- 5.25. Electrolytic dissociation of substances in solutions

Dissociation of compounds with an ionic type of bonding

Ionic crystals consist of cations and anions bound to one another by electrostatic forces of attraction into an ordered crystal structure. Under the action of polar water molecules, ionic crystals break down and dissolve, forming hydrated ions (Fig. 57).

Study Figure 57 and note the relative sizes of Na+ cations, Cl– anions, and water molecules, as well as the signs of the partial charges on the H and O atoms in the water molecule. Note which atom of the water molecule is turned toward the Na+ cation, and which toward the Cl– anion. Why is the water molecule oriented toward the Na+ cation with its oxygen atom, and toward the Cl– anion with its hydrogen atom?

A feature of the dissociation of ionic compounds (salts and alkalis) is that these substances already consist of ions, and polar water molecules merely separate the cations and anions. The energy expended on dissociation is compensated by the energy released during hydration of the ions, and overall the process proceeds spontaneously.

Being bases that are readily soluble in water, alkalis dissociate in aqueous solutions into hydroxide anions and metal cations:

As a result of the dissociation of salts in water, their ionic crystals break down into metal (or ammonium) cations and anions of the acid residues:

Dissociation of compounds with a covalent polar type of bonding

Most organic and inorganic acids are soluble in water and are electrolytes. In acids, the hydrogen atoms are joined to the atoms of the acid residues by covalent polar bonds. Polar molecules do not contain ready-made ions. Let us consider the process of dissociation of acids using the molecule as an example.

First, the polar acid molecules are attracted to the polar water molecules and become even more polarized.

Hydrogen bonds form between the acid and water molecules (Fig. 58).

Then the covalent polar bond lengthens and becomes less strong. As a result, it breaks, that is, ions are formed:

The H+ cation formed upon dissociation of the acid attaches to a water molecule to form a hydronium cation H3O+ by a donor–acceptor mechanism, with which you became familiar while studying the material in § 13. The chlorine ion formed upon ionization is likewise hydrated, as shown in Figure 57.

To simplify equations of reactions written in ionic form, the hydrated hydrogen cation is usually written as H+.

In a similar manner, under the action of polar water molecules, the dissociation of other compounds with a covalent polar type of bonding also takes place.

Dissociation includes the processes of hydration of the initial molecules, their polarization, the breaking of bonds within the molecules, and their breakdown to form hydrated ions.

Acid molecules, upon dissociation in aqueous solutions, dissociate fully or partially into hydrogen cations and anions of the acid residues:

*The difference between the chemical properties of ions and those of atoms and molecules

Atoms and molecules are electrically neutral particles. Metal cations, for example K+, as you know, are formed as a result of metal atoms giving up electrons. Anions of nonmetals, for example Br–, are formed as a result of the addition of electrons to nonmetal atoms.

The properties of cations and anions differ from the properties of the corresponding electrically neutral atoms. Thus, in an electric field created by electrodes, an ordered movement of the electrolyte's cations and anions occurs — an electric current. Positive ions — metal cations, H+, and others — move toward the negative electrode (the cathode).

Negative ions — anions of acid residues and OH– — move toward the positive electrode (the anode) (Fig. 58.1).

Unlike neutral atoms (in particular, metal atoms), cations, for example Cu2+, Fe3+, and others, have fewer electrons in their outer electron levels and a smaller size, and can always act as an oxidizing agent, accepting electrons:

CuSO4 + Zn = ZnSO4 + Cu↓

or Cu2+ + Zn = Zn2+ + Cu↓.

Compared with neutral atoms (I, S, O), simple anions, for example I–, S2– , O2– , and others, have a greater number of electrons in their outer electron levels and a larger size, and can act as a reducing agent, giving up electrons:

2NaI + Cl2 = 2NaCl + I2

or 2I– + Cl2 = 2Cl– + I2.

Strong and weak electrolytes

Simultaneously with electrolytic dissociation in an electrolyte solution, ion association processes occur. Therefore, in electrolytes only a certain fraction of the substance exists in the form of free ions, that is, the substance may not fully break down into ions.

To quantitatively characterize electrolytic dissociation, the concept of the degree of dissociation α is used.

The degree of dissociation α — is the ratio of the number of molecules or formula units of the electrolyte that have broken down into ions to the total number of its dissolved molecules or formula units:

.

The number of particles and their amount (mol) are related by the equation:

, hence, ,

where n is the amount (mol) of the corresponding molecules or formula units.

The value α is expressed as a fraction of unity or as a percentage. The degree of dissociation shows what part of the electrolyte has broken down and exists in the solution in the form of free ions. For example, the degree of dissociation of acetic acid in a solution with a molar concentration of 0.1 mol/dm3 is 1.3 %. This means that out of 1000 acid molecules, only 13 dissociated into H+ cations and CH3COO– anions.

The probability of ion association decreases as the electrolyte concentration decreases, and consequently, the degree of dissociation in dilute solutions is greater than in concentrated ones. This means that the degree of dissociation depends not only on the nature of the substance but also on the concentration of the electrolyte.

Depending on the numerical value of α, electrolytes are conventionally divided into strong and weak.

Strong electrolytes — are chemical compounds that are practically completely dissociated into ions in dilute solutions. The degree of dissociation α of strong electrolytes is close to 1.

Strong electrolytes include soluble ionic compounds and some substances with a covalent polar type of bond; for these, the process of association of the formed ions is practically absent. Therefore, in the equations reflecting electrolytic dissociation, an arrow in one direction is used instead of an equal sign. As examples, let us give the dissociation equations of soluble salts (a), bases (b) and strong acids (c) — H2SO4, HCl, HBr, HI, HClO4, HNO3:

• a)
• b)
• c)

The degree of dissociation of weak electrolytes is, as a rule, less than 5 %. Weak electrolytes include water (), as well as:

• • most organic acids, phenol, a number of inorganic acids: HNO2, HCN, H2S, H3BO3, H2CO3, H2SiO3, etc.;
• • ammonia solution in water.

Weak electrolytes — are chemical compounds that dissociate into ions only slightly, even in dilute solutions. Ions of weak electrolytes are in mobile equilibrium with molecules that have not dissociated, or with a substance of non-molecular structure that has not dissolved.

The dissociation of weak electrolytes is depicted by a breakdown equation with arrows in both directions: .

It is quite difficult to draw a clear boundary between strong and weak acids. Hydrofluoric acid (HF) is conventionally considered weak, although in dilute solutions it dissociates to 8–25 %. Phosphoric acid H3PO4, sulfurous acid H2SO3, and oxalic acid HOOC—COOH are also considered weak. At the first step of dissociation, they dissociate to 10–30 % in concentrated solutions and more than 50 % in dilute ones.

For example, in a dilute solution H3PO4 dissociates according to the equation:

The degree of dissociation at this step can reach 50 %. At each subsequent step, the degree of dissociation decreases significantly, and dissociation proceeds according to the equations:

The decrease in the degree of dissociation is related to the strengthening of the interaction of hydrogen ions with acid residues in the series , and .

The degree of dissociation α is usually determined experimentally from the electrical conductivity of the solution.

The dissociation of weak electrolytes, as a reversible reaction, obeys Le Chatelier's principle. Thus, the H+ cations formed at the first dissociation step shift the equilibrium of the dissociation reaction at the second and third steps to the left, in accordance with Le Chatelier's principle. This decreases the degree of dissociation of the and anions and the concentration of the and anions, respectively.

As a rule, at each subsequent step the degree of dissociation decreases.

In a H3PO4 solution with a molar concentration of 1 mol/dm3, the fractions of H3PO4 molecules and of the anions , and formed at the various stages are:

H3PO4	H+
92 %	~8 %	~8 %	~6 · 10–6 %	~10–16 %

All polybasic weak acids dissociate practically only at the first step. Let us consider one more example — the dissociation of weak hydrosulfuric acid. At the first step, a small part of the H2S molecules dissociates to form hydrogen ions and hydrosulfide ions; at the second, sulfide ions are formed from hydrosulfide ions:

H2S H+ + HS–;

HS– H+ + S2–.

Dissociation at the second step proceeds significantly more weakly, because:

• a) the number of HS– particles that can dissociate at the second step is small;
• b) unlike the dissociation of the H2S molecule, in the second reaction the H+ cation departs from the S2– anion with a charge of –2, rather than from the HS– anion with a charge of –1 as in the first reaction;
• c) dissociation at the first step is accompanied by the formation of H+ cations, which shifts the dissociation equilibrium at the second step to the left and suppresses this process.

Questions, exercises, problems

1. Why do salts conduct electric current after melting?

2. How does electric current in metals differ from electric current in solutions or melts of electrolytes?

3. Write the dissociation equations for the following substances: KHCO3, HClO4, Sr(OH)2, LiOH, КHSO4, NH4H2PO4, HNO2.

4. Why does electrolytic dissociation in solutions proceed spontaneously?

5. Compare the oxidizing and reducing ability of:

• a) the aluminum atom and the Al3+ ion;
• b) the iron atom and the Fe2+ ion;
• c) the sulfur atom and the sulfide ion;
• d) the sulfur atom S0 and the sulfur atom in the composition of the acid residue

6. Why does dissociation at the second step proceed more weakly than at the first step for the anions of most polybasic acids, for example , , , ? How will the degree of dissociation change when hydrogen ions are added to solutions of these acids?

7. An aqueous solution contains 50 mol of hydrogen fluoride. What is the total number of ions formed upon its dissociation, if α(HF) = 9 %?

8. Determine the total number of cations and anions of the salt in a solution containing sodium sulfate with a mass of 2.84 g.

9. What is the molar concentration of H+ cations in an aqueous solution of acetic acid with a volume of 4 dm3, if the degree of dissociation of the acid is 2.6 %, and the mass of the acid is 6 g?

10. In a solution with a volume of 2.5 dm3, the mass of Ba(OH)2 is 14 g. Calculate the molar concentration of OH– ions in this solution.

*Self-check

1. The dissociation of both NaCl and HCl includes:

• a) hydration of chloride ions;
• b) elongation of the covalent bond and its cleavage;
• c) hydration of cations;
• d) polarization of molecules and their breakdown with the formation of ions.

2. The F– ion differs from the fluorine atom F in:

• a) the number of electron shells;
• b) the nuclear charge;
• c) the number of electrons in the outer shell;
• d) size.

3. A solution of a certain salt contains 2 mol of cations and 3 mol of anions. This salt could be:

• a) Аl2(SO4)3;
• b) СаСl2;
• c) Fe2(SO4)3;
• d) Na2CO3.

4. Dissociation equations of weak electrolytes:

• a) H2SO3 2H++ ;
• b) H2O H+ + ОH–;
• c) С6Н5ОН С6Н5О– + Н+;
• d) Са(OH)2 → Са2+ + 2ОН–.

5. In a solution of nitrous acid, the number of undissociated molecules is 4 times greater than the number of dissociated ones. The degree of dissociation of the acid is equal to:

• a) 0.02;
• b) 20 %;
• c) 0.25;
• d) 25 %.