Read also:
|
FORMATION AND DISSOLUTION OF SEDIMENTS
IN ANALYSIS
Plan:
1. Solubility product and solubility. Precipitation condition
2. Solubility product for incomplete dissociation of a sparingly soluble compound
3. Factors affecting the completeness of sedimentation
4. Dissolution of precipitates
Solubility product and solubility.
Precipitation condition
heterogeneous called chemical and physico-chemical processes that occur in systems consisting of several phases. Phases can be liquid, solid and gaseous.
Phase - these are separate parts of a heterogeneous system delimited by interfaces.
We will consider the phase equilibrium of a liquid - solid, as having highest value for separation of elements, qualitative and quantitative determination.
Consider the equilibrium in the system sparingly soluble strong electrolyte A a B c, omitting the ion charges for simplicity:
A a B vtv ó aA + bB
This equilibrium is described by the thermodynamic equilibrium constant:
The activity of the solid phase is a practically constant value, the product of two constants will give a new constant, which is called thermodynamic solubility product (PR) :
In a solution above a precipitate of a poorly soluble strong electrolyte, the product of ion activities in powers of the corresponding stoichiometric coefficients is a constant value under given conditions (temperature, pressure, solvent).
PR T = f(T, p, solvent nature)
In German literature, PR is designated Lp (löslichkeitsprodukt), in English - Sp (solubility product).
R solubility S is the ability of substances to form a homogeneous system with a solvent.
Solubility is measured in mol/l, g/100ml, g/ml, etc.
The lower the solubility, the more difficult it is for the electrolyte to dissolve.
BaSO 4 (PR \u003d 1.05 10 -10) dissolves with difficulty during boiling only in concentrated sulfuric acid,
CaSO 4 (PR \u003d 9.1 10 -6) is quite soluble in water - gypsum water
For a poorly soluble strong electrolyte AB, which consists of ions of the same charge, the solubility S is the equilibrium concentration of ion A or ion B.
If we denote this concentration by x, then
PR \u003d [A] [B] \u003d x 2
S=x=
For an electrolyte precipitate consisting of ions of different charge A a B b, equilibrium in saturated solution
A a B b TV ó aA + bB
From here [A] = a S And [B] = b S
PR (A a B b)= [A] a [B] b = a [ b S] b = a a b b S a+b. From here
Knowing the numerical value of the solubility product makes it possible to predict whether a precipitate of a poorly soluble compound will precipitate as a result of an exchange reaction. For example, in order for the AgNO 3 and K 3 PO 4 solutions to precipitate Ag 3 PO 4 as a result of the exchange reaction
3 Ag + + PO 4 3– ® Ag 3 PO 4 ¯
it is necessary that the resulting solution be supersaturated with silver ions and phosphate ions. It is important to understand that PR is a characteristic related to a saturated solution, so precipitation will occur if, in the resulting solution the concentration product (PC) of the ions forming the precipitate is greater than the solubility product (SP) or, more briefly, precipitation condition: PC > PR.
Example 5.2. Determine whether Ag 3 PO 4 will precipitate when pouring 1 liter of Na 3 PO 4 solution with a concentration of 5·10 -5 mol/l and 1 liter of AgNO 3 solution with a concentration of 2·10 -3 mol/l. PR (Ag 3 PO 4) \u003d 1.3 10 -20.
When solving such problems, it is necessary first of all to find in the initial solutions the number of moles of those ions that can form a precipitate (in this case, these are Ag + and PO 4 3– ions).
In a solution of Na 3 PO 4: n (Na 3 PO 4) \u003d C (Na 3 PO 4) V solution (Na 3 PO 4);
n (Na 3 PO 4) \u003d 5 10 -5 mol / l 1 l \u003d 5 10 -5 mol \u003d n (PO 4 3-).
In AgNO 3 solution: n(AgNO 3) = C(AgNO 3) V AgNO 3 solution;
n(AgNO 3) \u003d 2 10 -3 mol / l 1 l \u003d 2 10 -3 mol \u003d n (Ag +).
In the solution formed after mixing, the number of moles of Ag + and PO 4 3– ions before the formation of a precipitate will be the same as in the initial solutions, and the volume of the solution will become 2 liters:
V total ≈ V solution Na 3 PO 4 + V solution AgNO 3 \u003d 1 l + 1 l \u003d 2 l.
C (Ag +) \u003d n (Ag +) / V total \u003d 2 10 -3 mol / 2 l \u003d 1 10 -3 mol / l;
C (PO 4 3–) \u003d n (PO 4 3–) / V total \u003d 5 10 -5 mol / 2 l \u003d 2.5 10 -5 mol / l.
Precipitation occurs as a result of the reaction proceeding according to the equation 3 Ag + + PO 4 3– ® Ag 3 PO 4 ¯, therefore, the product of concentrations (PC) of Ag + and PO 4 3– ions in the resulting solution should be calculated using the equation:
PC \u003d C 3 (Ag +) C (PO 4 3–) \u003d (1 10 -3) 3 2.5 10 -5 \u003d 2.5 10 -14.
Since PC \u003d 2.5 10 -14 > PR (Ag 3 PO 4) \u003d 1.3 10 -20, the solution is supersaturated with Ag + and PO 4 3– ions, therefore, a precipitate of Ag 3 PO 4 is formed.
Example 5.3. Determine whether a precipitate of PbCl 2 will precipitate when 200 ml of a 0.005 M solution of Pb (NO 3) 2 and 300 ml of a 0.01 M solution of NaCl PR (PbCl 2) = 1.6 10 -5 are drained.
Calculation of the amounts of Pb 2+ and Cl - ions in the initial solutions:
In a solution of Pb (NO 3) 2: n (Pb (NO 3) 2) \u003d C (Pb (NO 3) 2) V solution (Pb (NO 3) 2);
n (Pb (NO 3) 2) \u003d 0.005 mol / l 0.2 l \u003d 0.001 mol \u003d n (Pb 2+).
In NaCl solution: n(NaCl) = C(NaCl) V NaCl solution;
n (NaCl) \u003d 0.01 mol / l 0.3 l \u003d 0.003 mol \u003d n (Cl -).
In the solution formed after mixing, the number of moles of Pb 2+ and Cl - ions before the formation of a precipitate will be the same as in the initial solutions, and the volume of the solution will become 0.5 liters:
V total ≈ V solution Pb (NO 3) 2 + V solution NaCl \u003d 0.2 l + 0.3 l \u003d 0.5 l.
The concentrations of Ag + and PO 4 3– ions in the resulting solution will be as follows:
C (Pb 2+) \u003d n (Pb 2+) / V total \u003d 0.001 mol / 0.5 l \u003d 0.002 mol / l \u003d 2 10 -3 mol / l;
C (Cl -) \u003d n (Cl -) / V total \u003d 0.003 mol / 0.5 l \u003d 0.006 mol / l \u003d 6 10 -3 mol / l.
The formation of a precipitate occurs as a result of the reaction proceeding according to the equation Pb 2+ + 2 Cl - ® PbCl 2 ¯, therefore, the product of the concentrations (PC) of Pb 2+ and Cl - ions in the resulting solution should be calculated using the equation:
PC \u003d C (Pb 2+) C 2 (Cl -) \u003d 2 × 10 -3 × (6 10 -3) 2 \u003d 7.2 10 -8.
Since PC \u003d 7.2 10 -8< ПР(PbCl 2) = 1,6·10 –5 , образовавшийся раствор не насыщен ионами Pb 2+ и Cl – , и осадок PbCl 2 не образуется.
In task number 7 (table 5.2), students are asked to determine the possibility of precipitation when mixing two electrolyte solutions.
The solubility product is one of the main characteristics of a precipitate. Using this characteristic, one can change the solubility of the precipitate, calculate the optimal conditions for precipitation, and foresee which precipitation reactions are best used to determine certain ions.
Equation (3.1.) implies the conditions for the formation of a precipitate in a solution:
A precipitate of a sparingly soluble electrolyte is formed only when the product of the concentrations of its ions (P) in solution exceeds the value of the product of the solubility of this compound, those. when a solution becomes supersaturated with respect to a given poorly soluble compound. A precipitate does not separate from an unsaturated solution, the solid phase dissolves.
EXAMPLE 3.6 Determine whether a precipitate of PbCO 3 is formed when 400 ml of 0.001 M Pb(NO) 2 and 100 ml of 0.01 M K 2 CO 3 are mixed.
Solution: Let's find the molar concentrations of substances at the moment of mixing according to the formula:
The concentrations of the ions forming the precipitate are:
Cm 2 (Pb (NO) 2), because when dissociated from 1 mol of salt, 1 mol of lead ions is formed.
[CO 3 2-] \u003d Cm 2 (K 2 CO 3), because. upon dissociation from 1 mol of salt, 1 mol of CO 3 2- ions is formed.
Hence PR = ·[CO 3 2- ]= 0.0008·0.002=1.6·10 -5.
The value obtained is greater than PRPbCO 3 = 7.5·10 -14 , therefore, the solution is supersaturated with respect to lead carbonate and a precipitate forms.
EXAMPLE 3.7 At what ratio of the concentrations of Ba 2+ and Pb 2+ ions will their carbonates precipitate simultaneously with the introduction of CO 3 2- ions? PRVaCO 3 =7∙10 -9 , PRRBbCO 3 =1.5∙10 -13 .
Solution: We denote the concentration of carbonate ions introduced by CCO 3 2-, then:
So, barium and lead carbonates will precipitate simultaneously from the solution if CBa 2+ > СРb 2+ by 46,700 times. If the ratio of CBa 2+ / CPb 2+ > 46700, then BaCO 3 will be the first to fall out of the solution until the ratio of CBa 2+ / CPb 2+ is equal to 46700. And only after that will simultaneous precipitation begin. If the concentration ratio of barium and lead ions is less than 46700, then lead carbonate will begin to precipitate first. The precipitation of lead carbonate will proceed until the ratio of CBa 2+ / CPb 2+ reaches a value at which BaCO 3 and PbCO 3 will precipitate simultaneously.
Creating optimal precipitation conditions in quantitative determinations is even more important than in qualitative analysis, since any loss of matter is completely unacceptable here. Therefore, it is necessary to dwell on this - in more detail.
Let us first consider the process of precipitation formation. This process is certainly more complex than one would expect from the reaction equation. So, according to the equation
Ba 2+ + SO4 2- - BaSO4
one can think that for the formation of barium sulfate it is only necessary that two ions meet in solution: Ba2+ and SO2T. But this, of course, is not the case.
BaSO4 precipitates in the form of crystals, and a crystal lattice cannot be built from two ions. The process of formation of a solid phase in a solution is very complicated.
The so-called induction period is almost always observed, which lasts from the moment of mixing the reagent solutions containing the reactants until a visible precipitate appears. For different substances, the induction period is different; for example, during the precipitation of BaSO*, it is relatively large, while during the precipitation of AgCl, it is very short.
The presence of the induction period is explained by the fact that the formation of sediment passes through a series of stages. At the beginning, germinal, or primary crystals are formed. For their formation in space, a fairly large number of reacting ions must meet in a certain ratio and at a certain arrangement. In solution, ions are surrounded by a hydration shell, which must be destroyed during the formation of a precipitate.
The resulting primary crystals do not yet create an interface, i.e., the formation of these first particles of the solid phase and their combination (aggregation) into larger ones, consisting of tens or hundreds of molecules, does not yet cause the substance to precipitate. This stage of sediment formation corresponds to the existence of colloidal systems. Then the primary crystals or their aggregates form larger particles and precipitate. This process can proceed in two ways, which determine the form of the precipitate, i.e., the formation of a crystalline or amorphous precipitate. In the first case, when portions of the precipitating agent are added to the solution, no new crystallization centers or new aggregates appear. The solution remains in a supersaturated state for some time.
With the gradual introduction of a precipitant, the release of a substance from a supersaturated solution occurs mainly on the surface of the previously formed seed crystals, which gradually grow, so that in the end a crystalline precipitate is obtained, consisting of a relatively small number of relatively large crystals.
This is usually how precipitation proceeds when the solubility of the precipitate is not too low, especially if measures are taken to increase it by heating or by adding various reagents, such as acids.
Otherwise, the process of formation of amorphous precipitates occurs. In this case, the addition of each portion of the precipitant causes the rapid appearance in the liquid of a huge amount of tiny germinal crystals, which no longer grow as a result of the deposition of the corresponding substance on their surface, but as a result of their combination into larger aggregates, which settle under the influence of gravity to the bottom of the vessel. In other words, coagulation of the initially formed colloidal solution occurs.
Since the bond between the individual seed crystals in the resulting aggregates is relatively weak, these aggregates can again decompose with the formation of a colloidal solution.
As can be seen from what has been said, it is not entirely correct to call these sediments amorphous. It would be more correct to call them "cryptocrystalline", since they are formed from crystals, albeit the smallest ones. Indeed, the presence crystal lattice in amorphous precipitates can in most cases be proved experimentally by examining them with the help of x-rays and sometimes under a microscope.
The shape of the liberated precipitate depends on the individual properties of the substances. For example, polar, relatively well-soluble substances (BaSO4, AgCl, PbSO4, etc.) precipitate in the crystalline state.
But this or that form of sediment is not only associated with the individual properties of the substance, but also depends on the conditions of precipitation. For example, when precipitated from dilute aqueous solutions BaSO4 precipitates as a crystalline precipitate. If, however, it is precipitated from a mixture of water with 30-60% alcohol, which greatly reduces the solubility of barium sulfate, then colloid solution or amorphous precipitate. On the other hand, by precipitating sulfides in the presence of pyridine C5H5N, some of them are obtained in the form of crystals.
proved that any substance can be obtained both in the form of a crystalline and in the form of an amorphous precipitate. However, the formation of one of these forms is usually associated with the creation of conditions that are unacceptable for quantitative determinations. Therefore, depending on the individual properties of the compounds formed, some of them are obtained in the analysis in the form of crystalline, others - in the form of amorphous precipitates. The task of the analyst is to create conditions under which the resulting precipitate would be as pure as possible and convenient for further processing, i.e. for separation by filtration and washing.
In conclusion, it should be said that if the freshly precipitated precipitate is left for some time under the mother liquor, then the precipitate undergoes a series of changes, which are called "aging" of the precipitate.
The optimal conditions for deposition and aging are very different in the case of the formation of amorphous and crystalline precipitates.
To answer the question of whether precipitation will fall under these conditions, it is necessary to calculate the value products of PC concentrations and compare it with the tabular value of PR. In this case, 3 cases are possible:
1. PC< ПР. Такой раствор называется unsaturated. A precipitate does not form in this solution. The precipitate molecules immediately disintegrate into ions, because their concentration is below the equilibrium.
2. PC = PR. This solution is called rich . It has a moving balance. The precipitate does not fall out.
3. PC > PR. Precipitation is formed only in supersaturated solution. The formation of a precipitate will continue until the equality PC = PR is reached and the solution changes from supersaturated to saturated. Equilibrium sets in and further sediment formation stops.
The property of a saturated solution to keep the product of activities (concentrations) of ions constant in the appropriate powers is called solubility product rule.
According to this rule, the existence of such solutions is impossible, in which the product of activities would exceed the tabular value of PR at a given temperature. If the product of activities exceeds the PR by appropriate degrees, then a precipitate should form and the concentration of ions in the solution should decrease to such values that would satisfy the PR rule.
In accordance with the PR rule, if the concentration (activity) of one of the ions included in the PR expression increases, then the concentration (activity) of the other decreases.
This action of the ion of the same name underlies quantitative precipitation methods and is used in analytical chemistry.
Let us calculate, for example, the solubility of AgCl ( PR t= 1.78 10 -10). in water and in 0.01M KCl solution.
