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Substances that in solutions or melts are completely or partially disintegrated into ions are called electrolyte.

The degree of dissociation A.

- this is the ratio of the number of molecules that have broken into ions n ¢ to the total number of dissolved molecules N:

The degree of dissociation is expressed as a percentage or in the fractions of the unit. If a \u003d 0, then dissociation is missing and the substance is not an electrolyte. In case a \u003d 1, the electrolyte is completely disintegrated by ions.

Classification of electrolyte

According to modern representations of the theory of solutions, all electrolytes are divided into two classes: associated (weak) and non-associated (strong). Non-associated electrolytes in dilute solutions are almost completely dissociated on ions. For this class of electrolytes A close to one (to 100%). Unassocated electrolytes are, for example, HCl, NaOH, K 2 SO 4 in dilute aqueous solutions.

Associated electrolytes are divided into three types:

  1. Weak electrolytes exist in solutions both in the form of ions and in the form of unfinished molecules. Examples of associated electrolytes of this group are, in particular, H 2 S, H 2 SO 3, CH 3 La in aqueous solutions.
  2. Ion associates are formed in solutions by association of ordinary ions due to electrostatic interaction. Ion associates occur in concentrated solutions of well-soluble electrolytes. As a result, the solution is both simple ions and ion associates. For example, in a concentrated aqueous solution of KCl, simple ions K +, CL are formed - , as well as the formation of ion pairs (K + CL - ), ion tees (k 2 Cl +, KCL 2 - ) and ionic quadrupoles (K 2 Cl 2, KCl 3 2-, K 3 Cl 2+).
  3. Complex compounds
  4. (both ionic and molecular), the inner sphere of which is stepwise dissociate into ionic and (or) molecular particles.
    Examples of complex ions: 2+, 3+, +.

With this approach, the same electrolyte can relate to various types Depending on the concentration of the solution, the type of solvent and temperature. This is confirmed by the data given in the table.

Characteristic of solutions Ki. in various solvents

Approximately, for high-quality reasoning, you can use obsolete division of electrolytes to strong and weak. The allocation of the group of electrolytes of the "middle force" does not make sense. These electrolytes are associated. Weak electrolytes are usually attributed to electrolytes, the degree of dissociation of which is Mala A<<1.

Thus, the strong electrolytes includes diluted aqueous solutions of almost all well-soluble salts, many diluted aqueous solutions of mineral acids (NS1, HBr, NNO 3, NS1O 4 et al.), diluted aqueous solutions of alkali metal hydroxides. Weak electrolytes include all organic acids in aqueous solutions, some aqueous solutions of inorganic acids, for example, H 2 S, HCN, H 2 CO 3, HNO 2, H CLO and others. Water is also related to weak electrolytes.

Disconnection of electrolyte

The equation of the dissociation of strong electrolyte can be represented as follows. Between the right and left parts of the equation equation of the strong electrolyte, an arrow or a sign of equality is set:

HCl H + + Cl -.

Na 2 SO 3 \u003d 2NA + + SO 3 2-.

It is also allowed to put a means of reversibility, but in this case the direction indicates where the dissociation balance is shifted, or it is indicated that A 1. For example:

Na + + Oh -.

The dissociation of acidic and base salts in dilute aqueous solutions proceeds as follows:

NaHSO 3 Na + + HSO 3 -.

Anion salt will dissociate to a small extent, since it is an associated electrolyte:

HSO 3 - H + + SO 3 2. - .

Similarly, the dissociation of the main salts is occur:

Mg (OH) Cl Mgoh + + Cl-.

The cation of the main salt is subjected to further dissociation as a weak electrolyte:

MGOH + MG 2+ + OH -.

Double salts in diluted aqueous solutions are treated as non-associated electrolytes:

Kal (SO 4) 2 K + + Al 3+ + 2SO 4 2-.

Complex compounds in dilute aqueous solutions are dissociated by almost completely disused on the external and internal spheres:

K 3 3K + + 3-.

In turn, the complex ion is slightly subjected to further dissociation:

3- Fe 3+ + 6cn -.

Dissociation constant

When dissolved of weak electrolyte, the solution will establish a balance:

Ka K + + a - ,

which quantitatively describes the value of the equilibrium constant to d, called dissociation constant:

The dissociation constant characterizes the electrolyte's ability to dissociate on ions. The greater the dissociation constant, the greater the ions in the solution of weak electrolyte. For example, in a nitrate acid solution HNO 2 ions H + More than in HCN Sinyl Acid solution, because K (HNO 2) \u003d 4.6 · 10 - 4 , and K (HCN) \u003d 4.9 · 10 - 10.

For weak I-I electrolytes (HCN, HNO 2, CH 3 COOH ) The magnitude of the dissociation constant to D is associated with the degree of dissociation and electrolyte concentrationc. OSVald equation:

For practical calculations provided that <<1 используется приближенное уравнение

Since the process of dissociation of weak electrolyte is reversible, the principle of le chateel is applicable to it. In particular, adding CH 3 Coona to the water solution ch 3 Cooh will bring suppressing the proper dissociation of acetic acid and a decrease in the concentration of protons. Thus, adding substances containing the same name ions to the solution of associated electrolyte, reduces its degree of dissociation.

It should be noted that the dissociation constant of weak electrolyte is associated with a change in the energy of Gibbs in the process of dissociation of this electrolyte as a relation:

D G T 0 \u003d - RTLNK d. (13.5)

Equation (13.5) is used to calculate the constants of the dissociation of weak electrolytes by thermodynamic data.

Examples of solving problems

Determine the concentration of potassium ions and phosphate ions in 0.025 M solution of K 3 PO 4.

Decision. K 3 PO 4 - strong electrolyte and aqueous solution was dissociated by a focus:

K 3 PO 4 3 K + + PO 4 3-.

Consequently, the concentrations of ions K + and PO 4 3- equal, respectively, 0.075m and 0.025m.

Determine the degree of dissociation A D and the concentration of ions it - (mol / l) in 0.03 m NH 4 solution OH at 298 K, if at the specified temperature to d(NH 4 Oh) \u003d 1.76 × 10 - 5 .

Decision. The electrolyte dissociation equation: NH 4 OH NH 4 + + OH -.

Ion concentrations: \u003d Ca; \u003d. c. a. where c. - NH source concentration 4 Oh mol / l. Hence:

Insofar as a.<< 1, то К д » сa. 2 . Dissociation constant

depends on the temperature and on the nature of the solvent, but does not depend on the concentration of NH 4 solutions Oh. The law of dilution of ostelald expresses the dependence of the weak electrolyte from the concentration.

or 2.4%

Where \u003d 2.4 · 10 - 2 · 0.03 \u003d 7.2 · 10 -4 mol / l

Determine the constant dissociation of acetic acid, if the degree of dissociation CH 3 Cohh at 0.002 M solution is 9.4%.

Decision. Acid dissociation equation:

CH 3. Cooh CH 3 Soo - + H +.

,

From [n +] \u003d 9.4 · 10 - 2 · 0.002 \u003d 1.88 · 10 - 4 M.

.

As \u003d [n +] and » C. use (CH 3 cooh)

The dissociation constant can also be found according to the formula: K D "CA 2.

HNO Dissociation Constant 2 at 298k equal to 4.6 ×10 - 4 . Find the concentration of nitrogen acid at which the degree of dissociation of HNO 2 equal to 5%.

K d Ca 2 Where we getc. ex (hno 2) \u003d 4,6 · 10 - 4 / (5 · 10 - 2) 2 \u003d 0.184 M.

. Based on reference data, calculate the dissociation constant of formic acid at 298 K.

Decision. Muravic acid dissociation equation

Nson N + + Soam - .

In the "brief reference book of physico-chemical values" edited by A.A. Tallee and A.M. Ponomareva shows the values \u200b\u200bof the energy of Gibbs formations of ions in the solution, as well as hypothetically unfinished molecules. Gibbs energy values \u200b\u200bfor formic acid and ions H + and coxy - In an aqueous solution, below:

Changing the Gibbs energy of the dissociation process is equal

G T 0 \u003d - 351,5- (- 373.0) \u003d 21.5 kJ / mol.

To calculate the constant dissociation use equation (13.5). From this equation we get:
lNK d \u003d - d g T 0 / Rt \u003d - 21500 / (8.31 298) \u003d - 8.68.

Where we find: k d \u003d 1.7 × 10 - 4.

Tasks for self solutions