12th Chemistry - Vol 2 - English Medium - PDF

GOVERNMENT OF TAMIL NADU HIGHER SECONDARY SECOND YEAR CHEMISTRY VOLUME - II A publication under Free Textbook Programme of Government of Tamil Nadu...

GOVERNMENT OF TAMIL NADU HIGHER SECONDARY SECOND YEAR CHEMISTRY VOLUME - II A publication under Free Textbook Programme of Government of Tamil Nadu Department of School Education Untouchability is Inhuman and a Crime Introduction Pages.indd 1 16-12-2022 19:32:16  Government of Tamil Nadu First Edition - 2019 Revised Edition - 2020, 2021, 2022, 2023 Reprint - 2024 (Published under new syllabus) NOT FOR SALE Content Creation The wise possess all State Council of Educational Research and Training © SCERT 2019 Printing & Publishing Tamil NaduTextbook and Educational Services Corporation www.textbooksonline.tn.nic.in II Introduction Pages.indd 2 14/12/2023 11:58:08  Key features … Awareness about higher education avenues in Scope of Chemistry the field of Chemistry Describe the specific competency / Learning objectives performance capability acquired by the learner Additional information provided to relate the Do you know content to day-to-day life / development in the field Model problems worked out for clear-cut Example Problems comprehension by the learners To help the students to assess their own Evaluate yourself conceptual understanding Quick access to concepts, videos, animations Q.R code and tutorials opens up resources for learning; enables the ICT learners to access, extend transform ideas / informations Summary A glance on the substance of the unit Inter relating the concepts for enabling learners Concept map to visualize the essence of the unit To assess the level of understanding through Evaluation multiple choice question,numerical problems etc… Books for Reference List of relevant books for further reading To help the learners confirm the accuracy of Key answers the answers arrived and remedy the gaps in learning Important terms are enlisted with equivalent Glossary Tamil words III Introduction Pages.indd 3 16-12-2022 19:32:16  CONTENTS CHEMISTRY S.No. Topic Page.No. Month 8 Ionic Equilibrium 01 August 9 Electro Chemistry 33 October 10 Surface Chemistry 69 November 11 Hydroxy Compounds and Ethers 104 July August upto page 176. 12 Carbonyl Compounds and Carboxylic Acids 145 Remaining september 13 Organic Nitrogen Compounds 196 October 14 Biomolecules 237 November 15 Chemistry in Everyday Life 272 November ANSWERS 297 GLOSSARY 320 E-book Assessment IV Introduction Pages.indd 4 16-12-2022 19:32:16  UNIT IONIC 8 EQUILIBRIUM Peter Joseph William Debye Learning Objectives After studying this unit, the students will be able to Peter Joseph William Debye was ’ classify the substances into acids and Dutch-American physicist bases based on Arrhenius, Lowry – greatly contributed to the theory Bronsted and Lewis concepts. of electrolyte solutions. He also ’ define pH scale and establish studied the dipole moments relationship between pH and pOH of molecules, Debye won the ’ describe the equilibrium involved in Nobel Prize in Chemistry the ionisation of water. ’ explain Ostwald’s dilution Law and (1936) for his contributions derive a relationship between the to the determination of dissociation constant and degree of molecular structure through dissociation of a weak electrolyte. his investigations on dipole ’ recognise the concept of common ion moments and X-rays diffraction. effect and explain buffer action. ’ apply Henderson equation for the preparation of buffer solution ’ calculate solubility product and understand the relation between solubility and solubility product. ’ solve numerical problems involving ionic equilibria. 1 XII U8-Ionic equilibrium.indd 1 16-12-2022 19:33:17  INTRODUCTION We have already learnt the chemical equilibrium in XI standard. In this unit, we discuss the ionic equilibria, specifically acid – base equilibria. Some of the important processes in our body involve aqueous equilibria. For example, the carbonic acid – bicarbonate buffer in the blood. H3O+ (aq)+HCO3- (aq)  H2CO3 (aq)+H2O(l) We have come across many chemical compounds in our daily life among them acids and bases are the most common. For example, milk contains lactic acid, vinegar contains acetic acid, tea contains tannic acid and antacid tablet contains aluminium hydroxide / magnesium hydroxide. Acids and bases have many important industrial applications. For example, sulphuric acid is used in fertilizer industry and sodium hydroxide in soap industry etc… Hence, it is important to understand the properties of acids and bases. In this unit we shall learn the definitions of acids and bases and study, their ionisation in aqueous solution. We learn the pH scale and also apply the principles of chemical equilibrium to determine the concentration of the species furnished in aqueous solution by acids and bases. 8.1 Acids and bases The term ‘acid’ is derived from the latin word ‘acidus’ meaning sour. We have already learnt in earlier classes that acid tastes sour, turns blue litmus to red and reacts with metals such as zinc and produces hydrogen gas. Similarly base tastes bitter and turns red litmus to blue. These classical concepts are not adequate to explain the complete behaviour of acids and bases. So, the scientists developed the acid – base concept based on their behaviour. Let us, learn the concept developed by scientists Arrhenius, Bronsted and Lowry and Lewis to describe the properties of acids and bases. 8.1.1 Arrhenius Concept One of the earliest theories about acids and bases was proposed by swedish chemist Svante Arrhenius. According to him, an acid is a substance that dissociates to give hydrogen ions in water. For example, HCl, H2SO4 etc., are acids. Their dissociation in aqueous solution is expressed as  HCl(g)  H2 O  H (aq)+Cl (aq) + - The H+ ion in aqueous solution is highly hydrated and usually represented as H3O+ , the simplest hydrate of proton [H(H2O)]. We use both H+and H3O+ to mean the same. + Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example, substances like NaOH, Ca(OH)2 etc., are bases.  -  HO Ca(OH)2   Ca (aq)+2OH (aq) 2 2+ 2 XII U8-Ionic equilibrium.indd 2 16-12-2022 19:33:18  Limitations of Arrhenius concept i. Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents such as acetone, Tetrahydrofuran etc... ii. This theory does not account for the basic nature of the substances like ammonia ( NH3 ) which do not possess hydroxyl group. Evaluate yourself – 1 Classify the following as acid (or) base using Arrhenius concept i)HNO3 ii) Ba(OH)2 iii) H3PO4 iv) CH 3COOH 8.1.2 Lowry – Bronsted Theory (Proton Theory) In 1923, Lowry and Bronsted suggested a more general definition of acids and bases. According to their concept, an acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton from other substance. In other words, an acid is a proton donor and a base is a proton acceptor. When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCl behaves as an acid and H2O is base. The proton transfer from the acid to base can be represented as HCl+H2O  H3O+ +Cl - When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia ( NH3 ) acts as a base and H2O is acid. The reaction is represented as H2O+NH3  NH 4+ +OH- Let us consider the reverse reaction in the following equilibrium HCl + H2O  H 3 O+ + Cl - Proton donor (acid) Proton acceptor (Base) Proton donor (acid) Proton acceptor (Base) H3O+ donates a proton to Cl - to form HCl i.e., the products also behave as acid and base. In general, Lowry – Bronsted (acid – base) reaction is represented as Acid1 +Base 2  Acid 2 +Base1 The species that remains after the donation of a proton is a base (Base1 )and is called the conjugate base of the Bronsted acid ( Acid 1 ). In other words, chemical species that differ only by a proton are called conjugate acid – base pairs. 3 XII U8-Ionic equilibrium.indd 3 16-12-2022 19:33:21  Conjugate acid - base pair HCl + H 2O  H 3O + + Cl- Proton donar (acid) Proton acceptor (Base) Proton donar (acid) Proton acceptor (Base) Conjugate acid - base pair HCl and Cl- , H 2 O and H3O + are two conjugate acid – base pairs. i.e., Cl - is the conjugate base of the acid HCl. (or) HCl is conjugate acid of Cl-. Similarly H3O+ is the conjugate acid of H2O. Limitations of Lowry – Bronsted theory i. Substances like BF3 , AlCl 3 etc., that do not donate protons are known to behave as acids. Evaluate yourself – 2 Write a balanced equation for the dissociation of the following in water and identify the conjugate acid –base pairs. i) NH 4 + ii) H2SO4 iii) CH3COOH. 8.1.3 Lewis concept In 1923, Gilbert. N. Lewis proposed a more generalised concept of acids and bases. He considered the electron pair to define a species as an acid (or) a base. According to him, an acid is a species that accepts an electron pair while base is a species that donates an electron pair. We call such species as Lewis acids and bases. A Lewis acid is a positive ion (or) an electron deficient molecule and a Lewis base is a anion (or) neutral molecule with at least one lone pair of electrons. Les us consider the reaction between Boron tri fluoride and ammonia F H F H F B + : NH3 ammonia B N F Boron tri fluoride F F H electron deficient - electron pair donar - adduct (Lewis acid) - (Lewis base) Here, boron has a vacant 2p orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond. We have already learnt that in coordination compounds, the Ligands act as a Lewis base and the central metal atom or ion that accepts a pair of electrons from the ligand behaves as a Lewis acid. 4 XII U8-Ionic equilibrium.indd 4 16-12-2022 19:33:23  Lewis acids Lewis bases Electron deficient molecules such as Molecules with one (or) more lone pairs of BF3 ,AlCl 3 ,BeF2 etc... electrons. NH3 ,H2O,R-O-H,R-O-R, R - NH2 All metal ions All anions Examples: Fe2+ ,Fe3+ ,Cr 3+ ,Cu 2+ etc... F- ,Cl - ,CN- ,SCN - ,SO2-4 etc... Molecules that contain a polar double bond Molecules that contain carbon – carbon Examples : SO2 ,CO2 ,SO3 etc... multiple bond Examples: CH 2 = CH 2 , CH ≡ CH etc... Molecules in which the central atom can All metal oxides expand its octet due to the availability of CaO,MgO,Na 2O etc... empty d – orbitals Example: SiF4 ,SF4 ,FeCl 3 etc.. Carbonium ion Carbanion (CH ) 3 3 C+ CH3− Example Identify the Lewis acid and the Lewis base in the following reactions. Cr 3+ + 6 H2O → [Cr(H2O)6 ]3+ In the hydration of ion, each of six water molecules donates a pair of electron to Cr 3+ to from the hydrated cation, hexaaquachromium (III) ion, thus, the Lewis acid is Cr 3+ and the Lewis base H2O. Evaluate yourself – 3 Identify the Lewis acid and the Lewis base in the following reactions. i. CaO+CO2 → CaCO3 CH3 ii. Cl CH3 O CH3 + AlCl3 O Al Cl CH3 Cl 5 XII U8-Ionic equilibrium.indd 5 16-12-2022 19:33:26  Evaluate yourself – 4 H3BO3 accepts hydroxide ion from water as shown below H3BO3 (aq)+H2O(l)  B(OH)4 - +H+ Predict the nature of H3BO3 using Lewis concept 8.2 Strength of Acids and Bases The strength of acids and bases can be determined by the concentration of H3O+ (or) OH- produced per mole of the substance dissolved in H2O. Generally we classify the acids / bases either as strong or weak. A strong acid is the one that is almost completely dissociated in water while a weak acid is only partially dissociated in water. Let us quantitatively define the strength of an acid (HA) by considering the following general equilibrium. HA + H2O  H3O+ + A- acid 1 base 2 acid 2 base 1 The equilibrium constant for the above ionisation is given by the following expression [H3O+ ][A - ] K= .....(8.1) [HA][H2O] We can omit the concentration of H2O in the above expression since it is present in large excess and essentially unchanged. [H3O+ ][A - ] Ka = .....(8.2) [HA] Here, K a is called the ionisation constant or dissociation constant of the acid. It measures the strength of an acid. Acids such as HCl,HNO3 etc... are almost completely ionised and hence they have high K a value (K a for HCl at 25o C is 2×106 ) Acids such as formic acid (K a =1.8 × 10-4at 25oC) , acetic acid (1.8 × 10-5at 25oC) etc.. are partially ionised in solution and in such cases, there is an equilibrium between the unionised acid molecules and their dissociated ions. Generally, acids with K a value greater than ten are considered as strong acids and less than one are considered as weak acids. Let us consider the dissociation of HCl in aqueous solution, HCl + H - OH  H3O+ + Cl - acid 1 base 2 acid 2 base 1 As discussed earlier, due to the complete dissociation, the equilibrium lies almost 100% to the right. i.e., the Cl - ion has only a negligible tendency to accept a proton form H3O+. It means that the conjugate base of a strong acid is a weak base and vice versa. The following table illustrates the relative strength of conjugate acid – base pairs. 6 XII U8-Ionic equilibrium.indd 6 16-12-2022 19:33:29  HClO4 ClO 4 - HCl Strong Cl- acids Very weak H 2 SO4 - bases HSO 4 HNO3 NO3- H 3 O+ H 2O HNO2 NO 2 - Weak Weak HF acids F- bases CH 3COOH CH3 COO - NH 3 NH 2 - Strong O 2- OH- Very weak bases acids H- H2 8.3 Ionisation of water We have learnt that when an acidic or a basic substance is dissolved in water, depending upon its nature, it can either donate (or) accept a proton. In addition to that the pure water itself has a little tendency to dissociate. i.e, one water molecule donates a proton to an another water molecule. This is known as auto ionisation of water and it is represented as below... HO - H+H 2O  H 3O + + OH - acid 1 base 2 acid 2 base 1 Conjugate acid - base pairs In the above ionisation, one water molecule acts as an acid while the another water molecule acts as a base. The dissociation constant for the above ionisation is given by the following expression [H3O+ ][OH- ] K= .....(8.3) [H2O]2 The concentration of pure liquid water is one. i.e, [H 2 O]2 = 1  K w =[H3O+ ][OH- ] .....(8.4) Here, K w represents the ionic product (ionic product constant) of water It was experimentally found that the concentration of H3O+ in pure water is 1×10-7 at 25oC. Since the dissociation of water produces equal number of H3O+ and OH- , the concentration of OH- is also equal to 1×10-7 at 25o C. 7 XII U8-Ionic equilibrium.indd 7 16-12-2022 19:33:31  Therefore, the ionic product of water at 25oC is K w =[H3O]+[OH- ]......(8.4) K w =(1 × 10-7 )(1 × 10-7 ) = 1 × 10-14. K w values at different temperatures Like all equilibrium constants, K w is also are given in the following below a constant at a particular temperature. The dissociation of water is an endothermic reaction. Temperature ( o C ) Kw With the increase in temperature, the concentration of H3O+ and OH - also increases, and hence the 0 1.14 × 10−15 ionic product also increases. 10 −15 2.95 × 10 In neutral aqueous solution like NaCl solution, the concentration of H3O+ is always 25 1.00 × 10−14 equal to the concentration of OH- whereas in case 40 2.71 × 10−14 of an aqueous solution of a substance which may behave as an acid (or) a base, the concentration of 50 5.30 × 10−14 - H 3O + will not be equal to [OH ]. We can understand this by considering the aqueous HCl as an example. In addition to the auto ionisation of water, the following equilibrium due to the dissociation of HCl can also exist. HCl+H2O  H3O+ +Cl - In this case, in addition to the auto ionisation of water, HCl molecules also produces H3O+ ion by donating a proton to water and hence [H3O+ ]>[OH- ]. It means that the aqueous HCl solution is acidic. Similarly, in basic solution such as aqueous NH 3 , NaOH etc….. [OH–]>[H3O+]. Example 8.1 Calculate the concentration of OH- in a fruit juice which contains 2 × 10-3 M, H3O+ ion. Identify the nature of the solution. Given that H3O+ =2 × 10-3M K w =[H3O+ ][OH- ] Kw 1 × 10-14 ∴[OH- ]= = - =5 × 10-12 M [H3O ] 2 × 10 + 3 2 × 10-3 > > 5 × 10-12 i.e., [H 3O + ]>>[OH- ], hence the juice is acidic in nature 8 XII U8-Ionic equilibrium.indd 8 16-12-2022 19:33:37  Evaluate yourself - 5 At a particular temperature, the K w of a neutral solution was equal to 4 × 10-14. Calculate the concentration of [H3O+ ] and [OH- ]. 8.4 The pH scale We usually deal with acid / base solution in the concentration range 10-1 to 10-7 M. To express the strength of such low concentrations, Sorensen introduced a logarithmic scale known as the pH scale. The term pH is derived from the French word ‘Purissance de hydrogene’ meaning, the power of hydrogen. pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution. pH = - log 10[H3O+ ] .....(8.5) The concentration of H 3O + in a solution of known pH can be calculated using the following expression. -pH [H3O+ ]=10 (or) [H3O+ ]= antilog of (-pH) .....(8.6) Similarly, pOH can also be defined as follows pOH = - log 10 [OH- ] .....(8.7) As discussed earlier, in neutral solutions, the concentration of [H3O ] as well as [OH + ] is + equal to 1 × 10-7 M at 25C. The pH of a neutral solution can be calculated by substituting this H3O+concentration in the expression (8.5) pH = - log 10 [H3O+ ] = - log 10 10 -7 = (-7)(-1) log1010= +7 (1)= 7 [ log 1010 =1] Similary, we can calculate the pOH of a neutral solution using the expression (8.7), it is also equal to 7. The negative sign in the expression (8.5) indicates that when the concentration of [H 3O + ] increases the pH value decreases. For example, if the [H3O+ ] increases from to 10 to10 M , -7 -5 the pH value of the solution decreases from 7 to 5. We know that in acidic solution, [H 3O + ] > [OH - ] , i.e., [H 3O+ ] > 10-7. Similarly in basic solution [H 3O+ ] < 10-7. So, we can conclude that acidic solution should have pH value less than 7 and basic solution should have pH value greater than 7. 8.4.1 Relation between pH and pOH A relation between pH and pOH can be established using their following definitions + pH=-log pH=-log10 [H [H3O O+ ]] .....(8.5) 10 3 - pOH = - log pOH = - log10 [OH- ]] [OH 10 .....(8.7) Adding equation (8.5) and (8.7) 9 XII U8-Ionic equilibrium.indd 9 16-12-2022 19:33:42  Figure 8.1 The pH scale pH + pOH =-log10[H3O+ ]-log10 [OH- ] ( =- log10 [H3O+ ]+log10[OH- ] ) pH+pOH =-log10[H3O+ ][OH- ] [ log a+logb = logab ] We know that [H3O+ ][OH- ]=K w ⇒ pH+pOH= - log 10 K w ⇒ pH+pOH=pK w  pK w = −log10 K w  .....(8.8) at 25 C, the ionic product of water, K w =1 × 10-14  pK w = - log10 10-14 = 14 log10 10 = 14 ∴ (8.7) ⇒ ∴ At 25C, pH + pOH= 14 10 XII U8-Ionic equilibrium.indd 10 16-12-2022 19:33:43  Example 8.2 Calculate the pH of 0.001M HCl solution H2O + -  HCl  H3O + Cl 0.001 M 0.001 M 0.001M H3O+ from the auto ionisation of H2O (10-7 M) is negligible when compared to the H3O+ from 10-3M HCl. Hence [H3O+ ] = 0.001 mol dm–3 pH=-log [H O+ ] 10 3 =-log10 (0.001) -3 =-log10 (10 ) = 3 Note: If the concentration of the acid or base is less than 10–6 M, the concentration of H3O+ produced due to the auto ionisation of water cannot be negleted and in such cases [H3O+ ]=10-7 (from water) + [H3O+ ] (from the acid) similarly, [OH- ]=10-7 M (from water) + [OH- ] (from the base) Example 8.3 -7 Calculate pH of 10 M HCl If we do not consider [H3O+ ] from the ionisation of H2O, then [H3O+ ] =[HCl]=10-7 M i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10-7 M) Hence, the H3O+ (10-7 M) formed due to the auto ionisation of water cannot be neglected. so, in this case we should consider [H3O+ ] from ionisation of H 2 O [H3O+ ] = 10-7 (from HCl) +10-7 (from water) = 10-7 (1+1) = 2 × 10-7 pH=-log10[H3O+ ] =-log10 (2 × 10-7 ) = - log 2 + log 10-7  =- log 2-(-7).log10 10 =7-log 2 =7-0.3010 = 6.6990 = 6.70 11 XII U8-Ionic equilibrium.indd 11 16-12-2022 19:33:46  Evaluate yourself - 6 a) Calculate pH of 10-8 M H2SO4 B) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4 c) Calculate the pH of an aqueous solution obtained by mixing 50ml of 0.2 M HCl with 50ml 0.1 M NaOH 8.5 Ionisation of weak acids We have already learnt that weak acids are partially dissociated in water and there is an equilibrium between the undissociated acid and its dissociated ions. Consider the ionisation of a weak monobasic acid HA in water. HA+H2O  H3O+ +A - Applying law of chemical equilibrium, the equilibrium constant K c is given by the expression [H3O+ ][A - ] KC =    [HA][H2O] .....(8.9) The square brackets, as usual, represent the concentrations of the respective species in moles per litre. In dilute solutions, water is present in large excess and hence, its concentration may be taken as constant say K. Further H3O+ indicates that hydrogen ion is hydrated, for simplicity it may be replaced by H+. The above equation may then be written as, [H+ ][A - ] KC = [HA] × K .....(8.10) The product of the two constants K C and K gives another constant. Let it be K a [H+ ][A - ] Ka = [HA] .....(8.11) The constant K a is called dissociation constant of the acid. Like other equilibrium constants, K a also varies only with temperature. Similarly, for a weak base, the dissociation constant can be written as below. [B+ ][OH- ] Kb = [BOH] ....(8.12) 8.5.1 Ostwald’s dilution law Ostwald’s dilution law relates the dissociation constant of the weak acid (K a ) with its degree of dissociation (α) and the concentration (c). Degree of dissociation (α) is the fraction of the total number of moles of a substance that dissociates at equilibrium. 12 XII U8-Ionic equilibrium.indd 12 16-12-2022 19:33:49  Number of moles dissociated α= total number of moles We shall derive an expression for ostwald's law by considering a weak acid, i.e. acetic acid (CH3COOH). The dissociation of acetic acid can be represented as CH3COOH  H + + CH3COO- The dissociation constant of acetic acid is, [H+ ][CH3COO- ] ka = [CH3COOH] .....(8.13) CH3COOH H+ CH3COO- Initial number of 1 - - moles Degree of dissociation of α - - CH3COOH Number of moles at 1- α α α equilibrium Equilibrium (1 - α) C αC αC concentration Substituting the equilibrium concentration in equation (8.13) (αC)(αC) Ka = (1-α)C α 2C Ka = 1-α .....(8.14) We know that weak acid dissociates only to a very small extent. Compared to one, α is so small and hence in the denominator (1 - α)  1. The above expression (8.14) now becomes, K a = α2C Ka ⇒ α2 = C Ka α= C .....(8.15) Let us consider an acid with K a value 4 × 10-4 and calculate the degree of dissociation of that acid at two different concentration 1 × 10-2 M and 1 × 10-4 M using the above expression (8.15) For 1 × 10-2 M , 13 XII U8-Ionic equilibrium.indd 13 16-12-2022 19:33:51  4 × 10-4 α= 10-2 = 4 × 10-2 = 2 × 10-1 = 0.2 For 1 × 10-4 M acid, 4 × 10-4 α= 10-4 =2 i.e, When the dilution increases by 100 times, (Concentration decreases from 1 × 10-2 M to 1 × 10-4 M ), the dissociation increases by 10 times. Thus, we can conclude that, when dilution increases, the degree of dissociation of weak electrolyte also increases. This statement is known as Ostwald’s dilution Law. The concentration of H+ (H3O+ ) can be calculated using the K a value as below. [ H+ ]= α C  (Refer table).....(8.16) Equilibrium molar concentration of [H ] is equal to αC +  Ka  ∴[H+ ]=  C [ equation (8.15)]  C K aC2 = C [H+ ] = K a C .....(8.17) Similarly, for a weak base Kb K b = α2C and α = C [OH- ] = αC (or) [OH- ]= K bC .....(8.18) Example 8.4 A solution of 0.10M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25oC. Find the dissociation constant of the acid. 1.20 K a =α2 c Given that α=1.20%= =1.2 × 10-2 100 = (1.2 × 10-2 )2 (0.1)=1.44 × 10-4 × 10-1 = 1.44 × 10-5 14 XII U8-Ionic equilibrium.indd 14 16-12-2022 19:33:53  Example 8.5 Calculate the pH of 0.1M CH3COOH solution. Dissociation constant of acetic acid is 1.8 × 10-5. pH=-log[H+ ] For weak acids, [H+ ]= K a × C = 1.8 × 10-5 × 0.1 =1.34 × 10-3M pH=-log (1.34 × 10−3 ) =3 - log1.34 = 3 - 0.1271 = 2.8729  2.87 Evaluate yourself - 7 K b for NH 4 OH is 1.8 × 10-5. Calculate the percentage of ionisation of 0.06M ammonium hydroxide solution. 8.6 Common Ion Effect When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. For example, the addition of sodium acetate to acetic acid solution leads to the suppression in the dissociation of acetic acid which is already weakly dissociated. In this case, CH3COOH and CH3COONa have the common ion, CH 3COO- Let us analyse why this happens. Acetic acid is a weak acid. It is not completely dissociated in aqueous solution and hence the following equilibrium exists. CH 3COOH(aq)  H + (aq)+CH 3COO- (aq) However, the added salt, sodium acetate, completely dissociates to produce Na + and CH3COO- ion. CH3COONa(aq) → Na + (aq)+CH3COO- (aq) Hence, the overall concentration of CH 3COO- is increased, and the acid dissociation equilibrium is disturbed. We know from Le chatelier's principle that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, inorder to maintain the equilibrium, the excess CH3COO- ions combines with H+ ions to produce much more unionized CH 3COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH3COOH is suppressed. Thus, the dissociation of a weak acid (CH3COOH) is suppressed in the presence of a salt (CH3COONa) containing an ion common to the weak electrolyte. It is called the common ion effect. 15 XII U8-Ionic equilibrium.indd 15 16-12-2022 19:33:57  8.7 Buffer Solution Do you know that our blood maintains a constant pH, irrespective of a number of cellular acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action. Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases, and this ability is called buffer action. The buffer containing carbonic acid (H2CO3 ) and its conjugate base HCO-3 is present in our blood. There are two types of buffer solutions. 1. Acidic buffer solution : a solution containing a weak acid and its salt. Example : solution containing acetic acid and sodium acetate 2. Basic buffer solution : a solution containing a weak base and its salt. Example : Solution containing NH 4OH and NH 4 Cl 8.7.1 Buffer action To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other. Let us explain the buffer action in a solution containing CH3COOH and CH 3COONa. The dissociation of the buffer components occurs as below. + CH3COOH (aq) CH3 - COO(aq) + H3O (aq) - If an acid is added to this mixture, it will be consumed by the conjugate base CH3COO- to form the undissociated weak acid i.e, the increase in the concentration of H+ does not reduce the pH significantly. CH3COO (aq) + H (aq) CH3COOH (aq) If a base is added, it will be neutralized by H3O+, and the acetic acid is dissociated to maintain the equlibrium. Hence the pH is not significantly altered. OH (aq) + H3O+ ((aq) H2O (l) H2O (l) CH3COOH (aq) CH3COO (aq) + H3O+ ((aq) OH (aq) + CH3COOH (aq) CH3COO (aq) + H2O (l) These neutralization reactions are identical to those reactions that we have already discussed in common ion effect. 16 XII U8-Ionic equilibrium.indd 16 16-12-2022 19:33:59  Let us analyse the effect of the addition of 0.01 mol of solid sodium hydroxide to one litre of a buffer solution containing 0.8 M CH3COOH and 0.8 M CH3COONa. Assume that the volume change due to the addition of NaOH is negligible. (Given: K a for CH 3COOH is 1.8 × 10-5 )  CH 3 -COOH(aq) ←  → CH 3COO- (aq) + H + (aq) H2O  0.8-α α α CH3COONa(aq) → HO CH3COO- (aq)+ Na + (aq) 2 0.8 0.8 0.8 The dissociation constant for CH3COOH is given by [CH3COO- ][H+ ] Ka = ; [CH3COOH] [CH3COOH] [H+ ]=K a [CH3COO- ] The above expression shows that the concentration of H+ is directly proportional to [CH3COOH]. [CH3COO- ] Let the degree of dissociation of CH3COOH be α then, [CH3COOH]=0.8-α and [CH3COO- ]=α+0.8 (0.8-α) ∴[H+ ]=K a (0.8+α) α R C NH2 The above order of reactivity can be explained in terms of i) Basicity of the leaving group    ii) Resonance effect (i) Basicity of the leaving group Weaker bases are good leaving groups. Hence acyl derivatives with weaker bases as leaving groups (L) can easily rupture the bond and are more reactive. The correct order of the basicity of the leaving group is H2N : > : OR > RCOO : > : Cl Hence the reverse is the order of reactivity. (ii) Resonance effect Lesser the electronegativity of the group, greater would O O be the resonance stabilization as shown below. This effect makes the molecule more stable and reduces the R C R C reactivity of the acyl compound. The order of electronegativity G G of the leaving groups follows the order – Cl > - OCOR > - OR > - NH2 Hence the order of reactivity of the acid derivatives with nucleophilic reagent follows the order acid halide > acid anhydride > esters > acid amides 12.14.1 Nomenclature Compound IUPAC Name (common name, Structural formula, Prefix with position Primary Secondary IUPAC Name) Root used number suffix Suffix Acetyl chloride CH3 C Cl ane/ oyl – eth chloride O Ethanoylchloride Propionyl chloride C2H5 C Cl ane/ oyl – prop chloride O Propanoylchloride Benzoyl chloride C6H5 C Cl ane/ oyl – Benz chloride O Benzoylchloride 180 XII U12-Carbonyl Compounds.indd 180 08-12-2021 22:50:47  Acetic anhydride CH3 C O C CH3 ane/ oic – eth O O anhydride Ethanoic anhydride Propionic anhydride CH3 CH2 C O C CH2 CH3 oic – prop ane/ O O anhydride Propanoic anhydride Benzoic anhydride C6H5 C O C C6H5 oic – Benz anhydride O O Benzoic anhydride Esters Methyl acetate CH3 C O CH3 Methyl eth ane/ oate O Methyl ethanoate Ethyl acetate CH3 C O C2H5 Ethyl eth ane/ oate O Ethyl ethanoate Phenyl acetate CH3 C O C6H5 Phenyl eth ane/ oate O Phenyl ethanoate Acid Amides Acetamide CH3 C NH2 – eth ane/ amide O Ethanamide 181 XII U12-Carbonyl Compounds.indd 181 08-12-2021 22:50:48  Propionamide C2H5 C NH2 – prop ane/ amide O Propanamide Benzamide C6H5 C NH2 – benz – amide O Benzamide 12. 14. 2. Acid Halides: Methods of Preparation of acid chloride: Acid chlorides are prepared from carboxylic acid by treating it with anyone of the chlorinating agent such as SOCl2, PCl5, or PCl3 1) By reaction with thionyl Chloride (SOCl2) O O CH3 C OH + SOCl2 CH3 C Cl + HCl + SO2 Acetic acid Acetyl chloride This method is superior to others as the by products being gases escape leaving the acid chloride in the pure state. Physical properties: They emit pale fumes of hydrogen chloride when exposed to air on account of their reaction with water vapour. They are insoluble in water but slowly begins to dissolve due to hydrolysis. Chemical properties: They react with weak nucleophiles such as water, alcohols, ammonia and amines to produce the corresponding acid, ester, amide or substituted amides. 1) Hydrolysis. Acyl halides undergo hydrolysis to form corresponding carboxylic acids O O CH3 C Cl + HOH CH3 C OH + HCl Acetyl chloride Acetic acid 182 XII U12-Carbonyl Compounds.indd 182 08-12-2021 22:50:48  2) Reaction with Alcohols (Alcoholysis) gives esters. O O CH3 C Cl + HOC2H5 CH3 C OC2H5 + HCl Acetyl chloride Ethyl alcohol Ethyl acetate 3) Reaction with Ammonia (Ammonolysis) gives acid amides. O O CH3 C Cl+ H NH2 CH3 C NH2 + HCl Acetyl chloride Ammonia Acetamide 4) Reaction with 1o and 2o Amines gives N-alkyl amides. O O R C Cl+ H NHR' R C NHR' + HCl 1o amine N-alkylamide O O R C Cl+ H NR'2 R C NR'2 + HCl 2o amine N,N-dialkylamide (5) Reduction. (a) When reduced with hydrogen in the presence of ‘poisoned’ palladium catalyst, they form aldehydes. This reaction is called Rosenmund reduction. We have already learnt this reaction under the preparation of aldehydes O O Pd - BaSO4 CH3 C Cl + H 2 CH3 C H + HCl 2[H] Acetyl chloride Acetaldehyde (b) When reduced with LiAlH4 gives primary alcohols. O LiAlH4 CH3 C Cl + 4(H) CH3 CH2 OH + HCl Ethyl alcohol 12.14.3 Acid anhydride Methods of preparation 1. Heating carboxylic acid with P2O5 We have already learnt that when carboxylic acids are heated with P2O5 dehydration takes place to form acid anhydride. 183 XII U12-Carbonyl Compounds.indd 183 08-12-2021 22:50:49  2. By reaction of acid halide with a salt of carboxylic acids. Acid chlorides on heating with sodium salt of carboxylic acids gives corresponding anhydride. O O CH3 C Cl CH3 C Acetylchloride + O + NaCl CH3 C O Na CH3 C O O Sodium acetate Acetic anhydride Chemical properties 1. Hydrolysis Acid anhydride are slowly hydrolysed, by water to form corresponding carboxylic acids. O O O CH3 C O C CH3 + H OH 2CH3 C OH Acetic anhydride Acetic acid 2. Reaction with alcohol Acid anhydride reacts with alcohols to form esters. O O O O CH3 C O C CH3 + H OC2H5 CH3 C OC2H5 + CH3 C OH Acetic anhydride Ethylalcohol Ethyl acetate Acetic acid 3. Reaction with ammonia Acid anhydride reacts with ammonia to form amides. O O O O CH3 C O C CH3 + H NH2 CH3 C NH2 + CH3 C OH Acetamide Acetic acid Acetic anhydride Ammonia 4. Reaction with PCl5 Acid anhydride reacts with PCl5 to form acyl chlorides. O O O CH3 C O C CH3 + PCl5 2CH3 C Cl + POCl3 Acetic anhydride Acetyl chloride 184 XII U12-Carbonyl Compounds.indd 184 08-12-2021 22:50:49  12.14.4 Esters Methods of preparation 1. Esterification We have already learnt that treatment of alcohols with carboxylic acids in presence of mineral acid gives esters. The reaction is carried to completion by using an excess of reactant or by removing the water from the reaction mixture. 2. Alcoholysis of Acid chloride or Acid anhydrides ii) Treatment of acid chloride or acid anhydride with alcohol also gives esters Physical Properties Esters are colour less liquids or solids with characteristic fruity smell. Flavours of some of the esters are given below. S.No Ester Flavour 1 Amyl acetate Banana 2 Ethyl butyrate Pineapple 3 Octyl acetate Orange 4 Isobutyl formate Raspberry 5 Amyl butyrate Apricot Chemical Properties 1. Hydrolysis We have already learnt that hydrolysis of esters gives alcohol and carboxylic acid. 2. Reaction with alcohol ( Transesterification) Esters of an alcohol can react with another alcohol in the presence of a mineral acid to give the ester of second alcohol. The interchange of alcohol portions of the esters is termed transesterification O O H+ CH3 C OC2H5 + HOC3H7 CH3 C OC3H7+ C2H5OH Ethyl acetate Propyl acetate Ethyl alcohol Propyl alcohol The reaction is generally used for the preparation of the esters of a higher alcohol from that of a lower alcohol. 3. Reaction with ammonia (Ammonolysis) Esters react slowly with ammonia to form amides and alcohol. O O CH3 C OC2H5 + H NH2 CH3 C NH2 + C2H5OH Ethyl acetate Acetamide Ethyl alcohol 185 XII U12-Carbonyl Compounds.indd 185 08-12-2021 22:50:50  4. Claisen Condensation Esters containing at least one ∝- hydrogen atom undergo self condensation in the presence of a strong base such as sodium ethoxide to form b- keto ester. O O O O C2H5ONa CH3 C OC2H5 + H CH2 C OC2H5 CH3 C CH2 C OC2H5 + C2H5OH Ethyl acetate Ethyl acetate Ethyl aceto acetate Ethyl alcohol 5. Reaction with PCl5 Esters react with PCl5 to give a mixture of acyl and alkyl chloride O O CH3 C OC2H5 + PCl5 CH3 C Cl + C2H5Cl + POCl3 Ethyl acetate Acetyl chloride Ethyl chloride Evaluate yourself Why is acid anhydride preferred to acyl chloride for carrying out acylation reactions ? 12.14.5 Acid Amides Acid amides are derivatives of carboxylic acid in which the – OH part of carboxyl group has been replaced by – NH2 group. The general formula of amides are given as follows. O Now, we shall focus our attention mainly on the study of chemistry of acetamide. R C NH2 Methods of Preparation 1. Ammonolysis of acid derivatives Acid amides are prepared by the action of ammonia with acid chlorides or acid anhydrides. O O CH3 C Cl + H NH2 CH3 C NH2 + HCl Acetyl chloride Acetamide O O O O CH3 C O C CH3 + H NH2 CH3 C NH2 + CH3 C OH Acetic anhydride Acetamide 2) Heating ammonium carboxylates Ammonium salts of carboxylic acids (ammonium carboxylates) on heating, lose a molecule of water to form amides. O O  CH3 C O NH+4 CH3 C NH2 + H2O Ammonium acetate Acetamide 186 XII U12-Carbonyl Compounds.indd 186 08-12-2021 22:50:50  3) Partial hydrolysis of alkyl cyanides (Nitriles) Partial hydrolysis of alkyl cyanides with cold con HCl gives amides Conc HCl CH3 C N CH3 C NH2 H2O / OH O Methyl cyanide Acetamide Chemical Properties 1. Amphoteric character Amides behave both as weak acid as well as weak base and thus show amphoteric character. This can be proved by the following reactions. Acetamide (as base) reacts with hydrochloric acid to form salt O O + CH3 C NH2 + HCl CH3 C NH3 Cl Acetamide Acetamide hydrochloride Acetamide (as acid) reacts with sodium to form sodium salt and hydrogen gas is liberated. O O 2CH3 C NH2 + 2Na 2CH3 C NHNa + H2 Acetamide Sodium acetamide 2) Hydrolysis Amides can be hydrolysed in acid or in alkaline solution on prolonged heating O O dil HCl CH3 C NH2 + H2O CH3 C OH + NH4Cl Acetamide Acetic acid O O NaOH CH3 C NH2 CH3 C ONa + NH3 Acetamide Sodium acetate 3) Dehydration Amides on heating with strong dehydrating agents like P2O5 get dehydrated to form cyanides. O P2O5 CH3 C NH2 CH3 C N + H2O Acetamide Methyl cyanide (aceto nitrile) 187 XII U12-Carbonyl Compounds.indd 187 08-12-2021 22:50:51  4) Hoff mann’s degradation Amides reacts with bromine in the presence of caustic alkali to form a primary amine carrying one carbon less than the parent amide. O CH3 C NH2 + Br2 + 4 KOH CH3NH2 + K2CO3 + 2KBr + 2H2O Acetamide Methyl amine 5) Reduction Amides on reduction with LiAlH4or Sodium and ethyl alcohol to form corresponding amines. O LiAIH4 CH3 C NH2 + 4 (H) CH3 CH2 NH2 + H2O Acetamide Ethyl amine 12.15 Uses of carboxylic acids and its derivatives Formic acid It is used i) for the dehydration of hides. ii) as a coagulating agent for rubber latex iii) in medicine for treatment of gout iv) as an antiseptic in the preservation of fruit juice. Acetic acid It is used i) as table vinegar ii) for coagulating rubber latex iii) for manufacture of cellulose acetate and poly vinylacetate Benzoic acid It is used i) as food preservative either in the pure form or in the form of sodium benzoate ii) in medicine as an urinary antiseptic iii) for manufacture of dyes Acetyl Chloride It is used i) as acetylating agent in organic synthesis ii) in detection and estimation of – OH, - NH2 groups in organic compounds Acetic anhydride It is used i) acetylating agent ii) in the preparation of medicine like asprin and phenacetin iii) for the manufacture plastics like cellulose acetate and poly vinyl acetate. Ethyl acetate is used i) in the preparation of artificial fruit essences. ii) as a solvent for lacquers. iii) in the preparation of organic synthetic reagent like ethyl acetoacetate. 188 XII U12-Carbonyl Compounds.indd 188 08-12-2021 22:50:51  EVALUATION Choose the correct answer: 1. The correct structure of the product ‘A’ formed in the reaction (NEET) O H2 (gas, 1 atm) A is Pd / C, ethanol OH O OH OH a) b) c) d) 2. The formation of cyanohydrin from acetone is an example of a) nucleophilic substitution b) electrophilic substitution c) electrophilic addition d) Nucleophilic addition 3. Reaction of acetone with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is a) Grignard reagent b) Sn / HCl c) hydrazine in presence of slightly acidic solution d) hydrocyanic acid 4. In the following reaction, H2SO4 HC CH X Product ‘X’ will not give HgSO4 a) Tollen’s test b) Victor meyer test c) Iodoform test d) Fehling solution test i) O3 NH3 5. CH2 CH2 X Y ‘Y’ is ii) Zn / H2O a) Formaldelyde b) di acetone ammonia c) hexamethylene tetraamine d) oxime 6. Predict the product Z in the following series of reactions Ethanoic acid PCl C H → X AnhydrousAlCl  5 → Y i)CH  MgBr ii)H O 6 →Z. 6 3 3 3 + a) (CH3 )2C(OH)C 6 H5 b) CH3CH(OH)C 6 H5 CH2 - OH c) CH3CH(OH)CH2 - CH3 d) 7. Assertion: 2,2 – dimethyl propanoic acid does not give HVZ reaction. Reason: 2 – 2, dimethyl propanoic acid does not have α - hydrogen atom 189 XII U12-Carbonyl Compounds.indd 189 08-12-2021 22:50:52  a) if both assertion and reason are true and reason is the correct explanation of assertion. b) if both assertion and reason are true but reason is not the correct explanation of assertion. c) assertion is true but reason is false d) both assertion and reason are false. 8. Which of the following represents the correct order of acidity in the given compounds a)FCH2COOH > CH3COOH > BrCH2COOH > ClCH2COOH b)FCH2COOH > ClCH2COOH > BrCH2COOH > CH3COOH c) CH3COOH > ClCH2COOH > FCH2COOH > Br-CH2COOH d) Cl CH2COOH > CH3COOH > BrCH2COOH > ICH2COOH i) NH 9. Benzoic acid ii) D → A  NaOBr 3 → B  NaNO /HCl → C ‘C’ is 2 a) anilinium chloride b) O – nitro aniline c) benzene diazonium chloride d) m – nitro benzoic acid 10. Ethanoic acid P/Br  → 2 – bromoethanoic acid. This reaction is called 2 a) Finkelstein reaction b) Haloform reaction c) Hell – Volhard – Zelinsky reaction d) none of these H3 O+ → (B)  → (C) product (C) is 11. CH3Br  → (A)   KCN PCl 5 a) acetylchloride b) chloro acetic acid c) α - chlorocyano ethanoic acid d) none of these 12. Which one of the following reduces tollens reagent a) formic acid b) acetic acid c) benzophenone d) none of these i) Mg, ether H3O+ 13. a) BrCOOH A COOH B ‘B’ is ii) CO2 b) COOH COOH a) b) c) O d) c) O C d) O C 14. The IUPAC name of OH 190 XII U12-Carbonyl Compounds.indd 190 08-12-2021 22:50:53  a) but – 3- enoicacid b) but – 1- ene-4-oicacid c) but – 2- ene-1-oic acid d) but -3-ene-1-oicacid O C N2H4 15. Identify the product formed in the reaction CH3 C2H5 ONa a) b) NH2 C) d) O C O - C2H5 16. In which case chiral carbon is not generated by reaction with HCN O OH a) b) O O O C) d) Ph Ph    OH 17. Assertion : p – N, N – dimethyl aminobenzaldehyde undergoes benzoin condensation Reason : The aldehydic (-CHO) group is meta directing a) if both assertion and reason are true and reason is the correct explanation of assertion. b) if both assertion and reason are true but reason is not the correct explanation of assertion. c) assertion is true but reason is false d) both assertion and reason are false. 18. Which one of the following reaction is an example of disproportionation reaction a) Aldol condensation b) cannizaro reaction c) Benzoin condensation d) none of these 19. Which one of the following undergoes reaction with 50% sodium hydroxide solution to give the corresponding alcohol and acid a) Phenylmethanal b) ethanal c) ethanol d) methanol 191 XII U12-Carbonyl Compounds.indd 191 08-12-2021 22:50:53  20. The reagent used to distinguish between acetaldehyde and benzaldehyde is a) Tollens reagent b) Fehling’s solution c) 2,4 – dinitrophenyl hydrazine d) semicarbazide 21. Phenyl methanal is reacted with concentrated NaOH to give two products X and Y. X reacts with metallic sodium to liberate hydrogen X and Y are a) sodiumbenzoate and phenol b) Sodium benzoate and phenyl methanol c) phenyl methanol and sodium benzoate d) none of these 22. In which of the following reactions new carbon – carbon bond is not formed? a) Aldol condensation b) Friedel craft reaction c) Kolbe’s reaction d) Wolf kishner reduction 23. An alkene “A” on reaction with O3 and Zn - H2O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene “A” gives “B” as the major product. The structure of product “B” is CH3 CH2Cl a) Cl CH2 CH2 CH b) H3C CH2 CH CH3 CH3 CH3 CH3 c) H3C CH2 C CH3 d) H3C CH CH Cl Cl 24. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their (NEET) a) more extensive association of carboxylic acid via van der Waals force of attraction b) formation of carboxylate ion c) formation of intramolecular H-bonding d) formation of intermolecular H – bonding Short Answer Questions 1. How is propanoic acid is prepared starting from (a) an alcohol (b) an alkylhalide (c) an alkene 2. A Compound (A) with molecular formula C 2 H3 N on acid hydrolysis gives(B) which reacts with thionylchloride to give compound(C). Benzene reacts with compound (C) in presence of anhydrous AlCl 3 to give compound(D). Compound (D) on reduction with Zn/Hg and Conc.HCl gives (E). Identify (A), (B), (C), (D) and (E). Write the equations. 3. Identify X and Y. + CH3COCH2CH2COOC 2 H5 CH MgBr → X H 3O →Y 3 192 XII U12-Carbonyl Compounds.indd 192 08-12-2021 22:50:54  4. Identify A, B and C PCl5 Benzene benzoic acid A B Anh