicting the direction of the OM reactions proceeding from the OM potential difference. Influence of the ligand environment of the central atom on the value of the redox potential. Influence of the environment and external conditions on the direction of redox reactions and the nature of the resulting products.
17. Classification and essence of redox titration methods. Permanganatometry, iodometry.
Questions for self-control of knowledge:
Complete the phrases:
1. Complex compounds are …… ..
2. Complex compounds consist of …… and …… .., forming an inner sphere, and an outer sphere.
3. From the standpoint of the theory of valence bonds, the chemical bond between the complexing agent and the ligand is carried out ………….
4. Complexing agents - atoms or ions, …… electron pairs.
5. The role of the complexing agent is more often performed by …… .. and… ..…. elements.
6. Ligands are molecules and ions - ……… electron pairs.
7. Formulas of ligands with names: aqua -… ..; ammin -… ..; hydroxo - ... ..; cyano - …… .; thiosulfato - …… .; nitro - ..… .; chloro - ... ...; thiocyanato - .. ... ..
8. The charge of the inner sphere is defined as an algebraic sum ……….
9. The outer sphere of a complex compound is …… of the opposite sign, neutralizing …… .. of a complex ion and related ………. communication.
10. Cationic are complex compounds, the inner sphere of which has….…. charge.
11. Anionic are complex compounds, the inner sphere of which has ... ... ... charge.
12. The mathematical expression K n (3+) has the form: …………
13. The smaller the instability constant, the complex ……. stable.
14. Dentity - the number of connections, …………
15. The complexing agent in chlorophyll is an ion ... ..., in a cyanocobalamin molecule - an ion ... ... ..., in hemoglobin - an ion ... .., in cytochromes - an ion ... .. ..., in catalase - an ion ... .....
16. The ligand in hemoglobin is ……… ...
17. The main physiological forms of hemoglobin: …… ..
18. The biological role of hemoglobin - transport ………
19. Chelation therapy - ………. organism with the help of …… .. based on the formation of stable ……… .. compounds with ……………. - toxicants.
20. A precipitate is formed if the product of ion concentrations in a solution is equal to their stoichiometric coefficients …… .. solubility constant.
21. In an unsaturated solution, K s…. P with.
22. Necessary condition for the dissolution of the precipitate: K s…. P with.
23. The lower the solubility constant of the poorly soluble electrolyte, the ……. its solubility.
24. If K s (PbSO 4) \u003d 1.6 ∙ 10 -8; K s (SrSO 4) \u003d 3.2 ∙ 10 -7; K s (CaSO 4) \u003d 1.3 ∙ 10 -4, then the solubility is lower for …….
25. If K s (BaSO 4) \u003d 1.1 ∙ 10 -10; K s (SrSO 4) \u003d 3.2 ∙ 10 -7; K s (CaSO 4) \u003d 1.3 ∙ 10 -4, then the solubility is greater for ……
26. Crystals of calcium carbonate were introduced into a saturated solution of silver carbonate. Solubility Ag 2 CO 3 in this case …….
27. The solubility of electrolytes in the sequence: CaHPO 4 → Ca 4 H (PO 4) 3 → Ca 5 (PO 4) 3 OH gradually decreases, therefore a more stable form of calcium phosphate in the body is ………
28. The composition of the tooth enamel includes Ca 5 (PO 4) 3 F. The use of fluoride toothpastes leads to …… .. P s, K s ………….
29. The destruction of dental tissue, which includes Ca 5 (PO 4) 3 OH, will lead to: ……. pH of saliva, ……… Ca 2+ concentration in saliva.
30. Oxidizing agent (Ox) - a particle, …………… ...
31. Reducing agent (Red) - a particle, ………….
32. Reduction - a process during which the oxidant ……… .. and passes into the conjugated ……… form.
33. Oxidation is a process during which a reducing agent ………. and goes into the conjugate ……. shape.
34. Oxidation state - ……………… ...
35. Fill in the table.