2.5.1. The Arrhenius Groundbreaking Theory. Dissociation of Salts, Acids, and Bases. In 1884, Svante Arrhenius (1859–1927), then a 24-year-old Swedish researcher, presented his Doctoral Thesis titled "Investigations on the Galvanic Conductivity of Electrolytes". The dissertation contained a section that was immediately stamped "HERESY" by most scientists at that time. Less than 20 years later, in 1903, Arrhenius received the Nobel Prize in Chemistry for that "heresy", currently known as the Arrhenius electrolytic theory of dissociation.

On the basis of his experimental results, Arrhenius speculated in his Thesis that molecules of salts, acids, and bases dissociate (break down) into positively and negatively charged ions upon dissolution in water. This type of transformation, which was unheard of and hardly imaginable back in those days, is now a long-established fact. Figure 2-58 shows dissociation of a salt (NaCl), an acid (HCl), and a base (NaOH) to ions in aqueous solution. This process does not require any special conditions but rather occurs spontaneously at room temperature and atmospheric pressure.

Figure 2-58. Examples of electrolytic dissociation of a salt (NaCl), an acid (HCl), and a base (NaOH).


Each time we put table salt (NaCl) in our soup or chili, we immediately get sodium cations (Na⁺) and chloride anions (Cl^-) in the dish. Each time we use citric acid or acetic acid in the form of vinegar in our cooking, we sure add hydrogen ions (H⁺) to what we are cooking. It is the H⁺ that is responsible for the characteristic sour taste of citric acid, vinegar, and other acids. The soapy slippery feel of solutions of an alkali such as NaOH or KOH is due to the presence of the hydroxide anions (OH^-).

How do we know that many salts, acids, and bases dissociate into ions when dissolved in water? We know that from experiments demonstrating that such solutions conduct electricity (see, for example: Video 2-5). While pure water and solid NaCl are poor electrical conductors, a solution of NaCl in water conducts electricity well due to the movement of the Na⁺ and Cl^- ions produced on dissolution and dissociation of the salt.

Video 2-5. While pure water does not conduct electricity, a solution of NaCl in water does (source).


A compound that dissolves in water to give an electroconductive solution is called an electrolyte. Consequently, the phenomenon of the formation of ions from compounds on dissolution is referred to as electrolytic dissociation, or ionic dissociation. We will learn how electrolyte solutions conduct electricity in section 2.8 below. An electrolyte can be a solid such as NaCl and most other salts, a liquid such as sulfuric acid, H₂SO₄, or a gas such as HCl. However, not all water-soluble compounds are electrolytes. For example, aqueous solutions of sucrose (sugar, a solid), ethanol (alcohol, a liquid), and oxygen (O₂, a gas) do not conduct electricity. Sugar, ethyl alcohol, and oxygen are not electrolytes.

2.5.2. How Water Dissolves Ionic Compounds. Sodium chloride, NaCl, is a typical ionic compound. To understand how crystalline NaCl dissolves and dissociates in water, we need to familiarize ourselves with (i) the structure of NaCl and (ii) the dipolar nature of the water molecule. Crystals of NaCl (Figure 2-59) are made up of alternating sodium cations and chloride anions. Each sodium cation in the crystal is surrounded by six chloride anions (top-bottom; left-right; front-back) and vice versa. The Na⁺ and Cl^- ions are held together in this highly ordered 3-dimensional structure by ionic bonds (electrostatic forces).

Figure 2-59. Two representations of the structure of NaCl in the crystal (left image: source; right image: source).

A molecule of water is a dipole (from di and pole), meaning that there is an accumulation of electric charges on different sides of the molecule. Figure 2-60 provides a simplified illustration of the dipolar nature of water molecules. As we know from the previous section, the covalent O-H bond is highly polar. The partial negative and positive charges (Figure 2-60) create the dipole. Note that the + and – symbols used in the image on the right of Figure 2-60 do not imply full charges (+1 and -1) and are used just to illustrate that one end of a water molecule is charged positively and the other negatively.

Figure 2-60. Simple representation of a water molecule as a dipole.


Now let us pour water onto NaCl. Water molecules cannot go into the crystal lattice of NaCl because there is no room for them inside the tightly packed orderly crystal cage (Figure 2-59). However, the H₂O dipoles get attracted to the Na⁺ and Cl^- ions on the surface by electrostatic forces. As shown in Figure 2-61, the negatively charged ends of the water dipoles stick to the positively charged sodium ions, while the positively charged ends of the dipoles are attracted to the negatively charged chloride ions.

Figure 2-61. Water dipoles are attracted to the surface of NaCl by electrostatic (Coulomb) forces.


When enough water dipoles approach an ion on the surface, the attractive electrostatic forces between that ion and the dipoles overpower those between the same ion and its crystal cage neighbors. The ionic bonds holding the ion within the crystal first weaken (Figure 2-62) and, finally, the ion gets plucked off the crystal lattice (Figure 2-63). The separated ion is surrounded by molecules of water or, as chemists say, is hydrated (Figure 2-63). It is not just one ion on the surface that gets attacked by the water dipoles. Many of them do simultaneously. As a result, the ions constantly leave the surface of the crystal, exposing the ions beneath to electrostatic interactions with water dipoles until the solid has all dissolved. Video 2-6 displays an illustrative animated model for the process. The same sequence of events operates in the process of dissolution of other salts in water.

Figure 2-63. Eventually, the water dipoles pull the ion off the surface of the crystal.


The abstraction of ions from the surface by water dipoles may well be compared to the tug of war game (Figure 2-64). The team of the neighboring ions in the crystal loses to the individually weaker but more numerous and jointly stronger water dipoles.

Figure 2-64. Competition for an ion on the surface of a crystal is lost by the stronger attracting neighboring ions to the weaker attracting but more numerous and jointly stronger water dipoles (adapted from source).

Video 2-6. How water dissolves NaCl (source).


2.5.3. How Covalent Compounds Dissociate in Water. Heterolytic and Homolytic Cleavage of Covalent Bonds. Some acids such as HCl and H₂SO₄ dissociate in water as easily as ionic NaCl. However, acids are covalent, not ionic compounds. How do they dissociate? An answer to this question is given by Figure 2-65 showing dissociation of HCl in water as an example.

Figure 2-65. Dissociation of HCl in water.


The H-Cl bond is polar covalent. The shared electron pair in HCl is polarized away from the less electronegative H atom toward the more electronegative Cl atom. Due to this bond polarization, the H atom bears a partial positive charge and the Cl atom a partial negative charge. These partial charges attract water dipoles (Figure 2-65). As the number of water dipoles surrounding the HCl molecule increases, the overall electrostatic attraction between these dipoles and the H and Cl atoms grows stronger. The H-Cl bond becomes more and more polarized and eventually the dipoles pull the HCl molecule apart to give rise to the hydrated H⁺ and Cl^- ions (Figure 2-65).

When HCl dissociates, both electrons of the shared electron pair go entirely to one of the two atoms, the more electronegative chlorine. This type of covalent bond cleavage is called heterolytic (from Ancient Greek ἕτερος, meaning "different"). There is another type of covalent bond splitting, where each of the two bonded atoms keeps one electron of the originally shared electron pair when parting company. This type of covalent bond breaking is called homolytic (from Greek όμοιος meaning "same"). Figure 2-66 illustrates the two types of covalent bond fission. Naturally, heterolytic cleavage is characteristic of polar covalent bonds such as O-H, Cl-H, and Br-H, whereas nonpolar and weakly polar covalent bonds such as Cl-Cl, Br-Br, and C-H are more inclined to break homolytically.

Figure 2-66. Heterolytic cleavage of a polar covalent bond (H-Cl) and homolytic cleavage of a nonpolar covalent bond (Br-Br).


2.5.4. They Come Together and Part Non-Stop. Chemical Equilibrium. In solution, oppositely charged ions come together to produce an ion pair and then separate again (Figure 2-67). This process is dynamic, meaning that it occurs nonstop. The forward and backward arrows in Figure 2-67 indicate that the process of ion pair formation is reversible. (We briefly touched on reaction reversibility earlier in subsection 2.2.5 and used the double arrow sign in Figure 2-14.)

Figure 2-67. Reversible formation of an ion pair from a cation and an anion.


Chemists say that an ion pair and the ions it comprises equilibrate, or exist in equilibrium. If the rate of the ion pair formation is faster than the rate of its dissociation, we can say that the equilibrium is shifted to the ion pair (to the right in Figure 2-67). If an ion pair dissociates faster than it is formed, then the equilibrium is shifted to the ions (to the left in Figure 2-67).

2.5.5. Strong and Weak Electrolytes. Degree of Dissociation. Different compounds dissociate to a different degree. Let us consider an imaginary molecule that consists of atom A and atom B. The bond between A and B can be either ionic or covalent. In water, our molecule AB dissociates to cation A⁺ and anion B^- (Figure 2-68). As in the case of NaCl considered above, the dissociation of AB is reversible and all three species involved (AB, A⁺, and B^-) are hydrated (surrounded by water molecules). For the sake of simplicity and clarity, we omit the hydrating molecules of water from our considerations.

Figure 2-68. Reversible dissociation of a molecule of compound AB to cation A⁺ and anion B^-.


The dissociation of molecule AB and the backward process of reassociation of the ions A⁺ and B^- produced (Figure 2-68) occur nonstop. At any given moment, there are a certain number of dissociated AB molecules in the form of A⁺ and B^- and a certain number of non-dissociated molecules AB. The percentage of dissociated molecules of a compound in solution represents the degree of dissociation (α) of that compound. For instance, there are 100 molecules of compound AB in water. Of these 100 molecules in the solution, 10 exist as non-dissociated AB and 90 in the form of ions A⁺ and B^-. The degree of dissociation of AB is the percentage of 90 out of 100, which is 90%. The general formula for degree of dissociation α is as follows.

The degree of dissociation parameter may be compared to a divorce rate. Men and women meet and get married to form couples. Cations and anions meet to form molecules or ion pairs. A certain percentage of married couples get divorced. This percentage is the divorce rate. A certain percentage of ion pairs dissociate. This percentage is the degree of dissociation, α. Note that divorced men and women can and do remarry to new partners to form new couples. Likewise, the separated ions do find new oppositely charged partners to form new ion pairs or molecules.

The higher the α value, the more dissociated a given compound is in solution. All electrolytes are often divided into three groups: strong (30% < α < 100%), medium strength (3% < α < 30%) and weak (α < 3%). We will use this classification in our course, although some chemistry courses suggest other numbers, such as α ≈ 100% for strong electrolytes and α < 10% for weak electrolytes. A small number of strong, medium strength, and weak electrolytes need to be memorized (Table 4).


Table 4. Selected example of strong, medium strength, and weak electrolytes.

Electrolytic strength must not be confused with solubility. Some compounds are soluble in water while being weak electrolytes. Such compounds dissolve in water but dissociate only to a small extent. One example is mercury (II) chloride, HgCl₂, which is quite soluble in water, about 7.5 g in 100 mL at room temperature. However, only a small fraction of the HgCl₂ molecules in solution dissociate to ions, which makes HgCl₂ a weak electrolyte in spite of its solubility. Conversely, there are also compounds that are very poorly soluble in water while being strong electrolytes. For example, BaSO₄ is virtually insoluble (only about 0.00025 g in 100 mL of water) yet is a strong electrolyte because of the tiny amount that dissolves, almost all breaks up into the Ba²⁺ and SO₄^2- ions. Although BaSO₄ is approximately 30,000 times less soluble in water than HgCl₂, BaSO₄ is a strong electrolyte, wheres HgCl₂ is a weak electrolyte.

2.5.6. Water, Acids, and Bases from the Perspective of Electrolytic Dissociation. The degree of dissociation of water at room temperature is 0.00000018% (Figure 2-69). Obviously, water is a weak electrolyte (Table 4), but how weak?

Imagine that all adults currently living in our world are married, except for just one man and one woman. The percentage of the singles will then be roughly the same as that of dissociated molecules in water or, in other words, its degree of dissociation α (Figure 2-69). Being a very weak electrolyte, pure water does not conduct electricity. Sea water is a good electrical conductor because it contains considerable quantities of dissolved salts that are dissociated to ions. Even tap water usually conducts electricity due to the presence of mineral salts. As we know from Volume 1, mineral salts present in naturally occurring waters can be removed by distillation. Distilled water is an insulator, not a conductor.

Figure 2-69. Electrolytic dissociation of water.

In Volume 1, we defined an acid as a substance whose molecules consist of one or more hydrogen atoms and an acid remainder. From the perspective of electrolytic dissociation, acids are compounds that dissociate to one or more protons, H⁺. So, what we previously called an acid remainder is, in fact, the anion that is left after a proton (H⁺) has been lost from an acid molecule. Monobasic acids lose one proton (Figure 2-70).

Figure 2-70. Dissociation of monobasic acids.


Dibasic and tribasic acids can lose two and three protons, respectively. Electrolytic dissociation of polybasic acids always occurs stepwise (Figure 2-71), the first step being considerably more facile than the second and the second more facile than the third.

Figure 2-71. Stepwise dissociation of H₂SO₄, a dibasic acid.


It is the degree of dissociation α of an acid that determines its strength. The higher the α value, the larger quantity of the H⁺ ions an acid produces on dissociation and the stronger the acid is.

The chemical equations presented in Figures 2-70 and 2-71 are simplified. A proton (H⁺) cannot and does not exist in aqueous solutions "as is", but rather forms a coordinate bond to the oxygen atom of a water molecule (Figure 2-72).

Figure 2-72. A proton forms a coordinate bond to the O atom of a water molecule.


Compare the scheme in Figure 2-72 with that in Figure 2-54 to see that the formation of the ammonium cation, NH₄⁺, from NH₃ and H⁺ and that of the hydronium ion, H₃O⁺, from H₂O and H⁺ have the same nature. In both cases a proton uses a lone electron pair on the N or O atom to form a covalent bond, N-H or O-H, respectively. Likewise, the equation of dissociation of water (Figure 2-69) would be more correct in the following form.

2 H₂O ⇄ H₃O⁺ + OH^-

Once a molecule of H₂O dissociates, the H⁺ produced and and another, undissociated water molecule come together to form a hydronium ion, H₃O⁺.

Although it will not hurt to always keep in mind that the H⁺ ion exists in water as H₃O⁺, you do not have to for this introductory course. We can, do, and in many cases should simplify things, so long as we do not compromise on the adequacy of scientific models. A good example of such a simplification is the way chemists write the formula of water, H₂O, in chemical equations and schemes. We know that there are no individual H₂O molecules in liquid water, ice, or even steam at the boiling point of water. As described above, molecules of water are involved in an extended network of hydrogen bonds (Figure 2-56). For the sake of clarity and simplicity, however, we conventionally write the formula of water as H₂O. Likewise, in our introductory course we simplify equations for dissociation of water and acids in water by writing the formula of the cation as H⁺ rather than H₃O⁺.

Let us now learn how to draw structures of oxygen-containing acids, conventionally referred to as oxoacids, such as H₂SO₄, H₂CO₃, HPO₃, H₃PO₄, etc. Figure 2-73 shows how to do that step by step.

1. Write the symbol of the central atom (S, P, C, etc.).

2. Connect the central atom to the total number of oxygen atoms by single bonds (single lines).

3. Connect each of the H atoms to an O atom with a single bond.

4. Make the single bonds double bonds for those O atoms that are not connected to an H atom. Remember that hydrogen atoms of oxoacids are bonded to oxygen atoms, not to the central atom (S, P, C, etc.).

Figure 2-73. Drawing structures of oxygen-containing acids step by step.


In Volume 1, we defined bases as compounds composed only of a metal atom and one or more OH groups attached to it. From the perspective of the electrolytic dissociation theory, a base is a chemical compound that produces OH^- on dissociation (Figure 2-74). The name of the OH^- anion is hydroxide. The higher the degree of dissociation of a base, the stronger the base is. The strongest conventional bases, NaOH, KOH, and Ba(OH)₂, are all strong electrolytes. An example of a medium strength base is Ca(OH)₂, which is a medium strength electrolyte.

Figure 2-74. Electrolytic dissociation of NaOH and Ba(OH)₂, strong electrolytes and, consequently, strong bases.


Note the important difference between the dissociation of Ba(OH)₂, a strong dibasic base (Figure 2-74, bottom) and H₂SO₄, a strong dibasic acid (Figure 2-71). As shown in Figure 2-71, the dissociation of H₂SO₄ is a two-step process, whereas Figure 2-74 suggests that both OH^- anions dissociate from Ba(OH)₂ simultaneously. The reason for this difference is quite interesting and educational.

Dibasic and polybasic acids always dissociate stepwise, each dissociation step being much less facile than the previous one. The element-hydrogen bond of even the strongest acids such as H₂SO₄, HNO₃, and HCl is covalent. When one H⁺ leaves a covalent molecule of H₂SO₄, the anion HSO₄^- is produced, in which the remaining H atom is still covalently bonded to the rest of the ion. The new species formed, HSO₄^-, and the initial acid, H₂SO₄ are two different compounds, naturally having different chemical properties. The propensity to lose a proton on dissociation is no exception, being always higher for the initial acid.

In contrast with acids that are covalent compounds, strong bases such as NaOH, KOH, and Ba(OH)₂ are ionic substances. The ions are already there in the crystal even before the base is exposed to water. Consequently, on dissociation of the first OH^- from a dibasic base such as Ca(OH)₂ or Ba(OH)₂ no new species is formed that would have a different predisposition to dissociation. Actually, electrolytic dissociation of Ca(OH)₂ and Ba(OH)₂ in water is still stepwise, but the second step is virtually as effortless as the first one. Being very small, the difference in the rates of the two steps is conventionally ignored.

For practicing purposes, use the values presented in Figure 2-52 to calculate the electronegativity difference between O and H (H₂SO₄, HNO₃) and between Cl and H (HCl) to see that the O-H and Cl-H bonds are polar covalent. Then do similar calculations for O and Na, O and K, and O and Ba to confirm that the metal - hydroxide bond in NaOH, KOH, and Ba(OH)₂ is ionic.

There are ways to generate hydroxide (OH^-) other than from a metal hydroxide. For example, hydroxide ions are produced on dissolution of ammonia (NH₃) in water. Although the molecule of ammonia does not dissociate, it has a lone electron pair that can be used to form a covalent bond to a proton, H⁺ (Figure 2-54). Consequently, a molecule of ammonia can and does abstract a proton from water or, as chemists say, deprotonates H₂O to give ammonium hydroxide (Figure 2-75). Since it is the OH^- that determines the basicity of an aqueous solution, NH₃ is a base. It is a weak base though, because the equilibrium in Figure 2-75 is shifted to the left and, consequently, the concentration of the OH^- in aqueous ammonia is quite low.

Figure 2-75. Generation of OH^- from ammonia and water.


2.5.7. The pH Scale. Acid-Base Indicators. Have Fun with Flowers, Vinegar, and Ammonia. The acidity of an aqueous solution is determined by the concentration of hydrogen ions (H⁺), the amount of the H⁺ per unit volume. A lemon tastes more sour than an orange because the concentration of H⁺ in lemons is higher than in oranges. The concentration of a molecule or ion in solution is denoted by enclosing its formula in square brackets. For example, [NaCl], [H⁺], and [OH^-] mean concentrations of NaCl, H⁺, and OH^-, respectively.

The most widely used unified scale for acidity, as well as basicity, is pH (pronounced /piːˈeɪtʃ/). The pH scale ranges from 0 to 14 and is logarithmic, pH = - log [H⁺]. It is not a problem if you do not know logarithms. The center of the pH scale is at 7, which represents the concentration of H⁺ in pure water (Figure 2-76). Since water is a very weak electrolyte, [H⁺] in pure water is very small and, consequently, water does not have a sour taste. In pure water [H⁺] = [OH^-] because the source of both is just water that produces the H⁺ and OH^- in equal quantities on dissociation (Figure 2-69). If [H⁺] = [OH^-] for a given solution, we say that the solution is neutral or that it has a neutral pH of 7.

Figure 2-76. The pH scale (source).


Lower pH values indicate a more acidic solution. Figure 2-76 shows that lemon (pH = 2) is more acidic than vinegar (pH = 3) but less acidic than concentrated HCl (pH = 0). Since the pH scale is logarithmic, a pH one unit lower than another represents 10 times the acidity. Going one, two, three, etc. pH units below a whole pH value corresponds to an increase in [H⁺] by a factor of 10, 100, 1,000, etc. Let us now work through a few examples showing how to calculate the difference in [H⁺] using the pH data in Figure 2-76.

1. The pH values for a lemon (lemon juice) and vinegar are 2 and 3, respectively. Which of the two is more acidic and by how much? As the pH of lemon juice is lower than that of vinegar, lemon is more acidic (higher [H⁺]). The pH value difference of 3 – 2 = 1 indicates that the concentration of H⁺ in lemon juice is 10 times higher than in vinegar.

2. How does the acidity of lemon juice compare to that of concentrated HCl? The difference in the pH values of lemon juice and concentrated hydrochloric acid is: 2 – 0 = 2, meaning that [H⁺] in concentrated HCl is 100 times higher than in lemon juice.

3. What is the difference in [H⁺] between water and vinegar? The pH of vinegar is 3 and that of water is 7. The difference of 4 pH units (7 – 3 = 4) indicates that [H⁺] in vinegar is 10 x 10 x 10 x 10 = 10,000 higher than in water.

If the pH value for a solution is above 7, the concentration of H⁺ in that solution is lower than in pure water. Such solutions are called alkaline. To make a solution more acidic (to lower its pH), we just need to add more acid to it. But how can we make a solution less acidic (more alkaline)? We need to add a base, so that the OH^- from the base sequesters the H⁺ to give water, H⁺ + OH^- = H₂O. High concentrations of H⁺ (very low pH, strongly acidic) and OH^- (very high pH, strongly alkaline) are hazardous to living organisms.

As we know from Volume 1, there are chemical compounds that change color in the presence of an acid or a base. Such materials, called acid-base indicators, are widely used to probe chemical solutions, foods and drinks, pharmaceuticals, and other products for acidity/alkalinity. Two simple indicators that we have learned about previously are phenolphthalein and litmus paper. You are expected to remember that phenolphthalein is colorless in acidic solutions and purple pink in alkaline ones. Litmus is red in the presence of an acid and blue in the presence of a base. There are also universal indicators that not only allow us to tell acids from bases, but also measure the pH of aqueous solutions. Particularly convenient are pH test strips (Figure 2-77). The color change on contact of a pH test strip with a solution or a wet sample is compared against the provided color chart to determine the pH.

Figure 2-77. pH test strips (source).


Consider getting some pH test strips from a pharmacy or store to determine the pH of many things you can find in your home, such as vinegar, baking soda solutions, soap, various fruits and vegetables, juices, detergents, cosmetic products, etc.

Many natural dyes are acid-base indicators that change color at different pH's. As shown in Video 2-7, the dye of red cabbage is one such indicator.

Video 2-7. Red cabbage pigment as a pH indicator (source).


Many natural dyes that give the color to flower petals are acid-base indicators. You may try vinegar (aqueous acetic acid) and household ammonia on various flowers of different colors. Maybe you will be amazed how the colors change when some flowers are sprayed with ammonia or vinegar. If you decide to try such experiments, use regular white vinegar and regular household ammonia. It is advised to avoid using scented ammonia solutions such as "lemon ammonia" and those containing detergents such as "foaming ammonia". The cheapest no-name brands are usually best suited for acid-base experiments with flowers. Note that although vinegar is a weak acid and ammonia is a weak base, both have a pretty strong smell. Avoid inhaling vapors of vinegar and especially ammonia when doing experiments. Carrying out the tests outdoors is best. You may spray the flowers or immerse them into your ammonia solution or vinegar. Important: the concentration of NH₃ in household ammonia varies in a broad range of 0.5% to 10% depending on the brand. The more concentrated household ammonia smells stronger. Do not use concentrated NH₃ (~30%) that has a pungent odor and is a strong eye irritant!

2.5.8. Exercises.

1. An electrolyte is a chemical compound that (a) readily gives off electrons; (b) dissolves in water to produce a solution that conducts electricity; (c) always dissociates to give H⁺; (d) always dissociates to give OH^-. Answer

2. Define degree of dissociation (α) and strong, medium strength, and weak electrolytes.

3. Almost all salts and acids are strong electrolytes. True or false? Answer

4. Barium sulfate (BaSO₄) is very poorly soluble in water and is therefore a weak electrolyte. True or false? Answer

5. Identify strong, medium strength, and weak electrolytes among the following compounds: Ca(OH)₂, HCl, NaNO₃, K₂SO₄, H₂SO₄, H₃PO₄, Na₃PO₄, NaOH, Ba(OH)₂, HNO₃, H₂CO₃, MgBr₂. [Hint: Refer to Table 4 and text above]

6. Describe the process of dissolution and dissociation of NaCl. Draw sketches to illustrate the process.

7. Explain why a molecule of water is a dipole. Do you think a molecule of HBr is a dipole? [Tip: Consider the shape of HBr and the type of chemical bond between the Br and H atoms]

8. On dissociation in water, dibasic acids (a) lose both of their protons simultaneously; (b) lose the first proton much more readily than the second one; (c) lose only one proton and never both; (d) lose the protons step-wise, with the second proton coming off almost as easily as the first one. Answer

9. On dissociation in water dibasic bases such as Ca(OH)₂ and Ba(OH)₂ (a) lose both of their OH^- ions simultaneously; (b) lose the first OH^- much more readily than the second one; (c) lose only one OH^- and never both; (d) lose the OH^- ions step-wise, but the second OH^- comes off almost as easily as the first one. Answer

10. Provide an explanation for the difference in the correct answers to questions 8 and 9 above.

11. The pH value of an aqueous solution went up by two units after some chemical treatment. Consequently, the concentration of (a) H⁺ decreased by a factor of 2; (b) OH^- decreased by a factor of 2; (c) H⁺ increased by a factor of 2; (d) H⁺ decreased by a factor of 20; (e) H⁺ decreased by a factor of 100; (f) H⁺ increased by a factor of 100; (g) OH^- increased by a factor of 10; (h) OH^- increased by a factor of 100. Answer

12. Write chemical equations for electrolytic dissociation of Na₂SO₄, HBr, KOH, Mg(NO₃)₂, CaCl₂, H₂CO₃, and H₂SO₄. Indicate in the equations that the dissociation is reversible and specify what side of the equation the equilibrium is shifted to.

13. How is OH^- produced in aqueous solutions of ammonia? Is NH₃ a strong base?

14. A laboratory technician distilled about 100 mL of water, placed it immediately in a clean 500 mL glass beaker, and measured its pH. The pH was 7, as expected for pure water. The technician then covered the beaker with a sheet of clean paper towel to protect the water from dust and went for lunch. When he returned to the lab an hour later, he decided to measure the pH again. To his surprise, the new pH measurement was 6, indicating that the water became slightly acidic. He thought that something went wrong and repeated the distillation the following day. The freshly distilled water had a pH of 7, as it was supposed to. As before, he poured the water into a clean beaker, covered it with a paper towel, and left for lunch. This time, he locked up the lab to make sure no one could enter it to play a joke on him by somehow slightly acidifying the distilled water. When he came back from lunch and tested the pH of the water, it was 6 again. The technician was totally puzzled. Can you explain the reproducible slight increase in the acidity of the distilled water on standing? Answer

15. Would you expect the following bonds to be prone to homolytic or heterolytic cleavage: (a) F-H; (b) Si-H; (c) F-F; (d) O-H; (e) C-H; (f) Te-H. Answer