ACIDS, BASES AND SALTS
Acids, bases and salts find widespread occurrence in nature. Hydrochloric acid present in the gastric juice is secreted by the lining of our stomach in a significant amount of 1.2-1.5 L/day and is essential for digestive processes. Acetic acid is known to be the main constituent of vinegar. Lemon and orange juices contain citric and ascorbic acids, and tartaric acid is found in tamarind paste. As most of the acids taste sour, the word ‘acid’ has been derived from a latin word ‘acidus’ meaning sour. Acids are known to turn blue litmus paper into red and liberate dihydrogen on reacting with some metals. Similarly, bases are known to turn red litmus paper blue, taste bitter and feel soapy. A common example of a base is washing soda used for washing purposes. When acids and bases are mixed in the right proportion they react with each other to give salts. Some commonly known examples of salts are sodium chloride, barium sulphate, sodium nitrate. Sodium chloride (common salt ) is an important component of our diet and is formed by reaction between hydrochloric acid and sodium hydroxide. It exists in solid state as a cluster of positively charged sodium ions and negatively charged chloride ions which are held together due to electrostatic interactions between oppositely charged species (Fig.7.10). The electrostatic forces between two charges are inversely proportional to dielectric constant of the medium. Water, a universal solvent, possesses a very high dielectric constant of 80. Thus, when sodium chloride is dissolved in water, the electrostatic interactions are reduced by a factor of 80 and this facilitates the ions to move freely in the solution. Also, they are wellseparated due to hydration with water molecules.
 Faraday was born near London into a family of very limited means. At the age of 14 he was an apprentice to a kind bookbinder who allowed Faraday to read the books he was binding. Through a fortunate chance he became laboratory assistant to Davy, and during 1813-4, Faraday accompanied him to the ontinent. During this trip he gained much from the experience of coming into contact with many of the leading scientists of the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories, and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first important work was on analytical chemistry. After 1821 much of his work was on electricity and agnetism and different electromagnetic phenomena. His ideas have led to the establishment of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to work alone and never had any assistant. He disseminated science in a variety of ways including his Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers. |
Comparing, the ionization of hydrochloric acid with that of acetic acid in water we find that though both of them are polar covalent molecules, former is completely ionized into its constituent ions, while the latter is only partially ionized (< 5%). The extent to which ionization occurs depends upon the strength of the bond and the extent of solvation of ions produced. The terms dissociation and ionization have earlier been used with different meaning. Dissociation refers to the process of separation of ions in water already existing as such in the solid state of the solute, as in sodium chloride. On the other hand, ionization corresponds to a process in which a neutral molecule splits into charged ions in the solution. Here, we shall not distinguish between the two and use the two terms interchangeably.
7.10.1 Arrhenius Concept of Acids and Bases
According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions H+(aq) and bases are substances that produce hydroxyl ions OH-(aq). The ionization of an acid HX (aq) can be represented by the following equations:
HX (aq) → H+(aq) + X- (aq)
or
HX(aq) + H2O(l) → H3O+(aq) + X-(aq)
A bare proton, H+ is very reactive and cannot exist freely in aqueous solutions. Thus, it bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H3O+ {[H (H2O)]+} (see box). In this chapter we shall use H+(aq) and H3O+(aq) interchangeably to mean the same i.e., a hydrated proton.
Similarly, a base molecule like MOH ionizes in aqueous solution according to the equation:
MOH(aq) → M+(aq) + OH-(aq)
The hydroxyl ion also exists in the hydrated form in the aqueous solution. Arrhenius concept of acid and base, however, suffers from the limitation of being applicable only to aqueous solutions and also, does not account for the basicity of substances like, ammonia which do not possess a hydroxyl group.
| Hydronium and Hydroxyl Ions |
| Hydrogen ion by itself is a bare proton with very small size (~10–-15 m radius) and intense electric field, binds itself with the water molecule at one of the two available lone pairs on it giving H3O+. This species has been detected in many compounds (e.g., H3O+Cl-) in the solid state. In aqueous solution the hydronium ion is further hydrated to give species like H5O2+, H7O3+ and H9O4+. Similarly the hydroxyl ion is hydrated to give several ionic species like H3O2 -, H5O3- and H7O4- etc. |
 |
7.10.2 The Brönsted-Lowry Acids and Bases
The Danish chemist, Johannes Brönsted and the English chemist, Thomas M. Lowry gave a more general definition of acids and bases. According to Brönsted-Lowry theory, acid is a substance that is capable of donating a
hydrogen ion H+ and bases are substances capable of accepting a hydrogen ion, H+. In short, acids are proton donors and bases are proton acceptors. Consider the example of dissolution of NH3 in H2O represented by the following equation:
The basic solution is formed due to the presence of hydroxyl ions. In this reaction, water molecule acts as proton donor and ammonia molecule acts as proton acceptor and are thus, called Lowry-Brönsted acid and base, respectively. In the reverse reaction, H+ is transferred from NH4+ to OH-. In this case, NH4+ acts as a Bronsted acid while OH- acted as a Brönsted base. The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH- is called the conjugate base of an acid H2O and NH4+ is called conjugate acid of the base NH3. If Brönsted acid is a strong acid then its conjugate base is a weak base and viceversa. It may be noted that conjugate acid has one extra proton and each conjugate base has one less proton.
Consider the example of ionization of hydrochloric acid in water. HCl(aq) acts as an acid by donating a proton to H2O molecule which acts as a base.
It can be seen in the above equation, that water acts as a base because it accepts the proton. The species H3O+ is produced when water accepts a proton from HCl. Therefore, Cl- is a conjugate base of HCl and HCl is the conjugate acid of base Cl-. Similarly, H2O is a conjugate base of an acid H3O+ and H3O+ is a conjugate acid of base H2O.
It is interesting to observe the dual role of water as an acid and a base. In case of reaction with HCl water acts as a base while in case of ammonia it acts as an acid by donating a proton.
 Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he travelled extensively and visited a number of research centers in Europe. In 1895 he was appointed professor of physics at the newly formed University of Stockholm, serving its rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic solutions. In 1899 he discussed the temperature dependence of reaction rates on the basis of an equation, now usually known as Arrhenius equation.
He worked in a variety of fields, and made important contributions to immunochemistry, cosmology, the origin of life, and the causes of ice age. He was the first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development of chemistry.
|
Problem 7.12
What will be the conjugate bases for the following Brönsted acids: HF, H2SO4 and HCO3-?
Solution
The conjugate bases should have one proton less in each case and therefore the corresponding conjugate bases are: F-, HSO4- and CO32- respectively.
Problem 7.13
Write the conjugate acids for the following Brönsted bases: NH2-, NH3 and HCOO-.
Solution
The conjugate acid should have one extra proton in each case and therefore the corresponding conjugate acids are: NH3, NH4+ and HCOOH respectively.
Problem 7.14
The species: H2O, HCO3-, HSO4- and NH3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and conjugate base.
Solution
The answer is given in the following Table:
Species
|
Conjugate acid
|
Conjugate base
|
H2O
|
H3O+
|
OH -
|
HCO3-
|
H2CO3
|
CO32-
|
HSO4-
|
H2SO4
|
SO42-
|
NH3
|
NH4+
|
NH2-
|
7.10.3 Lewis Acids and Bases
G.N. Lewis in 1923 defined an acid as a species which accepts electron pair and base which donates an electron pair. As far as bases are concerned, there is not much difference between Brönsted-Lowry and Lewis concepts, as the base provides a lone pair in both the cases. However, in Lewis concept many acids do not have proton. A typical example is reaction of electron deficient species BF3 with NH3.
BF3 does not have a proton but still acts as an acid and reacts with NH3 by accepting its lone pair of electrons. The reaction can be represented by,
BF3 + :NH3 → BF3:NH3
Electron deficient species like AlCl3, Co3+, Mg2+, etc. can act as Lewis acids while species like H2O, NH3, OH- etc. which can donate a pair of electrons, can act as Lewis bases.
Problem 7.15
Classify the following species into Lewis acids and Lewis bases and show how
these act as such:
(a) HO- (b)F- (c) H+ (d) BCl3
Solution
(a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:OH- ).
(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.
(c) A proton is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
(d) BCl3 acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules.
7.11 IONIZATION OF ACIDS AND BASES
Arrhenius concept of acids and bases becomes useful in case of ionization of acids and bases as mostly ionizations in chemical and biological systems occur in aqueous medium. Strong acids like perchloric acid (HClO4), hydrochloric acid (HCl), hydrobromic acid (HBr), hyrdoiodic acid (HI), nitric acid (HNO3) and sulphuric acid (H2SO4) are termed strong because they are almost completely dissociated into their constituent ions in an aqueous medium, thereby acting as proton (H+) donors. Similarly, strong bases like lithium hydroxide (LiOH), sodium hydroxide (NaOH), potassium hydroxide (KOH), caesium hydroxide (CsOH) and barium hydroxide Ba(OH)2 are almost completely dissociated into ions in an aqueous medium giving hydroxyl ions, OH-. According to Arrhenius concept they are strong acids and bases as they are able to completely dissociate and produce H3O+and OH- ions respectively in the medium. Alternatively, the strength of an acid or base may also be gauged in terms of Brönsted- Lowry concept of acids and bases, wherein a strong acid means a good proton donor and a strong base implies a good proton acceptor. Consider, the acid-base dissociation equilibrium of a weak acid HA,
In section 7.10.2 we saw that acid (or base) dissociation equilibrium is dynamic involving a transfer of proton in forward and reverse directions. Now, the question arises that if the equilibrium is dynamic then with passage of time which direction is favoured? What is the driving force behind it? In order to answer these questions we shall deal into the issue of comparing the strengths of the two acids (or bases) involved in the dissociation equilibrium. Consider the two acids HA and H3O+ present in the above mentioned acid-dissociation equilibrium. We have to see which amongst them is a stronger proton donor. Whichever exceeds in its tendency of donating a proton over the other shall be termed as the stronger acid and the equilibrium will shift in the direction of weaker acid. Say, if HA is a stronger acid than H3O+, then HA will donate protons and not H3O+, and the solution will mainly contain A- and H3O+ ions. The equilibrium moves in the direction of formation of weaker acid and weaker base because the stronger acid donates a proton to the stronger base.
It follows that as a strong acid dissociates completely in water, the resulting base formed would be very weak i.e., strong acids have very weak conjugate bases. Strong acids like perchloric acid (HClO4), hydrochloric acid (HCl), hydrobromic acid (HBr), hydroiodic acid (HI), nitric acid (HNO3) and sulphuric acid (H2SO4) will give conjugate base ions ClO4 -, Cl, Br-, I-, NO3- and HSO4- , which are much weaker bases than H2O. Similarly a very strong base would give a very weak conjugate acid. On the other hand, a weak acid say HA is only partially dissociated in aqueous medium and thus, the solution mainly contains undissociated HA molecules. Typical weak acids are nitrous acid (HNO2), hydrofluoric acid (HF) and acetic acid (CH3COOH). It should be noted that the weak acids have very strong conjugate bases. For example, NH2-, O2- and H- are very good proton acceptors and thus, much stronger bases than H2O.
Certain water soluble organic compounds like phenolphthalein and bromothymol blue behave as weak acids and exhibit different colours in their acid (HIn) and conjugate base (In- ) forms.
Such compounds are useful as indicators in acid-base titrations, and finding out H+ ion concentration.
7.11.1 The Ionization Constant of Water and its Ionic Product
Some substances like water are unique in their ability of acting both as an acid and a base. We have seen this in case of water in section 7.10.2. In presence of an acid, HA it accepts a proton and acts as the base while in the presence of a base, B- it acts as an acid by donating a proton. In pure water, one H2O molecule donates proton and acts as an acid and another water molecules accepts a proton and acts as a base at the same time. The following equilibrium exists:
The dissociation constant is represented by,
K = [H3O+] [OH-] / [H2O] ——————————————————————————————-(7.26)
The concentration of water is omitted from the denominator as water is a pure liquid and its concentration remains constant. [H2O] is incorporated within the equilibrium constant to give a new constant, Kw, which is called the ionic product of water.
Kw = [H+][OH-] ————————————————————————————————(7.27)
The concentration of H+ has been found out experimentally as 1.0 x 10-7 M at 298 K. And, as dissociation of water produces equal number of H+ and OH- ions, the concentration of hydroxyl ions, [OH-] = [H+] = 1.0 x 10-7 M.
Thus, the value of Kw at 298K,
Kw = [H3O+][OH-] = (1 x 10-7)2 = 1 x 10-14 M2———————————————————————————(7.28)
The value of Kw is temperature dependent as it is an equilibrium constant. The density of pure water is 1000 g / L and its molar mass is 18.0 g /mol. From this the molarity of pure water can be given as, [H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.
Therefore, the ratio of dissociated water to that of undissociated water can be given as:
10-7/ (55.55) = 1.8 x 10-9 or ~ 2 in 10-9
(thus, equilibrium lies mainly towards undissociated water)
We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the H3O+ and OH- concentrations:
Acidic: [H3O+] > [OH- ]
Neutral: [H3O+] = [OH- ]
Basic : [H3O+] < [OH-]
7.11.2 The pH Scale
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the pH scale. The pH of a solution is defined as the negative logarithm to base 10 of the activity (aH+) of hydrogen ion. In dilute solutions (< 0.01 M), activity of hydrogen ion (H+) is equal in magnitude to molarity represented by [H+]. It should be noted that activity has no units and is defined as:
aH+ = [H+] / mol L-1
From the definition of pH, the following can be written,
pH = – log aH+ = – log {[H+] / mol L-1}
Thus, an acidic solution of HCl (10-2 M) will have a pH = 2. Similarly, a basic solution of NaOH having [OH-] =10-4 M and [H3O+] = 10-10 M will have a pH = 10. At 25 °C, pure water has a concentration of hydrogen ions, [H+] = 10-7 M. Hence, the pH of pure water is given as:
pH = -log(10-7) = 7
Acidic solutions possess a concentration of hydrogen ions, [H+] > 10-7 M, while basic solutions possess a concentration of hydrogen ions, [H+] < 10-7 M. thus, we can summarise that
Acidic solution has pH < 7 Basic solution has pH > 7 Neutral solution has pH = 7
Now again, consider the equation (7.28) at 298 K
Kw = [H3O+] [OH-] = 10-14
Taking negative logarithm on both sides of equation, we obtain
-log Kw = -log {[H3O+] [OH+]} = – log [H3O+] – log [OH-] = – log 10-14
pKw = pH + pOH = 14 —————————————————————————————–(7.29)
Note that although Kw may change with temperature the variations in pH with temperature are so small that we often ignore it.
pKw is a very important quantity for aqueous solutions and controls the relative concentrations of hydrogen and hydroxyl ions as their product is a constant. It should be noted that as the pH scale is logarithmic, a change in pH by just one unit also means change in [H+] by a factor of 10. Similarly, when the hydrogen ion concentration, [H+] changes by a factor of 100, the value of pH changes by 2 units. Now you can realise why the change in pH with temperature is often ignored.
Measurement of pH of a solution is very essential as its value should be known when dealing with biological and cosmetic applications. The pH of a solution can be found roughly with the help of pH paper that has different colour in solutions of different pH. Now-a-days pH paper is available with four strips on it. The different strips have different colours (Fig. 7.11) at the same pH. The pH in the range of 1-14 can be determined with an accuracy of ~0.5 using pH paper.
For greater accuracy pH meters are used. pH meter is a device that measures the pH-dependent electrical potential of the test solution within 0.001 precision. pH meters of the size of a writing pen are now available in the market. The pH of some very common substances are given in Table 7.5 (page 212).
Problem 7.16
The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10-3M. what is its pH ?
Solution
pH = – log[3.8 x 10-3 ]
= – {log[3.8] + log[10-3]}
= – {(0.58) + (- 3.0)} = – { – 2.42} = 2.42
Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic.
Problem 7.17
Calculate pH of a 1.0 x 10-8 M solution of HCl.
Solution
Kw = [OH-][H3O+] = 10-14
Let, x = [OH-] = [H3O+] from H2O. The H3O+ concentration is generated (i) from the ionization of HCl dissolved i.e.,
and (ii) from ionization of H2O. In these very dilute solutions, both sources of H3O+ must be considered:
[H3O+] = 10-8 + x
Kw = (10-8 + x)(x) = 10-14
or x2 + 10-8 x – 10-14 = 0
[OH- ] = x = 9.5 x 10-8
So, pOH = 7.02 and pH = 6.98
7.11.3 Ionization Constants of Weak Acids
Consider a weak acid HX that is partially ionized in the aqueous solution. The equilibrium can be expressed by:
Here, c = initial concentration of the undissociated acid, HX at time, t = 0. α = extent up to which HX is ionized into ions. Using these notations, we can derive the equilibrium constant for the above discussed aciddissociation equilibrium:
Ka = c2α2 / c(1-α) = cα2 / 1-α
Ka is called the dissociation or ionization constant of acid HX. It can be represented alternatively in terms of molar concentration as follows,
Ka = [H+][X-] / [HX] —————————————————————————————(7.30)
At a given temperature T, Ka is a measure of the strength of the acid HX i.e., larger the value of Ka, the stronger is the acid. Ka is a dimensionless quantity with the understanding that the standard state concentration of all species is 1M. The values of the ionization constants of some selected weak acids are given in Table 7.6.
Table 7.6 The Ionization Constants of Some
Selected Weak Acids (at 298K)
Acid
|
Ionization Constant,Ka
|
| Hydrofluoric Acid (HF) | 3.5 x 10-4 |
| Nitrous Acid (HNO2) | 4.5 x 10-4 |
| Formic Acid (HCOOH) | 1.8 x 10-4 |
| Niacin (C5H4NCOOH) | 1.5 x 10-5 |
| Acetic Acid (CH3COOH) | 1.74 x 10-5 |
| Benzoic Acid (C6H5COOH) | 6.5 x 10-5 |
| Hypochlorous Acid (HCIO) | 3.0 x 10-8 |
| Hydrocyanic Acid (HCN) | 4.9 x 10-10 |
| Phenol (C6 H5OH) | 1.3 x 10-10 |
The pH scale for the hydrogen ion concentration has been so useful that besides pKw, it has been extended to other species and quantities. Thus, we have:
pKa = -log (Ka) —————————————————————————————–(7.31)
Knowing the ionization constant, Ka of an acid and its initial concentration, c, it is possible to calculate the equilibrium concentration of all species and also the degree of ionization of the acid and the pH of the solution.
A general step-wise approach can be adopted to evaluate the pH of the weak
electrolyte as follows:
Step 1. The species present before dissociation are identified as Brönsted-Lowry
acids / bases.
Step 2. Balanced equations for all possible reactions i.e., with a species acting both as acid as well as base are written.
Step 3. The reaction with the higher Ka is identified as the primary reaction whilst the other is a subsidiary reaction.
Step 4. Enlist in a tabular form the following values for each of the species in the primary reaction
(a) Initial concentration, c.
(b) Change in concentration on proceeding to equilibrium in terms of α, degree of ionization.
(c) Equilibrium concentration.
Step 5. Substitute equilibrium concentrations into equilibrium constant equation for principal reaction and solve for α.
Step 6. Calculate the concentration of species in principal reaction.
Step 7. Calculate pH = – log[H3O+]
The above mentioned methodology has been elucidated in the following examples.
Problem 7.18
The ionization constant of HF is 3.2 x 10-4. Calculate the degree of
dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H3O+, F- and HF) in the solution and its pH.
Solution
The following proton transfer reactions are possible:
1)
Ka = 3.2 x 10-4
2)
Ka = 1.0 x 10-14
As Ka >> Kw, [1] is the principle reaction.
Substituting equilibrium concentrations in the equilibrium reaction for principal reaction gives:
Ka = (0.02α)2 / (0.02 – 0.02α) = 0.02 α2 / (1 -α) = 3.2 x 10-14
We obtain the following quadratic equation:
α2 + 1.6 x 10-2 α – 1.6 x 10-2 = 0
The quadratic equation in α can be solved and the two values of the roots are:
α = + 0.12 and – 0.12
The negative root is not acceptable and hence,
α = 0.12
This means that the degree of ionization, α = 0.12, then equilibrium concentrations of other species viz., HF, F - and H3O+ are given by:
[H3O+ ] = [F - ] = cα = 0.02 0.12
= 2.4 x 10-3 M
[HF] = c(1 – α) = 0.02 (1 – 0.12) = 17.6 10-3 M
pH = – log[H+] = -log(2.4 x 10-3 ) = 2.62
Problem 7.19
The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of species H+, A- and HA at equilibrium. Also, determine the value of Ka and pKa of the monobasic acid.
Solution
pH = – log [H+ ]
Therefore, [H+ ] = 10 -pH = 10-4.50 = 3.16 x 10 -5
[H+] = [A-] = 3.16 x 10-5
Thus, Ka = [H+][A+] / [HA]
[HA]eqlbm = 0.1 -(3.16 x 10-5) = 0.1
Ka = (3.16 x 10-5)2 / 0.1 = 1.0 x 10-8
pKa = – log(10-8) = 8
Alternatively, ‘Percent dissociation’ is another useful method for measure of strength of a weak acid and is given as:
Percent dissociation = [HA]dissociated/[HA]initial x 100% ———————————————————————————–(7.32)
Problem 7.20
Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization constant of the acid is 2.5 x 10-5.Determine the percent issociation of HOCl.
Solution
Ka = {[H3O+][ClO-] / [HOCl]}
= x2 / (0.08 – x)
As x<<0.08, therefore 0.08 – x = 0.08
x2 / 0.08 = 2.5 x 10-5
x2 = 2.0 x 10-6, thus, x = 1.41 x 10-34
[H+] = 1.41 x 10-3 M.
Therefore,
Percent dissociation = {[HOCl]dissociated / [HOCl]initial } x 100 = 1.41 x 10-3 / 0.08 = 1.76 %.
pH = -log(1.41 x 10-3) = 2.85.
7.11.4 Ionization of Weak Bases
The ionization of base MOH can be represented by equation:
In a weak base there is partial ionization of MOH into M+ and OH-, the case is similar to that of acid-dissociation equilibrium. The equilibrium constant for base ionization is called base ionization constant and is represented by Kb. It can be expressed in terms of concentration in molarity of various species in equilibrium by the following equation:
Kb = [M+][OH-] / [MOH] —————————————————————————————-(7.33)
Alternatively, if c = initial concentration of base and α = degree of ionization of base i.e. the extent to which the base ionizes. When equilibrium is reached, the equilibrium constant can be written as:
Kb = (cα)2 / c (1-α) = cα2 / (1-α)
The values of the ionization constants of some selected weak bases, Kb are given in Table 7.7.
Table 7.7 The Values of the Ionization Constant of Some Weak Bases a 298 K
| Base | Kb |
| Dimethylamine, (CH3)2NH | 5.4 x 10-4 |
| Triethylamine, (C2H5)3N | 6.45 x 10-5 |
| Ammonia, NH3 or NH4OH | 1.77 x 10-5 |
| Quinine, (A plant product) | 1.10 x 10-6 |
| Pyridine, C5H5N | 1.77 x 10-9 |
| Aniline, C6H5NH2 | 4.27 x 10-10 |
| Urea, CO (NH2)2 | 1.3 x 10-14 |
Many organic compounds like amines are weak bases. Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by another group. For example, methylamine, codeine, quinine and nicotine all behave as very weak bases due to their very small Kb. Ammonia produces OH- in aqueous solution:
The pH scale for the hydrogen ion concentration has been extended to get:
pKb = -log (Kb) ———————————————————————————–(7.34)
Problem 7.21
The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.
Solution
From the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:
[H+] = antilog (-pH) = antilog (-9.7) = 1.67 x 10-10
[OH-] = Kw / [H+] = 1 x 10-14 / 1.67 x 10-10 = 5.98 x 10-5
The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004M. Thus,
Kb = [NH2NH3 +][OH-] / [NH2NH2] = (5.98 x 10-5)2 / 0.004 = 8.96 x 10-7
pKb = -logKb = -log(8.96 x 10-7) = 6.04.
Problem 7.22
Calculate the pH of the solution in which 0.2M NH4Cl and 0.1M NH3 are present. The pKbof ammonia solution is 4.75.
Solution
The ionization constant of NH3, Kb = antilog (-pKb) i.e.
Kb = 10-4.75 = 1.77 x 10-5 M
Kb = [NH4+][OH-] / [NH3] = (0.20 + x)(x) / (0.1 – x) = 1.77 x 10-5
As Kb is small, we can neglect x in comparison to 0.1M and 0.2M. Thus,
[OH-] = x = 0.88 x 10-5
Therefore, [H+] = 1.12 x 10-9
pH = – log[H+] = 8.95.
7.11.5 Relation between Ka and Kb
As seen earlier in this chapter, Ka and Kb represent the strength of an acid and a base, respectively. In case of a conjugate acid-base pair, they are related in a simple manner so that if one is known, the other can be deduced. Considering the example of NH4+ and NH3we see,
Ka = [H3O+][ NH3] / [NH4+] = 5.6 x 10-10
Kb =[ NH4+][ OH-] / NH3 = 1.8 x 10-5
Kw = [H3O+][ OH- ] = 1.0 x 10-14 M
Where, Ka represents the strength of NH4+ as an acid and Kb represents the strength of NH3 as a base.
It can be seen from the net reaction that the equilibrium constant is equal to the product of equilibrium constants Ka and Kb for the reactions added. Thus,
Ka Kb = {[H3O+][ NH3] / [NH4+ ]} {[NH4+ ][ OH-] / [NH3]}
= [H3O+][ OH-] = Kw
= (5.6×10-10) (1.8 x 10-5) = 1.0 x 10-14 M
This can be extended to make a generalisation. The equilibrium constant for
a net reaction obtained after adding two (or more) reactions equals the product of the equilibrium constants for individual reactions:
KNET= K1 x K2 x …….…… ————————————————————————————————(3.35)
Similarly, in case of a conjugate acid-base pair,
Ka x Kb = Kw ——————————————————————————————-(7.36)
Knowing one, the other can be obtained. It should be noted that a strong acid will have a weak conjugate base and vice-versa. Alternatively, the above expression Kw = Ka x Kb, can also be obtained by considering the base-dissociation equilibrium reaction:
Kb = [BH+][OH-] / [B]
As the concentration of water remains constant it has been omitted from the
denominator and incorporated within the dissociation constant. Then multiplying and dividing the above expression by [H+], we get:
Kb = [BH+][OH-][H+] / [B][H+] ={[ OH-][H+]}{[BH+] / [B][H+]} = Kw / Ka
or Ka Kb = Kw
It may be noted that if we take negative logarithm of both sides of the equation, then pK values of the conjugate acid and base are related to each other by the equation:
pKa + pKb = pKw = 14 (at 298K)
Problem 7.23
Determine the degree of ionization and pH of a 0.05M of ammonia solution. The ionization constant of ammonia can be taken from Table 7.7. Also, calculate the ionization constant of the conjugate acid of ammonia.
Solution
The ionization of NH3 in water is represented by equation:
We use equation (7.33) to calculate hydroxyl ion concentration,
[OH-] = c α = 0.05 α
Kb = 0.05 α2 / (1 – α)
The value of α is small, therefore the quadratic equation can be simplified by neglecting α in comparison to 1 in the denominator on right hand side of the equation, Thus,
Kb = c α2 or α = √ (1.77 x 10-5 / 0.05)
= 0.018.
[OH-] = c α = 0.05 x 0.018 = 9.4 x 10-4M.
[H+] = Kw / [OH-] = 10-14 / (9.4 x 10-4) = 1.06 x 10-11
pH = -log(1.06 x 10-11) = 10.97.
Now, using the relation for conjugate cid-base pair,
Ka x Kb = Kw
using the value of Kb of NH3 from Table 7.7.
We can determine the concentration of conjugate acid NH4+
Ka = Kw / Kb = 10-14 / 1.77 x 10-5 = 5.64 x 10-10.
7.11.6 Di- and Polybasic Acids and Di- and Polyacidic Bases
Some of the acids like oxalic acid, sulphuric acid and phosphoric acids have more than one ionizable proton per molecule of the acid. Such acids are known as polybasic or polyprotic acids. The ionization reactions for example for a dibasic acid H2X are represented by the equations:
And the corresponding equilibrium constants are given below:
Ka1 = {[H+][HX-]} / [H2X]
and Ka2 = {[H+][X2-]} / [HX-]
Here, Ka1 and Ka2 are called the first and second ionization constants respectively of the acid H2 X. Similarly, for tribasic acids like H3PO4 we have three ionization constants. The values of the ionization constants for some common polyprotic acids are given in Table 7.8.
Table 7.8 The Ionization Constants of Some
Common Polyprotic Acids (298K)
| Acid | Ka1 | Ka2 | Ka3 |
| Oxalic Acid | 5.9 x 10-2 | 6.4 x 10-5 | |
| Ascorbic Acid | 7.4 x 10-4 | 1.6 x 10-12 | |
| Sulphurous Acid | 1.7 x 10-2 | 6.4 x 10-8 | |
| Sulphuric Acid | Very large | 1.2 x 10-2 | |
| Carbonic Acid | 4.3 x 10-7 | 5.6 x 10-11 | |
| Citric Acid | 7.4 x 10-4 | 1.7 x 10-5 | 4.0 x 10-7 |
| Phosphoric Acid | 7.5 x 10-3 | 6.2 x 10-8 | 4.2 x 10-13 |
It can be seen that higher order ionization constants (Ka2, Ka3 ) are smaller than the lower order ionization constant ( Ka1 ) of a polyprotic acid. The reason for this is that it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces. This can be seen in the case of removing a proton from the uncharged H2CO3 as compared from a negatively charged HCO3-.
Similarly, it is more difficult to remove a proton from a doubly charged HPO42- anion as compared to H2PO4-.
Polyprotic acid solutions contain a mixture of acids like H2A, HA- and A2- in case of a diprotic acid. H2A being a strong acid, the primary reaction involves the dissociation of H2A, and H3O+ in the solution comes mainly from the first dissociation step.
7.11.7 Factors Affecting Acid Strength
Having discussed quantitatively the strengths of acids and bases, we come to a stage where we can calculate the pH of a given acid solution. But, the curiosity rises about why should some acids be stronger than others? What factors are responsible for making them stronger? The answer lies in its being a complex phenomenon. But, broadly speaking we can say that the extent of dissociation of an acid depends on the strength and polarity of the H-A bond.
In general, when strength of H-A bond decreases, that is, the energy required to break the bond decreases, HA becomes a stronger acid. Also, when the H-A bond becomes more polar i.e., the electronegativity difference between the atoms H and A increases and there is marked charge separation, cleavage of the bond becomes easier thereby increasing the acidity.
But it should be noted that while comparing elements in the same group of the periodic table, H-A bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, H-A bond strength decreases and so the acid strength increases. For example,
Similarly, H2S is stronger acid than H2O.
But, when we discuss elements in the same row of the periodic table, H-A bond polarity becomes the deciding factor for determining the acid strength. As the electronegativity of A increases, the strength of the acid also increases. For example,
Typically, we recognize acids and bases by their simple properties, such as
taste. We know that a lemon is sour, so it is acidic. Bases tend to taste
bitter. Acids and bases also change the color of certain dyes, such as
phenolphthalein and litmus. Acids change litmus treated paper from blue to red.
Acids change basic phenolpthalein from red to colorless. Bases change litmus
treated paper from from red to blue and phenolphthalein from colorless to pink.
Acids and bases neutralize the action of each other. This is why we take
antacids for stomachaches, because the antacid is a base, and neutralizes the
acid in the stomach.
Arrhenius Concept of Acids and Bases
This guy named Arrhenius concocted the first successful concept of acids and
bases. He did this by defining acids and bases according to the effect these
substances have on water. The Arrhenius concept of acids and bases is as
follows: an acid is a substance that when dissolved in water increases the
concentration of the hydrogen ion, H+. A base is a substance that
when dissolved in water increases the concentration of the hydroxide ion,
OH-.
The hydrogen ion, is not just a bare proton, it is a proton bonded to a water
molecule, H2O. This results in a hydronium ion,
H3O+.
In Arrhenius's thoery, something that is a strong acid is a substance that
completely ionizes in aqueous solution to give a hydronium ion,
H3O+, and an anion. An anion is a negatively charged ion.
An example of a strong acid is perchloric acid:
What is going on above is that we have perchloric acid in an aqueous soluion.
This perchloric acid ionizes entirely and results in an hydronium ion and a
perchlorate anion. Some other examples of strong acids would be: HI, HBr, HCl,
HNO3, and H2SO4.
Now on to bases...A strong base is something that completely ionizes in
aqueous solution to give a hydroxide ion and a cation. A cation is a positively
charged ion. Strong bases are most of the hydroxides of Group IA elements and
Group IIA elements including LiOH, NaOH, KOH, Ca(OH)2,
Sr(OH)2 and Ba(OH)2.
Many of the acids and bases that we encounter in our everyday lives are not
strong acids, they are considered weak. Weak acids do not completely ionize in
solution, but exist in equilibrium. Let's look at the reaction for acetic
acid:
Bronsted-Lowery Concept of Acids and Bases
The Bronsted-Lowery concept of acids and bases is that acid-base reactions
can be seen as proton-transfer reactions. This results in acids and bases being
able to be defined in terms of this proton (H+) transfer. According
to the Bronsted-Lowery concept, acids donate a proton in a proton-transfer
reactions. Bases accept the proton in a proton-transfer equation. As an example,
lets look at the reaction of hydrochloric acid with ammonia. When we write it as
an ionic equation we get:
which reduces to:
because
there is two Cl-(aq) one each side. We now have the net ionic
equation after we cancel out the "spectator ions"(Cl-).
What happens in this reaction in aqueous solution is a proton transfer.
According to the Bronsted-Lowery concept, acids donate a proton in a
proton-transfer reactions. Bases accept the proton in a proton-transfer
equation. As an example, lets look at the reaction of hydrochloric acid with
ammonia shown above. What happens in this reaction in aquesous solution is that
a poton is transferred from H3O+ to NH3. This
results in H3O+ losing a (H+), resulting in
H2O. The NH3 gains the transferred proton, resulting in
NH4+. We call H3O+ the proton donor,
or acid. We call NH3 the proton acceptor, or base.
The
Bronsted-Lowery concept defines something as either an acid or base depending on
its function in the acid-base (proton transfer) reaction. Some things can act as
either an acid or a base. These are called amphiprotic species, they can either
lose or gain a proton, depending on the other reactant. An example of an
amphiprotic species would be HCO3-. In the presence of
OH-, it acts as an acid. In the presence of HF it acts as a base.
Water is also amphiprotic, as are most anions with ionizable hydrogens and
certain solvents. Water as an amphiprotic species is very important to the
acid-base reactions.
In the Bronsted-Lowery concept we have found that:
1. A base is a species that accepts protons, while an acid is a species that
dontates protons.
2. Acids and bases can be ions as well as molecular substances.
3. Some species can act as either acids or bases, depending on what the other
reactant is.
Lewis Concept of Acids and Bases
The Lewis concept of acids is generalized to include reactions of acidic and
basic oxides and many other reactions. A Lewis acid is something that can form a
covalent bond by accepting an electron air from another species. A Lewis base is
something that can form a covalent bond by donating an electron pair to
something else. The Lewis and Bronsted-Lowery concepts are different ways of
looking at the same chemical reactions. Here is a reaction in which an electron
pair is transferred. The proton (H+) is electron pair acceptor,
alewis acid. NH3 has a lone pair of electrons and is a Lewis Base.
Whether or not an aqueous solution is neutral, acidic or basic depends on the
hydrogen-ion concentration. We give the acidity of an aqueous solution in terms
of the pH. pH is defined as the negative logarithm of the molar hydrogen-ion
concentration. A pH of 7 means that a solution is neutral. A pH of below 7 means
that a solution is acidic; a pH of above 7 means that a solution is basic.
For example, let's say that we have a glass of frosty orange juice. This
orange juice has a hydrogen-ion concentration of 2.9 x 10-4 M. What
is the orange juice's pH?
The pH of this solution is less than 7 so this orange juice is acidic.
We can also find pH by solving for the hydroxide-ion concentration of a
solution. The measure of the hydroxide-ion concentration is called pOH.
Since we know that the pH scale goes from 0 to 14, we find that:
pH + pOH
= 14
Let's say that we want to find the pH of an ammonia solution that has a
hydroxide-ion concentration of 1.9 x 10-3 M. We start by finding the
pOH.
Now we want to find the pH by subtracting:
Equilibrium with Acids and Bases
Have you read the section about equilibrium yet? If you haven't this most
likely won't make any sense to you. If you have, lets join in on the fun of
acid-base equilibrium.
Remember Kc from the equilibrium section? It's back, and more
useful than ever. Now to distinguish between the Kc of acids and
bases we use Ka and Kb. (a for acids and b for bases) The
equilibrium that is calculated in acids is usually the disaccociation of the
H+ ions and the rest of the molecule. The weak acids and bases are
the only ones that have Ka's and Kb's because in the
strong acids dissociation is very close to 100%.
| Substance | Formula | Ka |
|---|---|---|
| Acetic Acid | HC2H3O2 | 1.7 x 10-5 |
| Benzoic Acid | HC7H5O2 | 6.3 x 10-5 |
| Boric Acid | H3BO3 | 5.9 x 10-10 |
| Carbonic Acid | H2CO3 | 4.3 x 10-7 |
| HCO3- | 4.8 x 10-11 | |
| Cyanic Acid | HCNO | 3.5 x 10-4 |
| Formic Acid | HCNO2 | 1.7 x 10-4 |
| Hydrocyanic Acid | HCN | 4.9 x 10-10 |
| Hydrofluric Acid | HF | 6.8 x 10 |
| Hydrogen Sulfate ion | HSO4- | 1.1 x 10-2 |
| Hydrogen Sulfide | H2S | 8.9 x 10-8 |
| HS- | 1.2 x 10-13 | |
| Hypochlorous acid | HClO | 3.5 x 10-8 |
| Nitrous Acid | HNO2 | 4.5 x 10-4 |
| Oxalic Acid | H2C2O4 | 5.6 x 10-2 |
| HC2O4- | 5.1 x 10-5 | |
| Phosphoric Acid | H3PO4 | 6.9 x 10-3 |
| H2PO4- | 6.2 x 10-8 | |
| HPO4-2 | 4.8 x 10-13 | |
| Phosphorous Acid | H2PHO3 | 1.6 x 10-2 |
| HPHO3- | 7 x 10-7 | |
| Propionic Acid | HC3H5O2 | 1.3 x 10-5 |
| Pyruvic Acid | HC3H3O3 | 1.4 x 10-4 |
| Sulfurous Acid | H2SO3 | 1.3 x 10-2 |
| HSO3- | 6.3 x 10-8 |
The base Kb's are as follows:
| Substance | Formula | Kb |
|---|---|---|
| Ammonia | NH3 | 1.8 x 10-5 |
| Aniline | C6H5NH2 | 4.2 x 10-10 |
| Dimethylamine | (CH3)3NH | 5.1 x 10-4 |
| Ethylamine | C2H5NH2 | 4.7 x 10-4 |
| Hydrazine | N2H4 | 1.7 x 10-6 |
| Hydroxylamine | NH2OH | 1.1 x 10-8 |
| Methylamine | CH3NH2 | 4.4 x 10-4 |
| Pyridine | C5H5N | 1.4 x 10-9 |
| Urea | NH2CONH2 | 1.5 x 10-14 |
Examples
Weak Acid Example:
Calculate the pH of a 0.100 M solution of HClO
| HClO | <--> | H+ | + | ClO- |
| .100 - x | x | x |
| Ka = | __x2__ |
| .100 - x |
The x in the denominator can be dropped because Ka/M is less than
10-3. If Ka/M is greater than 10-3 you have to
use the quadratic formula to solve the equation.
Therefore:
| 3.5 x 10-8 = | __x2__ |
| .100 |
3.5 x 10-9 = x2
x = 5.9 x 10-5
[H+] = 5.9 x 10-5 M
[ClO-] = 5.9 x
10-5 M
[HClO] = .100 M - 5.9 x 10-5 ~= .100 M
pH = -log(5.9 x 10-5) = 4.2
Weak Base Example:
What is the concentration of OH- of a .20 molar solution of
aniline?
| C6H5NH2 | <--> | C6H5NH3+ | + | OH- |
| .20 - x | x | x |
| Kb = | __x2__ |
| .20 - x |
The x in the denominator can be dropped because Kb/M is less than
10-3. If Kb/M is greater than 10-3 you have to
use the quadratic formula to solve the equation.
Therefore:
| Kb = | __x2__ |
| .20 |
x2 = 8.4 x 10-11
x = 9.2 x 10-6
[OH-] = 9.2 x 10-6
No comments:
Post a Comment