How to determine the pH of an acid. Gossip Dress for the summer! What's the best? And beautiful, and comfortable, and light! Calculation of pH in solutions of weak acids and bases
IPV– this is a constant value at a given temperature for water and any aqueous solutions, equal to the product of the concentration of hydrogen ions and hydroxide ions.
K(H2O) = *
K(H2O) = 1* (t = 25C)
The hydrogen index (pH) is a quantitative characteristic of the acidity of the medium, equal to the negative decimal logarithm of the concentration of free hydrogen ions in the solution.
Hydroxyl index (pOH) is a value equal to the negative decimal logarithm of the concentration of free hydroxide ions in solution
|
Neutral | ||||
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Alkaline |
Calculation of pH of solutions of strong and weak bases and acids.
Weak acid: pH=1/2pKk-1/2lgCk where pK= -lgK is the indicator of the dissociation constant of a weak acid or base.
Weak base: pH=14-1/2pKо+1/2lgCo
Strong acid: pH= -log(zCk) where z is the number of hydrogen ions.
Strong base: pH=14+lg(zCo) where z is the number of hydroxide ions.
Calculation of pH buffer systems. Basic equations. Henderson-Hasselbach equation.
Buffer solutions or systems are called solutions whose pH does not change when small amounts of a strong acid or alkali are added to them, or when diluted. The simplest buffer solution is a mixture of a weak acid and a salt that shares an anion with this acid. For example, a mixture of CH 3 COOH-acetic acid and sodium acetate CH 3 COONa.
Classification: distinguished by composition
1) acidic - consist of a weak acid and its salt. For example: oxyhemoglabin, phosphate bicarbonate.
2) basic ones consist of a weak base and its salt. For example, ammonia: amphoteric, ampholytic - consist of substances that exhibit the properties of both acids and bases (protein buffer). For a buffer system consisting of HAn mol/l of a weak acid and KtAn mol/l of its salt, the concentration of hydrogen ions H + =K Han =, - is called the Henderson-Hasselbach equation, hence H + =K HAn = where K Han is the electric constant. dissociation of a weak acid. Taking logarithms of both parts and changing the signs to the opposite, we arrive at an equation for calculating the pH of the buffer solution in question pH=p KHAn - log, where p KHAn is the decimal logarithm of the electrical dissociation constant of a weak acid. The ability of a buffer solution to maintain pH as a strong acid or alkali is added at approximately a constant level is far from unlimited and is limited by the value of the so-called buffer capacity B. A unit of buffer capacity is usually taken to be the capacity of a buffer solution whose pH change by one unit requires the introduction of a strong acid or alkali in an amount of 1 mol equivalent per 1 liter of solution. Buffer capacity B can be calculated using the formula B=. The total buffer capacity of arterial blood reaches 25.3 mmol/l; in venous blood it is slightly lower and usually does not increase 24.3 mmol/l.
The mechanism of buffer action using the example of ammonium chloride solution.
When adding strong acid (HCl)
A strong acid (HCl) reacts with a weak base (NH4OH)
A neutralization reaction occurs and the acid is replaced by an equivalent amount of salt.
The concentration of free hydroxide ions is replenished due to the potential basicity of ammonium hydroxide, and therefore the pH of the solution remains practically unchanged.
NH4OH+HCl=NH4Cl+H2O
NH4OH+H+Cl=NH4+Cl+H2O
By adding a strong base (NaOH)
Alkali (NaOH) reacts with salt (NH4Cl)
A weak base (NH4OH) is formed and the pH of the solution does not change.
NH4Cl+NaOH=NH4OH+NaCl
1. Calculation of pH in solutions of strong acids and bases.
Calculation of pH in solutions of strong monobasic acids and bases is carried out using the formulas:
pH = - log C to and pH = 14 + log C o
Where C k, C o – molar concentration of acid or base, mol/l
2. Calculation of pH in solutions of weak acids and bases
Calculation of pH in solutions of weak monobasic acids and bases is carried out using the formulas: pH = 1/2 (pK K - logC K) and pH = 14 - 1/2 (pK O - log C O)
3. Calculation of pH in solutions of hydrolyzing salts
There are 3 cases of salt hydrolysis:
a) hydrolysis of the salt by anion (the salt is formed by a weak acid and a strong base, for example CH 3 COO Na). The pH value is calculated using the formula: pH = 7 + 1/2 pK + 1/2 lg C s
b) hydrolysis of the salt by cation (the salt is formed by a weak base and a strong acid, for example NH 4 Cl). The pH in such a solution is calculated using the formula: pH = 7 - 1/2 pK o - 1/2 lg C s
c) hydrolysis of the salt by cation and anion (the salt is formed by a weak acid and a weak base, for example CH 3 COO NH 4). In this case, pH is calculated using the formula:
pH = 7 + 1/2 pK o - 1/2 pK o
If the salt is formed by a weak polybasic acid or a weak polyprotic base, then the values of pK k and pK o for the last step of dissociation are substituted into the above formulas (7-9) for calculating pH
4. Calculation of pH in solutions of various mixtures of acids and bases
When combining an acid and a base, the pH of the resulting mixture depends on the amounts of acid and base taken and their strength.
4.Buffer systems
Buffer systems include mixtures:
a) a weak acid and its salt, for example CH 3 COO H + CH 3 COO Na
b) a weak base and its salt, for example NH 4 OH + NH 4 Cl
c) a mixture of acid salts of different acidity, for example NaH 2 PO 4 + Na 2 HPO 4
d) a mixture of acidic and medium salts, for example NaHCO 3 + Na 2 CO 3
e) a mixture of basic salts of different basicities, for example Al(OH) 2 Cl + Al(OH)Cl 2, etc.
Calculation of pH in buffer systems is carried out using the formulas: pH = pK o – log C o /C s and pH = 14 – pK o + log C o /C s
Acid-base buffer solutions, Henderson-Hasselbach equation. General characteristics. Operating principle. Calculation of pH of buffer solution. Buffer capacity.
Buffer solutions – systems that maintain a certain value of a parameter (pH, system potential, etc.) when the composition of the system changes.
An acid-base solution is called a buffer solution , which maintains an approximately constant pH value when not too large quantities of a strong acid or strong base are added to it, as well as when diluted and concentrated. Acid-base buffer solutions contain weak acids and their conjugate bases. A strong acid, when added to a buffer solution, “turns” into a weak acid, and a strong base becomes a weak base. Formula for calculating the pH of a buffer solution: pH = pK O + lg C O /WITH With This equation Henderson–Hasselbach . From this equation it follows that the pH of the buffer solution depends on the ratio of the concentrations of a weak acid and its conjugate base. Since this ratio does not change upon dilution, the pH of the solution remains constant. Dilution cannot be unlimited. With a very significant dilution, the pH of the solution will change, since, firstly, the concentrations of the components will become so low that the autoprotolysis of water can no longer be neglected, and secondly, the activity coefficients of uncharged and charged particles depend differently on the ionic strength of the solution.
A buffer solution maintains a constant pH value with the addition of only small amounts of a strong acid or strong base. The ability of a buffer solution to “resist” changes in pH depends on the ratio of the concentrations of a weak acid and its conjugate base, as well as on their total concentration - and is characterized by buffer capacity.
Buffer capacity – the ratio of an infinitesimal increase in the concentration of a strong acid or a strong base in a solution (without a change in volume) to the change in pH caused by this increase (p. 239, 7.79)
In strongly acidic and strongly alkaline environments, the buffer capacity increases significantly. Solutions in which the concentration of a strong acid or a strong base is sufficiently high also have buffering properties.
Buffer capacity is maximum at pH=pKa. To maintain a certain pH value, you should use a buffer solution in which the pKa value of the weak acid included in its composition is as close as possible to this pH. It makes sense to use a buffer solution to maintain a pH in the range pKa + _ 1. This interval is called buffer work force.
19.Basic concepts related to complex compounds. Classification of complex compounds. Equilibrium constants used to characterize complex compounds: formation constants, dissociation constants (general, stepwise, thermodynamic, real and conditional concentration)
Most often, a complex is a particle formed as a result of the donor-acceptor interaction of a central atom (ion), called a complexing agent, and charged or neutral particles called ligands. The complexing agent and ligands must exist independently in the environment where complex formation occurs.
The complex compound consists of inner and outer spheres. K3(Fe(CN)6) - K3-outer sphere, Fe-complexing agent, CN-ligand, complexing agent+ligand=inner sphere.
Dentity is the number of ligand donor centers participating in the donor-acceptor interaction during the formation of a complex particle. Ligands are monodentate (Cl-, H2O, NH3), bidentate (C2O4(2-), 1,10-phenanthroline) and polydentate.
The coordination number is the number of donor ligand centers with which a given central atom interacts. In the above example: 6-coordination number. (Ag(NH3)2)+ has a coordination number of 2, since ammonia is a monodentate ligand, and in (Ag(S2O3)2)3- the coordination number is 4, since the thiosulfate ion is a bidentate ligand.
Classification.
1) Depending on their charge: anionic ((Fe(CN)6)3-), cationic ((Zn(NH3)4)2 +) and uncharged or nonelectrolyte complexes (HgCl2).
2)Depending on the number of metal atoms: mononuclear and polynuclear complexes. A mononuclear complex contains one metal atom, while a polynuclear complex contains two or more. Polynuclear complex particles containing identical metal atoms are called homonuclear (Fe2(OH)2)4+ or Be3(OH)3)3+), and those containing atoms of different metals are called heteronuclear (Zr2Al(OH)5)6+).
3) Depending on the character of the ligands: homogeneous and heteroligand (mixed ligand) complexes.
Chelates are cyclic complex compounds of metal ions with polydentate ligands (usually organic), in which the central atom is part of one or more cycles.
Constants. The strength of a complex ion is characterized by its dissociation constant, called the instability constant.
If reference data on stepwise instability constants is not available, use the general instability constant of the complex ion:
The general instability constant is equal to the product of the step instability constants.
In analytical chemistry, instead of instability constants, the stability constants of a complex ion have recently been used: 
The stability constant refers to the process of formation of a complex ion and is equal to the reciprocal of the instability constant: Kst = 1/Knest.
The stability constant characterizes the equilibrium of complex formation.
Thermodynamic and concentration constants see page 313.
20. The influence of various factors on the process of complex formation and the stability of complex compounds. The influence of the concentration of reacting substances on complex formation. Calculation of the mole fractions of free metal ions and complexes in an equilibrium mixture.
1) The stability of complex compounds depends on the nature of the complexing agent and ligands. The pattern of changes in the stability of many metal complexes with various ligands can be explained with help. Theories of hard and soft acids and bases (HAA): soft acids form more stable compounds with soft bases, and hard ones with hard ones (for example, Al3+, B3+ (liquid compounds) form complexes with O- and N-sod Ligands (liquid bases), and Ag+ or Hg2+ (m. bases) with S-solid Ligands (m. bases). Complexes of metal cations with polydentate ligands are more stable than complexes with similar monodentate ligands.
2) ionic strength. With increasing ionic strength and decreasing ion activity coefficients, the stability of the complex decreases.
3) temperature. If during the formation of a complex delta H is greater than 0, then with increasing temperature the stability of the complex increases, but if delta H is less than 0, it decreases.
4) side effects. The effect of pH on the stability of complexes depends on the nature of the ligand and the central atom. If the complex contains a more or less strong base as a ligand, then with a decrease in pH, protonation of such ligands occurs and the molar fraction of the ligand form involved in the formation of the complex decreases. The greater the strength of a given base and the less stable the complex, the stronger the influence of pH will be.
5) concentration. As the ligand concentration increases, the content of complexes with a large coordination number increases and the concentration of free metal ions decreases. If there is an excess of metal ions in the solution, the monoligand complex will dominate.
Mole fraction of metal ions not bound into complexes
Mole fraction of complex particles
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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 quite accurately equal to the total concentration of water, i.e. 1000/18 = 5.5 mol/dm 3.
In dilute aqueous solutions, the water concentration 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 ionic 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; are indicated by the symbols pH and pOH:
;
.
The decimal logarithm of the concentration of hydrogen ions, taken with the opposite sign, is called pH value(pH) .
Water ions in some cases can interact with solute ions, which leads to a significant change in the composition of the solution and its pH.
table 2
Formulas for calculating the hydrogen index (pH)
*Values of dissociation constants ( K) are indicated in Appendix 3.
p K= – lg K;
HAn – acid; KtOH – base; KtAn – salt.
When calculating the pH of aqueous solutions, you must:
1. Determine the nature of the substances included in the solutions and select a formula for calculating pH (Table 2).
2. If a weak acid or base is present in the solution, find it in the reference book or in the appendix 3 p K this connection.
3. Determine the composition and concentration of the solution ( WITH).
4. Substitute the numerical values of the molar concentration ( WITH) 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 the solution contains only a strong acid (HAn), which is a strong electrolyte and almost completely dissociates into ions 
, then the hydrogen index (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 (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 solution contains a mixture of strong and weak acids, then the ionization of the weak acid is practically suppressed by the strong acid, therefore, when calculating 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. pH calculations are carried out according to formula (2).
If the solution contains a mixture of strong acids or strong bases, then the pH is calculated using the formulas for calculating 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 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 (hydrolyzes by cation) or a weak acid and a strong base (hydrolyzes by an anion), a weak acid and a weak base (hydrolyzes by cation and anion), then these salts, undergoing hydrolysis, change pH value, and the calculation is carried out using 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 into a cation according to the equations:
Hydrogen ions (H+) remain in excess in the aqueous solution.
2. To calculate pH, we use the formula for calculating the pH value for a salt undergoing hydrolysis by cation:
.
Weak base dissociation constant 
(R K = 4,74).
3. Substitute the numerical values into the formula and calculate the hydrogen index:
.
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 hydrogen index will be determined by the sum of hydroxide ions. To do this, we summarize the concentrations 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 a 0.10 M solution of formic acid and a 0.10 M solution of 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 by a weak acid and a strong base; hydrolyzes at the anion, an excess of hydroxide ions appears in the solution:
3. To calculate pH, we will use the formula for calculating the hydrogen values of buffer solutions formed by a weak acid and its salt, according to formula (5)
Let's substitute the numerical values into the formula and get
4. The pH value of buffer solutions does not change when diluted. If the solution is diluted 10 times, its pH will remain equal to 3.76.
Example 4. Calculate the hydrogen index 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 added 20 cm 3 0.2
n solution of CH 3 COONa. Calculate the pH of the resulting solution if K(CH 3 COOH) = 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 a buffer solution of this composition using 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. Let us substitute the obtained values of acid and salt concentrations
into the formula
.
Example 6. To 200 cm 3 of 0.1 N hydrochloric acid solution, 200 cm 3 of 0.2 N solution of potassium hydroxide was added to 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 water, the pH value will be affected by potassium hydroxide in excess in this solution. KOH is a strong electrolyte; to calculate pH we use formula (2):
135. How many grams of potassium hydroxide are contained in 10 dm 3 of a solution whose pH value 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 by how many times?
137. Specify the reaction of the medium and find the concentration of ions in solutions for which the pH is equal to: a) 1.6; b) 10.5.
138. Calculate the pH of solutions in which the concentration is equal (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 ion concentration is equal (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 = 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 acid dissociation 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 an acetic acid solution whose pH is 5.2?
144. Determine the molar concentration of a solution of formic acid (HCOOH), whose pH is 3.2 ( K NCOOH = 1.76∙10 –4).
145. Find the degree of dissociation (%) of a 0.1 M CH 3 COOH solution 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/cm 3).
148. Calculate the pH of a solution of potassium hydroxide if 2 dm 3 of solution contains 1.12 g of KOH.
149. Calculate the pH of a 0.5 M ammonium hydroxide solution. = 1.76∙10 –5.
150. Calculate the pH of the solution obtained by mixing 500 cm 3 of 0.02 M CH 3 COOH with an equal volume of 0.2 M CH 3 COOK.
151. Determine the pH of a buffer mixture containing equal volumes of NH 4 OH and NH 4 Cl solutions with mass fractions of 5.0%.
152. Calculate in what ratio sodium acetate and acetic acid must be mixed to obtain a buffer solution with pH = 5.
153. In which 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 the pH of: a) acetate buffer mixture; b) ammonia buffer mixture.
155. Calculate the molar concentration of a HCOOH solution having pH = 3.
156. How will the pH change if the following 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 COONa?
157*. A 0.1 N solution of acetic acid was neutralized with a 0.1 N solution of sodium hydroxide to 30% of its original concentration. Determine the pH of the resulting solution. 
158*. To 300 cm 3 0.2 M solution of formic acid ( K= 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 diluted solution.
159*. To 500 cm 3 0.2 M acetic acid solution ( K= 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 equations of the chemical reaction.
160*. To maintain the required pH value, the chemist prepared a solution: to 200 cm 3 of 0.4 M formic acid solution he added 10 cm 3 of 0.2% KOH solution ( p= 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).
→ → →PHformula
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Story
The laboratories in which pHformula was created were founded at the end of the 19th century in Barcelona. They are currently run by the fourth generation of a family of pharmacists specializing in dermatology. The brand actively invests in research to scientifically substantiate and prove the effectiveness of its products, actively collaborating with the best medical institutions. All active components of the formulas are pharmaceutical-cosmeceutical ingredients, and studies demonstrating their effectiveness are published in the public domain.
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Battleship “Victory” – Legendary sailing ships
Formation of a peptide bond Two equivalent forms
Kasatkin Mikhail Alexandrovich (1902–1974) M Kasatkin