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Chapter 4 (Equilibrium) Dr.VikramPanchal Institute Of Chemistry Worksheet Sem–2 Section – B (1). What is meant by reversible reaction? In reversible reactions, forward and reverse reactions continuously occur. (2). What is called chemical equilibrium? The equilibrium established in chemical reactions is called chemical equilibrium. (3). How can be said that equilibrium is dynamic in nature? The equilibrium is dynamic and not steady as the forward and the reverse reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed vessel. (4). How many types of equilibrium are there? Which are they? There are two types of equilibrium. Physical (ii) Chemical (5). What is called reverse reaction? The change of product to reactant is called the reverse reaction. (6). What is meant homogeneous equilibrium? Give its example. When all reactants and products are in same physical state in a reversible reaction, it is called homogeneous equilibrium. E.g. Haber process (7). What is meant by heterogeneous equilibrium? Give its example. When all reactants and products are not in same physical state in a reversible reaction, it is called heterogeneous equilibrium. E.g. decomposition of CaCO3 (8). Mention requirements of chemical equilibrium. The reverse reaction should occur in a closed system with constant temperature and pressure. (9). Mention the law of active masses. ‘The driving force of chemical reaction is directly proportional to the active masses of reactants.’ (10). Which scientists gave the law of mass action? Norvegian scientist Guldberg and Waage in 1864 proposed the law of mass action. (11). Interpret Ke = Kc. Ke = [conc. Of products] / [conc. Of reactants] Then, Keq = Kc (12). Interpret Keq = Kp. Ke = partial pressure of products / partial pressure of reactants Keq = Kp (13). Write formula of Kp for NH4COONH2(s) 2NH3(g) + CO2(g) 3 Kp = 4/27 p (14). Give formula of equilibrium constant for decomposition of NH4HS(s). NH4HS(s) NH3(g) + H2S(g) Kp = p2/4 (15). Write formula showing relation between Kc and Kp. Kp = Kc (RT)∆n ∆n= np-nr (16). Give the law of chemical equilibrium. (17) Mention the formula for equilibrium constant for the reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) (18). Under which conditions Kp and Kc will be equal? When ∆n = np – nr = 0, then Kp = Kc. (19). When Kp > Kc? When ∆ng > 0, then Kp > Kc. (20). Write le- Chatelier’s principle. If from the factors determining the equilibrium state, any one factor is changed, there will be such a change in the system that the effect will be nullified or made negligible so that the value of equilibrium constant at that temperature will remain constant. (21). What idea can be obtained by Le- Chatelier’s principle? We can study in detail the effect of change in external factors like concentration, pressure and temperature at equilibrium. (22). Why concentrations of components remain constant at equilibrium? Equilibrium state is the end of reaction and at the end conc. Of each component is constant. (23). Which are the factors affecting equilibrium? The external factors like, concentration, pressure and temperature can affect equilibrium. (24). Give equation showing relation between ∆G and Qc ∆G = ∆G0 + RTlnQc (25). Give unit for the equilibrium constant of the reaction N2(g) + 3H2(g) 2NH3(g) M2-4 = M-2 Kp = p2NH3/(pN2 x p3H2) (atm)-2 (26). If Qc < Kc and Qc > Kc then in which directions the reaction will occur? If Qc < Kc, the reaction will occur in forward direction. If Qc > Kc, then the reaction will occur in reverse direction. (27). Give formula of Kc for the reaction Ag2O(s) + 2HNO3(aq) H2O(l) 2AgNO3(aq) + Because [Ag2O], [H2O] are constants. (28). Write relation between ∆G0 and equilibrium constant K. ∆G0 = -RT ln K or K = e-∆G0/RT (29). What will be the effect of catalyst on equilibrium? The catalyst reduces time required to complete the reaction but does not affect equilibrium state. (30). In the reaction A2 + B2 2AB + Q cal., what should be decreased to increase the proportion of AB? On decreasing temperature, the rate of endothermic i.e., forward reaction increases. (31). Mention function of catalyst. The catalyst decreases the energy of activation. (32). Complete the reaction Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq) Light yellow colour colourless colour like blood (33).In how many sections, Michael Faraday classified substances? (i) electrolyte (ii) Non- electrolyte. (34). What is called an electrolyte? Give example. When a substance dissolves in water and ionized so its aqueous solution can conduct electricity, it is known as electrolyte. Eg. : NaCl (35). What is called non electrolyte? Give example. When a substance dissolves in water or remain insoluble in water and cannot conduct electricity is known as non – electrolyte. Eg.: C6H12O6 (36). What is called strong electrolyte? Give example. An electrolyte whose solubility is more than 0.01 M or completely ionized in aqueous solution, known as strong electrolyte. (37). What is called dissociation? In the dissociation, the positive ion and the negative ion present in the original substance are separated showing dissociation, viz., Na+Cl-(s) + H2O(l) Na+(aq) + Cl-(aq) (38). What is called ionization? In ionization original substance is changed into ion form in aqueous solution , viz. CH3COOH(l) + H2O(l) CH3COO-(aq) + H3O+(aq) (39). Approximately how many litres of HCl is secreted in the stomach of human being? In the stomach of a human being about 1.2 to 1.5 litres HCl is secreted. (40). Which acid is included in vinegar? CH3COOH is included in vinegar. (41). What can be called acid and base according to Arrhenius theory? Substance which dissociate in water and give H+ are called acids and substance which dissociate in water and give out ions are called bases. (42). Mention two drawbacks of Arrhenius theory. (i) Proton is highly unstable. (ii) It cannot exist independently. (43). Give definitions of acid-base according to Lowry – Bronsted theory. The substance which gives proton is called the acid and substance which receives proton is called base. (44). Give definitions of Lewis acid – base. The substance which accept electron pair is acid and the substance which donate electron pair is base. (45). Which substances do not obey the definition of Lowry – Bronsted acid – base? BF3, SiO2, Ag+, Cu+2 etc do not obey definition of Lowry – Bronsted theory. (46). What is used to measure strength of different acid – base according to Lowry – Bronsted theory. According to Lowry – Bronsted theory, the substance has more capacity to donate proton is acid and more capacity to accept proton is base. (47). Write conjugate acids of water and dimethyl amine. H2O – H3O+ (conjugate acid of H2O) (CH3)2NH – (CH3)2N+H2 (conjugate acid of dimethyl amine) (48). What will be the conjugate acid of (C6H5)2NH ? (C6H5)2NH → (C6H5)2NH2+ is a conjugate acid. (49). Write conjugate acid – base of H2PO4-. H2PO4-1 - H3PO4 (conjugate acid) H2PO4-1 - HPO42- (conjugate base) (50). Give definition of conjugate base. When Lowry – Bronsted acid lose proton it gives conjugate acid. (51). Give definition of conjugate acid. When Lowry – Bronsted base accept proton, it gives conjugate acid. (52). What indicates the values of Ka and Kb ? Ka = ionization constant of weak acid. Kb = ionization constant of weak base. (53). CH3NH2 is stronger base in comparison to NH3. Why ? KbCH3NH2 > KbNH3 CH3NH2 has +I group with nitrogen which increases electron density on nitrogen and also increases basic strength. (54). Which substances act as Lewis acid – base? The substance which can accept electron pair is acid and the substance which can donate electron pair is base. (55). Classify the given substances in Lewis acid – base: AlCl3, BF3, CH3CH3OH, NH3, NH2-, Ag+ Lewis acid- AlCl3, BF3, Ag+, CH3CH2OH Lewis base- NH3, NH2-, CH3CH2OH (56). What is meant by self ionization of water? H2O + H2O H3O+(aq) + OH-1(aq) (57). What is meant by ionic product of water? Kw = [H3O+] = [OH-] = 1 x 10-14 at 298 K. A product of [H3O+] and [OH-] is known as ionic product of water. (58). What is concentration of pure water at 298 K temperature ? [H2O] = 55.5 M (59). Give definition of pH. A negative logarithm to the base 10 of H3O+ ion concentration is known as pH. (60). Give definition of pOH. A negative logarithm to the base 10 of OH- ion concentration is known as pOH. (61). What will be the sum of values of pH and pOH at 298 K temperature ? At 298 K, pH + pOH = 14. (62). On the basis of which values of pH, solutions can be said to be acidic, basic or neutral? pH = 7 neutral solution pH < 7 acidic solution pH > 7 basic solution (63). What is the pH of CH3COONa solution ? The pH of CH3COONa is greater than 7. (64). What is the pH of FeCl3 solution ? The pH of FeCl3 solution is less than 7. (65). What will be the effect of aqueous CaCl2 solution on litmus paper ? The aqueous solution of CaCl2 is acidic therefore it turns blue litmus paper to red. (66). What is meant by hydrolysis of salt ? When salt is added to water, ions of salt react with water and produce acid and base is known as hydrolysis. (67). Which instrument is used to determine accurate pH of solution? pH meter is used to measure pH. (68). What is meant by hydrolysis constant ? Hydrolysis reaction is an equilibrium reaction and so its corresponding equilibrium constant can be calculated which is known as hydrolysis constant. (69). What is called buffer solution ? The solution which resists the change in pH carried out by addition of acid or base in small proportion to them or are being diluted, and the values of their pH remain constant are called buffer solution. (70). Give one example each of acidic, basic and neutral buffer solutions. Acidic buffer – CH3COOH + CH3COONa Basic buffer – NH4OH + NH4Cl Neutral solution – CH3COOH + NH4OH (71). Give examples of sparingly soluble salts. AgCl, CuS, Al(OH)3, BaSO4 etc. are sparingly soluble salts. (72). Give definition of solubility product (Ksp). The product of concentration of ions of sparingly soluble salt in its aqueous solution is known as solubility product. (73). In which solvent, non polar substances like naphthalene dissolves ? Naphthalene dissolves in non- polar solvents like benzene, toluene etc. (74). Give formula for solubility product (Ksp) of sparingly soluble salt Sb2S3. Sb2S3(aq) 2Sb3+(aq) + 3S2-(aq) 2s 3s Ksp = [Sb3+]2[S2-]3 = [2S]2[3S]3 = 10855 (75). Mention the conditions between Ip and Ksp. When Ip > Ksp, salt will precipitate. Ip < Ksp, no precipitation. (76). Give four examples of sparingly soluble salts. AgCl, CuS, CdS, BaSO4. (77). What is indicated by the value of Ksp of sparingly soluble salt ? The value of Ksp indicates the solubility of a given salt. (78). What is not included in the equation of Ksp equilibrium constant of a sparingly soluble salt? Concentration of sparingly soluble salt and concentration of H2O because both are constant. (79). What is called common ion effect ? When concentration of one of the ions of sparingly soluble salt is increased, so that solubility of sparingly soluble salt decreases and the rate of reverse reaction increases. (80). The ions of which group are not precipitated in presence of NH4OH and NH4Cl in qualitative analysis? The ions of III B, IV, IV A groups are not precipitated in presence of NH4Cl and NH4OH. (81). The ions of which group are precipitated in presence of HCl and H2S in qualitative analysis? The ions of II groups are precipitated in presence of HCl and H2S. (82). What is meant by ionic product Ip? The product of cations and anions of sparingly soluble salt obtained from two different soluble salts is known as Ip. (83).How can acidic buffer solution be prepared? Give example. Acidic buffer solution can be prepared by mixture of weak acid and its salt with strong base. Eg: CH3COOH + CH3COONa (84). How can basic buffer solution be prepared? Give example. Basic buffer solution can be prepared by mixture of weak base and its salt with strong acid. Eg: NH4OH + NH4Cl (85). Mention the importance of buffer solutions. The pH of solution can be maintained by adding buffer solution . eg: pH of blood is 7.35. (86). How can neutral buffer solution be prepared? Neutral buffer solution can be prepared by neutralization of weak acid and weak base. Section-C 1. Mention the operational and conceptual definitions of acid and base. Ans. Acid means such a substance which is (1) Sour in taste (2) Turns wet Blue litmus paper red. (3) forms salt and water by reacting with base. (4) In certain circumstances produces hydrogen gas by reacting with metals. Similarly, base means such a substance which is (1) bitter in taste. (2) turns wet red litmus blue. (3) form salt and water by reacting with acid. These are operational definitions. The modern definitions used at present are called conceptual definitions in which Arrhenius, Bronsted-Lowry and Lewis definitions of acid base are included. Arrhenius concept about Acid and Base: According to Arrhenius concept, substances which dissociate in water and give hydrogen ion (H+) are called acids and substances which dissociate in water and give hydroxyl ions (OH-) are called bases. Concept of Bronsted – Lowry for acid and base: The substance which gives a proton or donates a proton is called the acid and the substance which receives the proton or accepts a proton is called the base. Lewis Concept of Acid and Base : Lewis mentioned that acid means a substance which can accept a pair of electrons and base means a substance which can donate a pair of electrons. 2. Derive the formula for dissociation constant of a weak acid. Ans. Ionization Constant of Weak Acid (Ka) : In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the equilibrium is obtained as below : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation is α, then following can be written : Reaction : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Initial Concentration(M) C 0 0 Degree of ionisation(α) 1-α α α Concentration at equilibrium(M) (1 - α)C αC αC [H3O+][A-] Equilibrium constant Ke= [HA][H2O] = (αC)(αC) ….1 (1 - α)C[H2O] But [H2O] is accepted as constant and so Ke[H2O] = (αC)(αC) (1 - α)C = α2C = Ka …2 (1 - α) Where Ka is the ionization constant or dissociation constant of the Weak acid HA. 3. Derive the formula for dissociation constant of a weak base. Ans. ionisation constant (Kb) of weak base: The ionization of monoacidic weak base MOH will take place in aqueous solution as follows: H2O MOH(aq) M+(aq) + OH-(aq) As base is weak, incomplete ionization will occur and so equilibrium will be obtained and it can be expressed as below: [M+] [OH-] …………1 and Kc = [MOH] [H2O] [M+] [OH-] =Ke ………. 2 Kex[H2O]= [MOH] Where Kb is the ionization or dissociation constant of monoacidic weak base. If we know the initial concentration of weak base and its degree of ionization. 4. Obtain the relation between equilibrium constants Kc and Kp. Ans. Relation between Kp and Kc As seen earlier the equilibrium constant of a gaseous reaction can be written as …. 1 But we know that according to simple gas equation PV = nRT. Hence, it can be written as P = (n/V)RT = CRT (where n/V = C = concentration in mol lit-1) Substituting the values of p in the above equation 1, it can be written as ….2 ….3 = Kc X (RT)ng …. 4 Where ∆ng = (c+d) – (a+b) Means number of total moles of gaseous products minus number of total moles of gaseous reactants. Hence, it can be written as Kp = Kc X (RT)ng…. 5 5. Derive the formula for solubility product of sparingly soluble salt CaF2. Ans. CaF2 Ca2+ + 2FS 2S 2+ - 2 Ksp = [Ca ][F ] = [S][2S]2 = 4S3. 6. Mention the use of effect of common ion in qualitative analysis. Ans. The use of effect of common ion can be made to separate one ion from the other in presence of other ion in qualitative analysis. It can also be used for decrease in solubility of the components in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of second group are less in comparison to solubility products of sulphides of metal of III B group ions, therefore, HCl is added before adding H2S water to test the second group ions. H2S(aq) 2H+(aq) + S2-(aq) HCl(aq) H+(aq) + Cl-(aq) The common ion available from HCl creates common ion effect on the equilibrium and decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group can only be precipitated because their solubility products are less. In the same way, for precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH. The concentration of OH- available from ionization of NH4OH gets decreased due to common ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group only will be precipitated because the values of solubility products of the hydroxides of III A group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions. It is necessary to note that under certain situations the solubility increases instead of decreasing. The solubility of salt like phosphates increase when acid is added to their solutions or pH of the solution decreases. The reason for this is that, phosphate ion combines with H+ available from acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases. 7. Give definition of buffer, explain acidic, basic and neutral buffer, giving suitable examples. Ans. “ The solution which resists the change in pH carried out by addition of acid or base in small proportion to them or are being diluted, and the values of their pH remain constant are called buffer solution.” (i) Acidic buffer solution: Acidic buffer solution can be prepared by mixture of weak acid and its salt with strong base. (ii) Basic buffer solution: Basic buffer solution can be prepared by mixture of weak base and its salt with strong acid. (iii) Neutral buffer solution: Neutral buffer solution can be prepared by neutralization of weak acid and weak base. These type of buffer solutions are shown below: Type Substances Value of pH Acidic CH3COOH + CH3COONa <7 Basic NH4OH + NH4Cl >7 Neutral CH3COOH + NH4OH =7 8. Bronsted-Lowry acid can be a Lewis acid but all Lewis acids cannot be Bronsted-Lowry acid. Explain giving suitable example. Ans. As far as bases are concerned, there is not much difference between Lowry-Bronsted and Lewis concepts as a base provides a lone-pair in both the cases. Further, the compounds which are classified as acids by Lowry-Bronsted concepts also acts as a lewis acids. However, Lewis concept fro an acid includes many substances that do not have a proton. Some such compounds are given below Lewis acid + Lewis Base BF3 + :NH3 F3B NH3 BF3 + :F BF4 Ag+(aq) + 2:CN-(aq) Ag(CN)-2(aq) + Ag (aq) + 2:NH3(aq) Ag(NH3)+2(aq) AlCl3 + :Cl AlCl4In the above reactions, :NH3, :F , :CN-, :Cl- all contain a lone-pair of electrons (depicted as : ) Utilizing which they react with species like BF3, Ag+ and AlCl3. Here BF3, Ag+ AlCl3 accept a lone-pair of electrons so they are lewis acid. While species like : NH3, :F-, :CN-, :Cl- donate electron pair and hence they are lewis bases. Thus a molecule acting as a lewis acid can be a positive ion as a molecule having a deficit of an electron pair. Similarly, a molecule acting as a Lewis acid can be an anion or a molecule having a lone-pair of electrons. 9. Derive the formula of ionic product of water Ans. Ionic Product of Water Water is an amphoteric oxide when acid is added to it; it accepts the proton and act as a base and when base is added to it, it donates a proton and acts as acid. When reaction between two molecules of water takes place, one molecule donates proton and other molecule receives proton and shows conjugate acid-base reaction. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Acid-1 Base-2 Conjugate acid-2 Conjugate base-2 If we express the equilibrium constant of above reaction them. [H3O+][OH-] Ka = [H2O][H2O] Where Ka is the dissociation constant of acid. There is no significant change in concentration of water (55.5M) because H2O is a weak acid (possesses about 10-7M H+). Thus, if H2O is considered constant. Ka x [H2O]2 = [H3+O] [OH-] = Kw. Where Kw is ionic product of water. Water is neutral and so [H3+O] and [OH-] in it are 10-7 M. Hence, Kw = [H3+O] [OH-] = (10-7)(10-7) = 10-14 which is constant and equilibrium constant remains constant at constant temperature; so the value of Kw will be constant at constant temperature, viz., the value of Kw is 1 x 10-14 at 298 K. 10. Explain calculation of pH of a solution from dissociation constant Ans. In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the equilibrium is obtained as below : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation is α, then following can be written : Reaction : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Initial Concentration(M) C 0 0 Degree of ionisation(α) 1-α α α Concentration at equilibrium(M) (1 - α)C αC αC + [H3O ][A ] Equilibrium constant Ke= [HA][H2O] = (αC)(αC) …1 (1 - α)C[H2O] But [H2O] is accepted as constant and so Ke[H2O] = (αC)(αC) (1 - α)C = α2C = Ka …2 (1 - α) Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we consider both the terms same) Hence, for any weak monobasic acid, following can be written [H3O+][A-] Ke= [HA] …3 -1 The unit of Ka will be mollit . Similarly Kb = [OH-][M+] [MOH] As the values of Ka depend on [H3+O] the values of ka will be different for different. Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite temperature. According to the relations seen earlier, pH = -log10[H3+O] pOH = -log10[OH-] Similarly, pKa = -log10[Ka] PA- = -log10[A-] can be written. Where [A-] is the concentration of negative ion. 11. Discuss the dynamic nature of equilibrium. Ans. Dynamic Nature of Equilibrium The most important matter in the case of equilibrium is that there is a continuous transformation of reactant to product and product to reactant. This state appears to be steady but it is not so. This type of reaction which takes place in both the directions is called reversible reaction and it is expresses by the symbol of two half arrows(↔). This symbol indicates that such reaction occurs simultaneously in both (forward and reverse) directions. Generally, the change of reactant to product is called forward reaction and the change of product to reactant is called reverse reaction. Thus, in reversible reactions, forward and reverse reactions continuously occur and we find it as equilibrium state. The mixture of reactants and products obtained at equilibrium time is called equilibrium mixture. The decomposition reaction of solid calcium carbonate in a cold vessel, at high temperature can be shown as below. ∆ CaCO3(s) ↔CaO(s) + CO2(g) The equilibrium is dynamic and not steady or static as the forward and the reverse reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed vessels. In the above reaction obtaining CaO and CO2 by decomposing of CaCO3and obtaining CaCO3 by combination of CaO and CO2 continuously take place. Suppose, if we deposit, some amount in our bank account and withdraw the same amount, then balance in the account appears steady or static. But this can be considered operative or dynamic and not closed or static. It is very difficult to determine the dynamic nature of equilibrium, even then with the help of radioactive isotope, it can be proved, viz. 14CO2 gas containing radioactive isotope, 14C and CaCO3 are taken in two different flasks and CO2 obtained by decomposition is connected with vessel containing 14CO2 gas, after sometime, Ca14CO3 will be formed in the vessel of CaCO3 and CO2 will be obtained in the vessel containing 14CO2. Thus if the equilibrium would have been steady, there must not be exchange of 12C and 14C. With the help of suitable counter, the radioactivity can be measured and the proof for the dynamic nature of equilibrium can be obtained through the proportions of concentrations of reactants and products remain constant. The reaction can be fast or slow depending upon the nature of the reactant and the experimental conditions. Equilibrium reactions can be divided into following three categories: (i) Reactions which are almost at the extent of completion and concentration of reactants may be negligible. It is not possible to detect this experimentally. (ii) Reactions in which the products are formed in very less proportions and most part of the reactant remains unchanged at the equilibrium. (iii) Reactions in which the concentrations of reactants and products are in comparable proportions at equilibrium. 12. Discuss the dynamic nature of chemical equilibrium Ans. Dynamic Nature of Chemical Equilibrium The dynamic nature of chemical equilibrium can be demonstrated by taking example of the production of ammonia. By keeping known quantities of dinitrogen and dihydrogen at high temperature and pressure in a closed vessel, the amount of ammonia formed can be determined at constant intervals by a series of experiments. The quantities of unreacted dinitrogen and dihydrogen also can be determined. From this it is concluded that even if the reactants and products are in different proportions, their concentrations are same at equilibrium. This constancy in composition indicates dynamic nature of equilibrium. For this, in synthesis of ammonia, deuterium (D2) instead of dihydrogen, (H2) is used and ammonia gas is produced by Haber process and it is studied. The results obtained are similar to those obtained above. In the mixture proportions of N2, D and ND3 instead of N2, H2, NH3 can be determined and equilibrium can be obtained. If D2 is added after the formation of ammonia by reaction of N2 and H2, the reaction may not occur but H in NH3 is displaced by D and ND3 can be determined by mass spectrometer. Thus, it is proved that in the reaction That the reactions from reactants to products and products to reactants that is forward and reverse reactions continuously occur with the same rates and so ND3 instead of NH3is obtained. By the use of radioactive isotope, the dynamic nature of equilibrium can be proved viz. For the reaction, H2(g) + I2(g) 2HI(g) 131 Radioactive isotope I of iodine can be used to study the dynamic nature of chemical equilibrium. As the equilibrium is dynamic, certain properties or factors are found similar. E.g. Intensity of colour, constant pressure, constant concentration etc. 13. Explain effect of catalyst on equilibrium Ans. Effect of catalyst: The use of suitable catalyst always helps to increase the rate of reaction, viz. iron powder is used as the catalyst in the production of ammonia by Haber’s process. Use of catalyst is associated with chemical kinetics because it affects the rate of reaction. The function of catalyst is to decrease the energy of activation. Hence, the reaction easily moves forward towards product. During this reaction, the energy of activation decreases but has no effect on equilibrium constant, that is, more proportion of product cannot be obtained. Let us examine the case of reaction of ammonia gas obtained by combination of dinitrogen gas and dihydrogen gas with the help of Haber’s process. N2(g) + 3H2(g) 2NH3(g) ∆H = -92.38 KJ molIn the above reaction, total 2 moles product are obtained from total 4 moles of reactants. Hence, according to Le Chatelier’s principle, increase in pressure is advantageous to get more ammonia but for the reactions occurring in closed vessel and so the pressure has to be kept limited. In addition, this reaction is exothermic and so according to Le Chatelier’s principle, the decrease in temperature is advantageous but decrease in temperature affects the rate of reaction. Hence, more time is required for completion of the reaction, which is not economically advantageous in industry. Hence, temperature is also to be restricted. By making compromise with these two and using catalyst whereby the energy of activation is decreased and increasing rate of reaction, more possible product is obtained in less time. Hence, Haber used iron powder as catalyst in the production reaction of ammonia and satisfactory results were obtained. In the production of ammonia, the values of optimum pressure and temperature respectively and iron powder is used as catalyst. If the value of equilibrium constant of the reaction Kc is very low then the use of catalyst is not fruitful or helpful. 14. Explain effect of concentration of equilibrium Ans. Effect of change in concentration: If we add or remove the reactant or the product from the reaction in equilibrium, its effect on equilibrium according to Le Chatelier’s principle will be as follows: (a)If the concentration of reactant or product is increased by addition of reactant or product, the reaction will occur in such a way that the increase in concentration will be taken for use, i.e. the increase in concentration of reactant, the concentration of product will increase and if concentration of product is increased the reaction will result in the direction of increase in concentration of the reactant. (b)If the concentration of reactant or product is decreased by removing reactant or product, the reaction will occur in such a way that product or reactant will be established again. Hence, if any change in concentration of reactant of product is carried out then the equilibrium will try to make this effect minimum and equilibrium will be established accordingly. If we take this as an example, in the reaction If concentration of reactants dinitrogen or dihydrogen is increased, the reaction will proceed towards right hand side and the product ammonia obtained will be more. If the concentration of nitrogen or hydrogen is decreased, the reaction will proceed towards left hand side and reactants of nitrogen or hydrogen will be obtained back i.e. production of ammonia will decrease. Here, it is necessary to remember that 1 mole dinitrogen combines with 3 moles of dihydrogen in this reaction and forms 2 moles of ammonia. Hence, increase in concentration of dihydrogen rather than dinitrogen, will give more product. We take another example of heterogeneous equilibrium. If solid CaCO3(s) is heated in closed vessel, the following decomposition reaction will occur. CaCO3(s) CaO(s) + CO2(g) Hence, if more CaO(s) is to be obtained then, the CO2(g) formed in the reaction should be removed because CO2(g), gas can combine with solid CaO(s) and carry out reverse reaction then proportion of product will decrease. Hence, by removing CO2(g) from the reaction vessel, more CaO(s) can be obtained. 15. Do the concentration of reactant and product remain constant at the time of equilibrium? Why? Ans. In the beginning of the reaction, the reactants which are in closed vessels, slowly change into products and the concentration of reactant decreases as the time proceeds. Along with this phenomenon, there is increase in concentration of product as the time proceeds. A certain point of time comes when there is no change in the concentration of reactant and product even if the time proceeds. This situation is called equilibrium state. 16. Deduce the formula of decomposition constant of solid NH4HS in a closed vessel. Ans. NH4HS(s) NH3(g) + H2S(g) x x = total 2x moles Kp = (PNH3) x (PH2S) = ( ½)P x (½)P = ( ¼)P2 atm2 17. Explain the effect of temperature on equilibrium by taking suitable example. Ans. Effect of temperature: The value of equilibrium constant is associated with temperature, i.e. the value of equilibrium is constant at constant temperature. If temperature changes its value also changes. Reactions can be of two types (1) Exothermic and (2) Endothermic, with the change in temperature, there is a change in absorbed or evolved heat. In exothermic reaction when heat is absorbed it works as a reactant. In exothermic reaction when heat is evolved, it works as a product. Hence, it can be said that the equilibrium constant increases with increase in temperature in endothermic reaction; equilibrium constant decreases with increase in temperature in exothermic reaction. Especially, let us make it clear that the increase or decrease in temperature affects the rate of reaction. Let us think about production of NH3(g). ∆H means change in enthalpy which is the difference in total enthalpies of products and reactants, its positive value indicates endothermic reaction and negative value indicates exothermic reaction. Hence, the above reaction is exothermic. As seen earlier that increase in temperature is not favourable for exothermic reaction because reaction will go in reverse direction and will decrease the product. Hence, decrease in temperature is advantageous to obtain more products but by decreasing the temperature the rate of reaction decreases. Hence, more time is required for completion of the reaction. Hence, as a compromise, at the lowest possible temperature and at the highest possible pressure, catalyst is used. Hence, more possible product is obtained in less possible time, this type of state is called optimization state. 18. Give Le-Chatelier’s principle and mention its importance. Ans. “ If from the factors determining the equilibrium state, any one factor is changed, there will be such a change in the system that the effect will be nullified or made negligible so that the value of equilibrium constant at that temperature will remain constant.” This principle can be applied to both physical and chemical equilibrium. 19. Explain the effect of change in pressure on equilibrium by taking suitable example. Ans. Effect of change of pressure The change in pressure can be carried out by increasing or decreasing the concentration of the gas or respectively decreasing or increasing the volume of the vessel. By carrying out this type of change, there will be change in proportions of gaseous reactants or products or total products. Le Chatelier’s principle can also be applied to such reactions. In heterogeneous equilibrium if we do not take into consideration the effect of pressure on solid or liquid substances in equilibrium, it can work because their volumes and concentration are independent of change of pressure. Let us take the following example: In this reaction 1 mole of reactant CO(g) reacts with 3 moles of reactant H2(g) and forms 1 mole product CH4(g) and 1 mole product H2O(g). thus 2 moles product is obtained from 4 moles of reactants. Hence, there is decrease in number of moles during reaction. Suppose, the pressure on the closed vessel in which the reaction is carried out at constant temperature and the volume of equilibrium mixture is made half, then what will happen? Total pressure will be double because PV = constant. The concentrations or pressures of reactants or products are increased. Hence, according to Le Chatelier’s principle, the equilibrium will try to attain the original state. As pressure is doubled ad 2 moles of products are obtained and 4 moles of reactants, there is decrease in number of moles and the reaction will go in forward direction, i.e. more product will be obtained. As a reverse of this, if in reaction CaCO3(s) CaO(s) + CO2(g) The number of moles of products increases (from 0 to 1). If pressure is increased by addition of CO2(g), then, the reaction will become reverse reaction and will decrease the product. 20. Explain the effect of increase in temperature on the system 2SO2(g) + O2(g) 2SO3(g) + heat Ans. The above reaction is an exothermic reaction. According to Le-Chatelier’s principle, increase in temperature shifts equilibrium to the left and decrease in concentration of products 21. What will be the effect on product in the system N2(g) + 3H2(g) 2NH3(g) by increasing the pressure? Explain on the basis of Le-Chatelier’s principle. Ans. N2(g) + 3H2(g) 2NH3(g) 1 mole 3 mole 2 mole In this reaction total 2 moles of products are obtained from 4 moles of reactants. According to Le-Chatelier’s principle, as pressure is increased and 2 moles of products are obtained and 4 moles of reactants, there is decrease in number of moles and the reaction will go in forward direction i.e more product will be obtained 22. Discuss homogeneous and heterogeneous equilibrium taking suitable examples Ans. (i) Homogeneous Equilibrium: In homogeneous equilibrium all the reactants and products will be in one similar phase. viz. The product ammonia produced by the reaction of reactants dinitrogen and dihydrogen, all are in gaseous phase. N2(g) + 3H2(g) 2NH3(g) Similarly, the hydrolysis of methyl acetate is also homogeneous equilibrium because, the reactants and the products in it are also in the same phase (liquid phase) CH3COOCH3(l) + H2O(l) CH3COOH(l) + CH3OH(l) In addition to this, the reaction between aqueous solution containing Fe3+ ion and aqueous solution containing SCN- ions is also an example of homogeneous equilibrium. But the only change in it is that this equilibrium is ionic equilibrium. Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq) All the equilibrium reactions and their equilibrium constants that we discussed earlier are the examples of homogeneous equilibrium and also derived from the relations between Kp, Kc and Kx related to them. For calculation of values of Kp pressures must be expressed in unit of bar because, bar is the unit in standard condition but in SI unit it is Pascal. The relation between them is as follows: 1 Pascal Pa = 1 Nm-2(Newton meter-2) and 1 bar = 105 Pa (ii) Heterogeneous Equilibrium : If reactants and products, posses more than one phase, the equilibrium is called heterogeneous equilibrium, The equilibrium between water (liquid) and water vapour(steam) in a closed vessel is the example of heterogeneous equilibrium. H2O(l) H2O(g) Similarly, Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Is also the example of heterogeneous equilibrium.(Equilibrium between solid and liquid). It is necessary to note here that this is an example for ionic equilibrium. Generally, in heterogeneous equilibrium, pure solid or liquid are associated, the concentrations of reactants and products can be separated, viz. The concentration of pure solid or liquid is its density and density is constant at constant temperature. Hence, concentration can also be taken as constant or the pure solid or liquid which are present will be independent of concentration. Suppose, some substance X is involved in this, then concentrations X(s) and X(l) will be taken as constant in whatever proportion they are present while concentrations of X(g) and X(aq) will change and will vary with volume. 23. What will be the effect on equilibrium system H2O(g) + CO(g) H2(g) + CO2(g) if water vapour is introduced ?Explain. Ans. Increase in vapour will lead to increase in reactants because H2O(g) is in vapour state. Hence, acc. to law, equilibrium is shifted in forward direction and hence, product increases and Kc also increases. 24. What will happen if Cl2 is removed from the equilibrium system PCl5(g) PCl3(g) + Cl2(g)? Explain. Ans. If Cl2(g) is removed (i.e product is removed) from the system, conc. Of product will decrease and acc. to law, equilibrium shifts in forward direction resulting in increased conc. Of products [i.e Cl2] and hence increases Kc. 25. Mention general characteristics of equilibrium involved in physical process. Ans. The following are the general characteristics of systems in equilibrium for the physical systems (i) At constant temperature, equilibrium is possible only in closed system. (ii) Both the processes (reactions) opposite to each other, that is, the forward and the reverse reactions occur at the same rates and the equilibrium involved in it remains dynamic but condition remains steady (static). (iii)All the properties f the system which can be measured remains constant. (iv)For physical processes, when equilibrium is obtained then at constant temperature, the value of one of the factors remains constant. The list of these properties is shown in Table 4.1 (v)At any stage magnitude of such quantities show to what extent the physical process has advanced before reaching to equilibrium. 26. Predict the direction of forward or reverse reaction Ans. R P If concentration of [R] increases, reaction is in forward direction. If concentration of [R] decreases, reaction is in reverse direction. If concentration of [P] increases, reaction is in reverse direction. If concentration of [P] decreases, reaction is in forward direction. 27. Give conjugate acid-base theory giving suitable example. Ans. Danish chemist Bronsted, an English chemist Lowry presented the concept of acid and base. They made H+ (proton) as a base. According, to their concept , the substance which gives a proton or donates a proton is called the acid and the substance which receives the proton or accepts a proton is called the base. Thus, acid is a proton donor and base is proton acceptor. Let us take the dissociation reaction of hydrogen chloride in water. Base Acid H3+O (aq) + Cl-(aq) HCl(g) + H2O(l) Acid Similarly, Base Acid NH3(g) + H2O(l) Base NH4+(aq)+ OH-(aq) Base Acid We shall understand in detail the first from the above reactions : HCl H+ + ClAcid-1 proton conjugate base-1 As it gives proton, HCl is an acid H2O + H+ H3O+ Base-2 Proton Conjugate acids-2 As it accepts a proton, H2O is a base. Total reaction: HCl + H2O H3O+(aq) + Cl-(aq). Acid-1 Base-2 Acid-2 Base-1. In the above reaction giving-taking of proton is not shown. Hence, it can be said that only transfer of proton takes place, it is not obtained free. Every acid will lose proton and so its conjugate base will be formed and every base will accept a proton and so its conjugate acid will be formed. Hence, this concept is known as proton transfer or conjugate acid-base concept. 28. Mention the relation between solubility product and ionic product Ans. Ksp > Ip then precipitation will take place Ksp < Ip then precipitation will not take place. Ksp = Ip then reaction will be in equilibrium. 29. Water is amphoteric. Explain it on the basis of Lowry-Bronsted theory Ans. Danish scientist Bronsted and English chemist Lowry presented the concept of acid and base. They made H+(proton) as a base. According to their concept, the substance which gives a proton or donates a proton is called the acid and the substance which receives a proton or accepts a proton is called the base. Water also donates H+ and also receives a proton hence water is amphoteric. 30. How can it be said that pure water is neutral? Ans. Ionic product of water is 10-14 and moreover [H3O+] = [OH-] = 10-7 which is constant and equilibrium constant remains constant at constant temperature. Hence, water is neutral. 31. What is the value of ionic product of water at 298 K temperature? What is the effect of temperature in its value? why? Ans. The value of ionic product of water at 298K is 1 x 10-14. As the process of self ionization of water is endothermic, with increase in temperature, reaction goes in forward direction resulting in increase concentration of [H+] and [OH-] ions. Hence, with increase in temperature, Kw increases. 32. Aqueous solution of FeCl3 is acidic. Explain Ans. FeCl3 is a salt of strong acid and weak base. FeCl3 Fe3+(aq) + 3Cl-(aq) HCl H+ + Cl- (acid is strong, so it gets fully dissociated) Fe(OH)3 Fe3+ + 3OH- (weak base so partially dissociates) Moreover, hydrolysis constant for strong acid and weak base is Kh = Kw/Kb = [OH-]/Co. Hence pH of solution is < 7. Hence it is acidic. 33. Aqueous solution of CH3COONa is basic. Explain Ans. CH3COONa is a salt of strong base and weak base. CH3COONa CH3COO- + Na+ CH3COOH CH3COO- + H+ (partial dissociates) + NaOH Na + OH- (completely dissociates) Moreover Kh of weak acid and strong base is Kh = Kw/Ka = [H3O]2/Co. Hence pH of solution is >7. Hence it is basic. 34. Deduce the formula of Ksp of sparingly soluble salt bismuth sulphide – Bi2S3. Ans. Bi2S3 2Bi3+ + 3S23+ 2 2- 3 Ksp = [Bi ] [S ] = (2S)2(3S)3 = 108S5 35. Deduce the formula of Ksp of sparingly soluble salt Zinc Phosphate- Zn3(PO4)2. Ans Zn3(PO4)2 3Zn2+ + 2PO433S 2S 2+ 3 3- 2 3 Ksp = [Zn ] [PO4 ] = [3S] [2S]2 = 27S3 x 4S2 = 108S5 36. Deduce the formula of Ksp of sparingly soluble salt aluminium hydroxide Al(OH)3 Ans. Al(OH)3 Al3+ + 3OHS 3S 3+ - 3 3 Ksp = [Al ][OH ] = [S][3S] = 27S4 37. The solubility of AgCl decreases in presence of NaCl. Explain giving reason Ans. A saturated aqueous solution of silver chloride has following equilibrium : AgCl(s) Ag+(aq) + Cl-(aq) If, to this solution, a small amount of NaCl is added, the concentration of Cl - will be increased. Therefore, some cl- ions will combine with Ag+ ions to form insoluble AgCl. Thus the solubility of AgCl decreases because of common ion effect 38. Derive formula of Ksp for sparingly soluble salt- Ca3(PO4)2. Ans. Ca3(PO4)2 3Ca2+ + 2PO433S 2S 2+ 3 3- 2 3 Ksp = [Ca ] [PO4 ] = [3S] [2S]2 = 27S3 x 4S2 = 108S5 39. The solubility of AgCl decreases by addition of AgNO3 to a saturated solution of AgCl. Explain Ans. AgCl(s) Ag+(aq) + Cl-(aq) If, to this solution, a small amount of AgNO3 is added, the concentration of Ag+ will be increased. Therefore, some Ag+ ions will combine with Cl- ions to form insoluble AgCl. Thus the solubility of AgCl decreases because of common ion effect 40. White precipitates are obtained when HCl gas is passed through a saturated solution of NaCl. Give reason Ans. In the solution of some electrolytes the concentration of undissociated electrolysis increases under the effect of common ion effect. For e.g, on passing HCl gas through a saturated solution of NaCl, the concentration of Cl- will increase. As a result of this NaCl is precipitated. Common ion effect is utilized for the purification of impure NaCl obtained from sea water. 41.After addition of small amount of HCl in a mixture of Cu2+ and Zn2+, if H2S gas is passed from the solution, CuS is precipitated but ZnS is not precipitated. Give reason Ans. Use is made of common ion effect in qualitative analysis. If H2S gas is passed through an acidified solution (acidified by HCl) containing Cu2+ and Zn2+ ions, only copper is precipitated as CuS. Zns is not precipitated. In the presence of HCl, the ionization of H2S is decreased. Consequently, the concentration of S2- ions being much low, only CuS is precipitated. 42. The value of Kp for the reaction NH2COONH4(s) atm3. What will be the value of Kc for this reaction? Ans. Data: Kp = 3.2 x 102 atm3 T = 298K R = 0.082 atm lit mol- k-. ∆n(g) = np- nr (in gaseous state) = 3- 0 = 3. Kp = Kc x (RT)∆n(g) So, Kc = Kp (RT)∆n(g) So, Kc = 3.2 x 102 (0.082 x 298)3 Kc = 3.2 x 102 (24.436)2 2NH3(g) + CO2(g) is 3.2 x 102 So, Kc = 2.193 x 10-2 (mol/litre)3. 43. The value of Kp = 600 atm3 is found for the reaction when solid NH4COONH2(s) is heated at 400 K temperature; What will be the value of total pressure of the system at equilibrium state? Ans. NH4COONH2(s) 2NH3(g) + CO2(g) From, the equation, use have at equilibrium, that PNH3 will be twice than that of Pco2. At equilibrium, let Pco2 be x, then PNH3 will be 2x KP = [NH3]2 x [CO2] = (2x)2 x X = 4x3 Kp = 600 atm3 (from data) = 4x3 x3 = 150 atm3 X = 5.172 atm (from log) Now, total pressure Ptotal = PNH3 + Pco2 = 2x + x = 3x = 3 (5.172) Ptotal = 15.516 atm. 44. The value of Kp = 0.05 atm for the reaction A(g) + 2B(g) 3C(g) + D(g). Write the value Kc terms of R at 1000 K temperature. Ans. Data: Kp = 0.05 atm T = 1000 K ∆n(g) = np – nR ( In gaseous state) =4–3 =1 Kp = Kc x (RT)∆n(g) Kc = Kp (RT)∆n(g) = 0.05 (R x 1000)-1 = 5 x 10-2 (R x 103)1 Kc = 5 x 10-5 atm K-1 R 45. What will be the value of Kp if the pressure at equilibrium is 3 atm for the reaction. NH3COONH2(s) 2NH3(g) + CO2(g) ? Ans. NH3COONH2(s) 2NH3(g) + CO2(g) From the equation, use have, at equilibrium PNH3 will be there than that of PCO2. At equilibrium, at PCO2 = x PNH3 = 2x Ptotal = PNH3 + PCO2 = 2x + x = 3x Nowwe are given, Ptotal = 3atm 3x = 3 atm x = 1 atm Now, Kp = (PNH3)2 x PCO2 = (2x)2 x X = 4x3 Kp = 4x3 = 4 x (1 atm)3 Kp = 4 atm3. 46. In the reaction H2(g) + I2(g) 2HI(g), 0.4 moles H2 and I2 are taken. At equilibrium 0.5 mole HI is formed. Find equilibrium constant Kp. Ans. Data: Reaction: H2(g) + I2(g) 2HI(g) Conc. at Equilibrium 0.15 0.15 0.15 [ Total no. of moles of reactant initially of H2 and I2 are 0.4 moles each, hence total no. of moles = 0.8 moles. While at equilibrium, 0.5 moles of product are formed and at equilibrium [H2] = [I2] → 0.3 moles are unreacted at equilibrium [H2] = [I2] = 0.15 moles (at equilibrium)] Kc = [HI]2 [H2] [I2] = (0.5)2 (0.15) (0.15) = 0.25 0.0225 Kc = 11.11 Now, Kp = Kc x (RT)∆n(g) ∆n(g) = np – nR (In gaseous state) = 2 – (H1) =0 Kp = Kc x (RT)0 Kp = Kc Kp = 11.11 [Kp has no unit because ∆n(g) = 0] 47. Calculate the pH of 0.5M HF(aq). Ka = 2 x 10-1. Ans. Data: Ka = 2 x 10-4 H2O HF H3O+ + F0.5M Ionisation constant, Ka = [H3O+] [F-] [HF] [As, ratio of H3O+ and F- ion is 1:1, then moles would be equal Ka = [H3O+]2 HF -4 2 x 10 = [H3O2] [H3O+]2 = 1 x 10-4 0.5 + 2 [H3O ] = 1 x 10-2 M PH = -log [H3O+] or –log [H+] = -log [1 x 10-2] = -log1 + -log(10-2) = 0 + 2 (log 10) PH = 2 [H3O+] = [F-]] log (axb) = log a + log b log ab = b log a log10 = 1 log1 = 0 48. KSP of Mg(OH)2 = 1 x 10-12 At what pH 0.01M Mg(OH)2 will be precipitated? Ans. Mg(OH)2(s) Mg2(aq) + 2OH-(aq) Let the solubility be S mol/litre [Mg2+] = S mol/litre [OH-] = 2S mol/litre KSP = [Mg2+][OH-]2 = [S] [2S]2 = 4S3 KSP = 1 x 10-12 given KSP = 4S3 = 1 x 10-12 → S3 = ¼ x 10-12 M3 S = (¼)1/3x 10-4 M [ (¼)1/3= log 1/3 (1-4) = 1/3 (log1 – log4) = 1/3 (0 – 0.6021) = - 0.2007 = anti log (- 0.2007) = 1 + antilog (0.7993) = 6.299 x 10-1 S = 6.299 x 10- M ] [OH-] = 2 x 6.299 x 10-5 M = 12.598 x 10-5 M = 1.2598 x 10-4 M POH = -log [OH-] = -log (1.2598 x 10-4) = 4 log 10 – log (1.2598) = 4 – 0.01 = 3.99 PH + POH = 14 → PH = 14 - POH = 14 – 3.99 PH = 10.01 49. What will be the concentration of H+ when 0.1M HCN is mixed with 0.2M NaCN ? Ka = 6.2 x 10-10 [Hint : [H+] = Ka x Concentration of Acid solution Concentration of Salt solution Ans. NaCN Na+ + CN- (Salt solution) 0.2M HCN H+ + CN- (Acid solution) 0.1M + [H ] = Ka x Concentration of Acid solution Concentration of Salt solution = 6.2 x 10-10 x 0.1 0.2 [H ] = 3.1 x 10 M. 50. Find the change in pH of 1.0ml 0.10M HCL when its volume is made 50 ml by adding water? Ans. HCl H+ + Cl- (As ratio of 1:1) 0.1M 0.1M 0.1M H P 1 = -log [H+] = -log [1 x 10-1] pH1 = 1 → (1) Now, as volume 50 mol is added to the solution, udarity of solution nill change. M1 V1 = N2 V2 Now, M1 = 0.10M V1 = 1ml V2 = 50ml N2 = ? (0.10) x (1) = N2 x (50) N2 = 0.1 50 -3 M2 = 2 x 10 M PH2 = - log10 [H+] 9after adding 50 ml to solution) = - log [2 x 10-3] = - [log 2 + -3 (log10) ] = - (0.3010 -3) [ log 10 = 1] PH2 = 2.699 → (2) H H H N our change in p = p 2 - p 1 = 2.699 – 1 [ from equation (1) and (2) ] = 1.699 Change in pH = 1.699 51. The dissociation constant of HCOOH is 1.8 x 10-4. What will be the diassociation constant of conjugate base-formate ion? Ans. Dissociation Constant, Ka = 1.8 x 10-4 HCOOH HCOO-(aq) + H2(aq) + -10 Weak Acid Conjugate Base HCOO- + H2O HCOOH(aq) + OH-(aq) Hydrolysin constant, Kn Now, Kn = Kw = Ionic product of water Ka Dissouolion constant of HCOOH -14 Kn = 1 x 10 1.8 x 10-4 Kn = 5.55 x 10-9 52. A saturated solution can be prepared by dissolving 0.08gram. CaF2 in 2.901 litre at 298 K tempareture. Find its KSP. (Molecular mass = 78 grammol-1). Ans. W = 0.08 gm CaF2 Molecula mass = 78 gm mol-1 Moles of CaF2 = 0.08 78 = 1.0256 x 10-3 moles Conc. of CaF2 = Moles Litre = 1.0256 x 10-3 2.901 = 3.53547 x 10-4 mol/litre → (1) CaF2 Ca2+ + 2FS mol/litre 2S mol/litre Let Dolubility Product of CaF2 be S mol/litre Then [Ca2+] = S mol/litre [ F-] = 2S mol/litre KSP = [Ca2+] [ F-]2 = [S] [2S]2 = 4S3 = 4 x (3.5347 x 10-4)3 M3 = 4 x 44.1629 x 10-12 M3 = 1.7665 x 10-10 (mol/litre)3 KSP = 1.76 x 10-10 M3 53. The value of Ka of a weak acid is 109 times more than the value of Kw. Calculate pH and pOH of 0.1 m aqueous solution of this acid. Ans. For weak acid, HA(1) + H2O H3O+(aq) + A-(aq) Ka = [H3O+] [A-] { Concentration of positive ion [H3O+] = concentration of [HA] negetive ion [A-]} Now, Weak acid will ionize to much less ectent and so concentration of its unionized part will remain nearly same as its original concentration i.e. O.1M [HA] = 0.1M Ka = 109 Kw (as given) But, Kw = 1 x 10-14 (Ionic produce of water) Ka = 109 x (1 x 10-14) Ka = 10.5 → (1) Now, for Ionisatin of weak acid, Ka = [H3O+]2 [Ha] 1.5 = [H3O+]2 1 x 10-1 + 2 [H3O ] = 10.6 [H3O+]2 = 1 x 10-3 M Now, PH = -log10 [H3O+] = -log [1 x 10-3] = -[log 1 + -3(log10)] = - (-3) ( log 1 = 0) H P =3 ( log 10 = 1) Now, pH + pOH = 14 pOH = 14 – pH = 14 – 3 pOH = 11 54 .In aqueous solution of 0.05 M C6H5OH the ionisation constant Ka = 1.28 x 10-10 calculate concentration of H3O+ ion in the solution. Ans. C6H5OH(1) + H2O H3O+(aq) + C6H5O-(aq) Phenol (phenoxy Ion) Now, Ka=[H3O+]2 [C6H5OH] [Concentration of H3O+ ion and C6H5O- ion will remain nearly equal. Moreover, weak acid will ionize to much less extant and hence, concentration of unionised part will remain nearly same as original concentration] Given, Ka = 1.28 x 10-10 [C6H5OH] = 5 x 10-2M Ka=[H3O+]2 [C6H5OH] [H3O+] = Ka. [C6H5OH] + -10 -2 1/2 [H3O ] = (1.28 x 10 x 5 x 10 ) = (6.4 x 10-12)1/2 + [H3O ] = 2.529 x 10-6M = 2.53 x 10-6M 55. Calculate pH of 0.002M HNO3 solution. Ans. HNO3 H+(aq) + NO3-(aq) 0.002M 0.002M [As ratio is 1:1] + -3 [H ] = 2 x 10 M PH = -log [H+] = -log [2 x 10-3] = -[log 2 + log (10-3)] [log(axb) = log a + log b] = -[0.3010 + -3(log 10)] [logab = b loga] = -[0.3010 – 3] [ log 10 = 1] H p = 2.6990 56. Calculate values of pH and POH of 0.06 M H2SO4 solution. Ans. H2SO4 2H+ + SO420.06M 0.12M 0.06M [ As ratio of H+ ion production is 2:1, hence, molar concentration will be doubled H+ = 2 x 0.06M = 0.12M] Now, pH = -log [H+] = -log [0.12] = -log [1.2 x 10-1] = -log 1.2 – log (10-1) = -0.0792 + 1 [ log 10 =1] = 0.9208 pH = 0.9208 Now, pH + pOH = 14 0.9208 + pOH = 14 OH p = 14 – 0.9208 = 13.0792 OH p = 13.0792 57. Whatwill be [OH-] in 100 ml. NaOH solution having pH = 10.0? Ans. Data: pH = 10.0 [OH-] = ? pH = -log [H+] = 10 = -log [H+] = 10 = -log [H+] = -10 [H+] = 1 x 10-10M Now, H2O(1) H+(aq) + OH-(aq) → (1) [ Self – lonization of water ] Weak electrolyte Kw = [H+] [OH-] = 1 x 10-14 ( found at 250 C) Now, from equation (1), [H+] = 1 x 10-10 1 x 10-10 x [OH-] = 1 x 10-14 [OH-] = 1 x 10-4M 58. [H3O+] in aqueous solution of CH3COOH is 1 x 10-3M. What will be its intial concentration if Ka = 1.25 x 10-5M ? Ans. [H3O+] = 1 x 10-3M Ka = 1.75 x 10-5M For weak acid, CH3COOH(1) + H2O(1) H3O+(aq) + CH3COO-(aq) + Ka = [H3O ] [CH3COO ] [CH3COOH] Now, the concentration of positive ion [H3O+] =concentration of positive ion [CH3COO-] → Moreover, weak acid (CH3COOH) will ionize to much less extent and so concentration of its unionized part will remain nearly same as its original concentration [CH3COOH] = Co Ka = [H3O+]2 Ka = [H3O+]2 Co Co = (1 x 10-3)2 1.75 x 10-5 Co = 5.714 x 10-2M 59. KSP of Mg(OH)2 = 1.2 x 10-11. Find solubility in pure water. (S = 1.442 x 10-4M) Ans. Suppose, the solubility of Mg(OH2) is S mol/litre. Then, the concentration of [Mg2+] and [OH-] ion in saturated solution will be as follows. Mg2+(aq) + 2OH-(aq) S mol/litre 2S mol/litre Now, for Mg (OH)2, KSP = [Mg2+] [OH-]2 = (S) (2S) = 4S3 → (1) KSP = 4S3 KSP = 1.2 x 10-11 (from data) From eqn(1) KSP = 1.2 x 10-11 = 4S3 S3 = 3 x 10-12 S = 1.442 x 10-4M [ log (S3) = log (3) 3 log S = 0.4771 log S = 0.4771 3 Mg(OH)2 = 0.1590 S = 1.442] Section –D 1. Explain dynamic nature of chemical and physical equilibrium. Ans. Dynamic Nature of Equilibrium The most important matter in the case of equilibrium is that there is a continuous transformation of reactant to product and product to reactant. This state appears to be steady but it is not so. This type of reaction which takes place in both the directions is called reversible reaction and it is expresses by the symbol of two half arrows(↔). This symbol indicates that such reaction occurs simultaneously in both (forward and reverse) directions. Generally, the change of reactant to product is called forward reaction and the change of product to reactant is called reverse reaction. Thus, in reversible reactions, forward and reverse reactions continuously occur and we find it as equilibrium state. The misture of reactants and products obtained at equilibrium time is called equilibrium misture. The decomposition reaction of solid calcium carbonate in a cold vessel, at high temperature can be shown as below. ∆ CaCO3(s) ↔CaO(s) + CO2(g) The equilibrium is dynamic and not steady or static as the forward and the reverse reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed vessels. In the above reaction obtaining CaO and CO2 by decomposing of CaCO3and obtaining CaCO3 by combination of CaO and CO2 continuously take place. Suppose, if we deposit, some amount in our bank account and withdraw the same amount, then balance in the account appears steady or static. But this can be considered operative or dynamic and not closed or static. It is very difficult to determine the dynamic nature of equilibrium, even then with the help of radioactive isotope, it can be proved, viz. 14CO2 gas containing radioactive isotope, 14C and CaCO3 are taken in two different flasks and CO2 obtained by decomposition is connected with vessel containing 14CO2 gas, after sometime, Ca14CO3 will be formed in the vessel of CaCO3 and CO2 will be obtained in the vessel containing 14CO2. Thus if the equilibrium would have been steady, there must not be exchange of 12C and 14C. With the help of suitable counter, the radioactivity can be measured and the proof for the dynamic nature of equilibrium can be obtained through the proportions of concentrations of reactants and products remain constant. The reaction can be fast or slow depending upon the nature of the reactant and the experimental conditions. Equilibrium reactions can be divided into following three categories: (i) Reactions which are almost at the extent of completion and concentration of reactants may be negligible. It is not possible to detect this experimentally. (ii) Reactions in which the products are formed in very less proportions and most part of the reactant remains unchanged at the equilibrium. (iii) Reactions in which the concentrations of reactants and products are in comparable proportions at equilibrium. Dynamic Nature of Chemical Equilibrium The dynamic nature of chemical equilibrium can be demonstrated by taking example of the production of ammonia. By keeping known quantities of dinitrogen and dihydrogen at high temperature and pressure in a closed vessel, the amount of ammonia formed can be determined at constant intervals by a series of experiments. The quantities of unreacted dinitrogen and dihydrogen also can be determined. From this it is concluded that even if the reactants and products are in different proportions, their concentrations are same at equilibrium. This constancy in composition indicates dynamic nature of equilibrium. For this, in synthesis of ammonia, deuterium (D2) instead of dihydrogen, (H2) is used and ammonia gas is produced by Haber process and it is studied. The results obtained are similar to those obtained above. In the mixture proportions of N2, Dand ND3 instead of N2, H2, NH3 can be determined and equilibrium can be obtained. If D2 is added after the formation of ammonia by reaction of N2 and H2, the reaction may not occur but H in NH3 is displaced by D and ND3 can be determined by mass spectrometer. Thus, it is proved that in the reaction That the reactions from reactants to products and products to reactants that is forward and reverse reactions continuously occur with the same rates and so ND3 instead of NH3is obtained. By the use of radioactive isotope, the dynamic nature of equilibrium can be proved viz. For the reaction, H2(g) + I2(g) 2HI(g) Radioactive isotope 131I of iodine can be used to study the dynamic nature of chemical equilibrium. As the equilibrium is dynamic, certain properties or factors are found similar. E.g. Intensity of colour, constant pressure, constant concentration etc. 2. Explain equilibrium solid in solution and gas in solution. Ans. Equilibrium Involving Dissolution of Solid in Solution For the study of this type of equilibrium, the example of sugar and its solution in water can be taken. At constant temperature, take some water. Add some sugar into it and stir. In the beginning sugar easily dissolves, but as more and more amount of sugar is added, it dissolves according to its solubility and then some amount of sugar remains in solid form without dissolution. We know this state as saturated solution but in this system the equilibrium is also establishes sugar(solid) and liquid(solution of sugar). This can be expressed as below: Sugar(s) Sugar solution(aq) Solid liquid As studies earlier, equilibrium is dynamic because the forward and the reverse reactions continuously occur in each system. In this system, the amount of sugar that dissolves in water, the same amount of sugar separates from solution of sugar. Hence, the number of molecules of sugar and number of aqueous molecules of sugar in the solution remain constant in this system. Equilibrium involving Gas and Solution At constant temperature and pressure, carbon dioxide can be dissolved in water in a closed vessel(system), so that a system containing gas and a solution of carbon dioxide can be formed. As temperature and pressure are constant, carbon dioxide dissolves according to its pressure and temperature and forms solution of carbon dioxide and the excess carbon dioxide gas remains in equilibrium with it. Equilibrium is dynamic and so molecules have carbon dioxide gas that dissolve in water is the same as the number of molecules of the gas that release from the solution in the system. Thus, in this closed system, the total number of molecules(amount) of carbon dioxide in gaseous form and those that have dissolved in water remain constant. This equilibrium can be shown as below: CO2(g) CO2(soln) All the reactions studies above are processes in which only physical change takes place and so they all are examples of physical process equilibrium. 3. Mention the characteristics of chemical equilibrium Ans. Characteristics of Chemical Equilibrium (i) During chemical equilibrium, the properties like colour, concentration, pressure or temperature of the system remain constant and they are similar in the total area of the system. (ii) When chemical equilibrium is attained, the rates of forward and reverse reactions become equal. (iii) If the factors like concentration, pressure, temperature etc. which affect the chemical equilibrium are changed, they produce effect on equilibrium. (iv) Even if the initial concentrations are different, the equilibrium constant remains constant at constant temperature. (v) The value of equilibrium constant changes if the temperature changes. (vi) For attainment of equilibrium, the reaction can be carried out from left to right (reactants to products) or from right to left (products to reactants) (vii) There is no effect of catalyst on the equilibrium constant and so the proportions of products remain same but the rate of reaction to attain equilibrium increases. 4. Give chemical equilibrium (active masses) law and derive its formula Ans. Law of Chemical Equilibrium and Equilibrium Constant The mixture of reactants and products at equilibrium is called equilibrium mixture. We shall study the relation between concentrations of reactants and products at equilibrium state. Let us take a simple reversible reaction as follows: A+B C + D In this reaction A and B are reactants and C and D are products. This means that in this reaction moles of reactants and products are one each but in all reactions this may not happen. Hence, it is necessary that their moles are expressed. Balanced reaction determines their moles. Viz. N2(g) + 3H2(g) 2NH3(g) From the experimental studies of many reversible reactions scientists of Norway, Guldberg and Waage mentioned in 1864 that the concentrations of substances in equilibrium mixture can be expressed by following equilibrium equation. Where Kc is equilibrium constant and [ ] bracket expresses concentration of reactant or product on mol/lit or M. The equilibrium equation is also known as law of active masses because in the early years of chemistry, concentration was said to be ‘active mass’. Now, we shall derive the equation for equilibrium constant of a general reaction. Suppose, if a reaction takes places, as given below in which the reactants and products are shown in balanced form with their proper moles (a, b, c or d) aA + bB cC + dD On the basis of Guldberg and Waage’s law the rate of forward reaction Vf [A]a [B]b or Vf = Kf [A]a [B]b Where Kf is the proportionality constant for forward reaction. The rate of reverse reaction Vr [C]c [D]d Vr= Kr [C]c [D]d Where Vr is the proportionality constant for reverse reaction. At equilibrium the rates of forward and reverse reaction will be equal and so Vf = Vr That is, Kf[A]a [B]b = Kr [C]c[D]d Thus, when equilibrium is attained if we determine the concentration of the reactants and the products in any reaction and their stoichiometric multiples, the equilibrium constant Kc can be obtained. 5. What is meant by Kc and Kp? Deduce relation between Kc and Kp. Ans. Relation between Kp and Kc As seen earlier the equilibrium constant of a gaseous reaction can be written as …. 1 But we know that according to simple gas equation PV = nRT. Hence, it can be written as P = (n/V)RT = CRT (where n/V = C = concentration in mol lit-1) Substituting the values of p in the above equation 4.10, it can be written as ….2 …3 = Kc X (RT)ng …. 4 Where ∆ng = (c+d) – (a+b) Means number of total moles of gaseous products minus number of total moles of gaseous reactants. Hence, it can be written as Kp = Kc X (RT)ng…. 4 6 Deduce the equilibrium constant Kp pCO2 for decomposition of solid CaCO3 in closed vessel Ans. Kc = [CaO][CO2] Now, Density of CaCO3 = Constant [CaCO3] Density of CaO Kc x Density of CaCO3 = [CO2] Density of CaO Kp = PCO2 7. Deduce the formula Kp = (P2)/4 for decomposition of solid NH4HS in closed vessel. Ans. NH4HS(s) NH3(g) + H2S(g) x x Total moles = 2x Total pressure = Patm P/2 P/2 Kp = [PNH3][PH2S] = (P/2)(P/2) = (P2/4) atm2 8. Derive the formula for decomposition constant of solid ammonium carbonate and prove that Kp = (4/27)P3. Ans. NH4COONH2(s) 2NH3(g) + CO2(g) Moles = 2x x Total moles = 3x Kp = [PNH3]2[PCO2] = (2P/3)2(P/3)2 = (4P3/27) atm3 9. Give Le-Chatelier’s principle and explain in detail the effect of concentration on equilibrium. Ans. Equilibrium constant is independent of initial concentration of reactant but, if change is carried out in concentration of pressure, there is an effect on equilibrium, the equilibrium tries to nullify this effect. Reaction can be endothermic or exothermic. Hence, the heat absorbed or evolved, functions like that of the reactant and its effect is on equilibrium which tries to nullify the effect. Le Chatelier studies the effect of concentration and temperature and his presentation is called Le Chatelier’s principle whose statement can be written as below: “ If from the factors determining the equilibrium state, any one factor is changed, there will be such a change in the system that the effect will be nullified or made negligible so that the value of equilibrium constant at that temperature will remain constant.” This principle can be applied to both physical and chemical equilibrium. We shall study in detail the factors like (1) change in concentration (2) change in pressure (3) addition of inert gases (4) change in temperature and (5) use of catalyst affecting the equilibrium (1)Effect of change in concentration: If we add or remove the reactant or the product from the reaction in equilibrium, its effect on equilibrium according to Le Chatelier’s principle will be as follows: (a)If the concentration of reactant or product is increased by addition of reactant or product, the reaction will occur in such a way that the increase in concentration will be taken for use, i.e. the increase in concentration of reactant, the concentration of product will increase and if concentration of product is increased the reaction will result in the direction of increase in concentration of the reactant. (b)If the concentration of reactant or product is decreased by removing reactant or product, the reaction will occur in such a way that product or reactant will be established again. Hence, if any change in concentration of reactant of product is carried out then the equilibrium will try to make this effect minimum and equilibrium will be established accordingly. If we take this as an example, in the reaction If concentration of reactants dinitrogen or dihydrogen is increased, the reaction will proceed towards right hand side and the product ammonia obtained will be more. If the concentration of nitrogen or hydrogen is decreased, the reaction will proceed towards left hand side and reactants of nitrogen or hydrogen will be obtained back i.e. production of ammonia will decrease. Here, it is necessary to remember that 1 mole dinitrogen combines with 3 moles of dihydrogen in this reaction and forms 2 moles of ammonia. Hence, increase in concentration of dihydrogen rather than dinitrogen, will give more product. We take another example of heterogeneous equilibrium. If solid CaCO3(s) is heated in closed vessel, the following decomposition reaction will occur. CaCO3(s) CaO(s) + CO2(g) Hence, if more CaO(s) is to be obtained then, the CO2(g) formed in the reaction should be removed because CO2(g), gas can combine with solid CaO(s) and carry out reverse reaction then proportion of product will decrease. Hence, by removing CO2(g) from the reaction vessel, more CaO(s) can be obtained. 10. Explain Arrhenius acid-base theory and mention it’s drawbacks Ans. Arrhenius concept about Acid and Base: According to Arrhenius concept, substances which dissociate in water and give hydrogen ion (H+) are called acids and substances which dissociate in water and give hydroxyl ions (OH-) are called bases. E.g; (1) HCl(g) +H2O H+ + Cl- Acid +H2O (2) NaOH(s) Na+ + OHBase The limitations of this concept are as follows: (i) proton (H+) is highly unstable. (ii) It can not exist independently. (iii) It immediately combines with molecules of solvent water and gives H+3O ions. Its addition in certain bases OH- is not present even then they show properties of base, viz., NH3. Similarly compounds like BF3 do not possess H+, even then they act as acid. (iv) this concept can only be applied to aqueous solution because salt like NH4Cl reacts as acid in solvent like liquid NH3. Hence, it is difficult to accept Arrhenius concept as the universal one. Because ionization is given importance. (In addition it is necessary to know that ionic radius of H+ is about 10-15 meter and so it is very small in size9. Hence it easily gets combined with molecules of water and forms H3+O ion which is called as hydronium ion. One estimate is that one H+ can be combined with four molecules of water showing H+ + 4H2O H9O4+ ion. 11. Explain with suitable example, Lowry-Bronsted acid-base theory(proton transfer theory) Ans. Concept of Bronsted – Lowry for acid and base: Danish chemist Bronsted, an English chemist Lowry presented the concept of acid and base. They made H+ (proton) as a base. According, to their concept , the substance which gives a proton or donates a proton is called the acid and the substance which receives the proton or accepts a proton is called the base. Thus, acid is a proton donor and base is proton acceptor. Let us take the dissociation reaction of hydrogen chloride in water. Base Acid H3+O (aq) + Cl-(aq) HCl(g) + H2O(l) Acid Similarly, Base Acid NH3(g) + H2O(l) Base NH4+(aq)+ OH-(aq) Base Acid We shall understand in detail the first from the above reactions : HCl H+ + ClAcid-1 proton conjugate base-1 As it gives proton, HCl is an acid H2O + H+ H3O+ Base-2 Proton Conjugate acids-2 As it accepts a proton, H2O is a base. Total reaction: HCl + H2O H3O+(aq) + Cl-(aq). Acid-1 Base-2 Acid-2 Base-1. In the above reaction giving-taking of proton is not shown. Hence, it can be said that only transfer of proton takes place, it is not obtained free. Every acid will lose proton and so its conjugate base will be formed and every base will accept a proton and so its conjugate acid will be formed. Hence, this concept is known as proton transfer or conjugate acid-base concept. We have earlier seen in limitations of Arrhenius concept that OH- is not present in NH3 even then it acts as a base but according to Bronsted-Lowry concept it can be explained. NH3 + H2O NH+4(aq) + OH-(aq) Base-1 Acid -2 Acid-1 Base-2 In the reaction, base NH3 accepts a proton and forms conjugate acid NH4+ ion and acid H2O loses proton and forms conjugate base OH-. Thus the concept of Bronsted - Lowry is found to be more applicable and acceptable than Arrhenius concept. Even then its limitations are also known. This difficulties are observed in the study of reactions in organic chemistry and complex salts. BF3 has no proton even then it acts as an acid. Hence, the third concept has come in to existence, which is known as Lewis acid- base concept. Proton is given importance in Bronsted – Lowry concept. 12. Explain in detail Lewis acid-base theory. Ans. Lewis Concept of Acid and Base : Lewis, in presenting this concept, in 1923, mentioned that acid means a substance which can accept a pair of electrons and base means a substance which can donate a pair of electrons. Thus, instead of the concepts of proton, ionization, conjugate acid or base, he made electrons associated with all reactions and substances and its arrangement as the base of the concept. As seen earlier BF3 can be said acid or not and NH3 can be said a base or not can be solved by this concept. It will be clear from the following reaction: Thus, BF3 accepts the pair of electron and so it is acid and NH3 donates a pair of electron, so it is acid and NH3 donates a pair of electron, so it is base. Electron deficient substances or ions like AlCl3, Co3+, Mg+2, will act as acid and substances like H2O, NH3, OH-, F- will act as base. They are respectively called Lewis acid and Lewis base. 13. Derive the formula of dissociation constant of Weak acid Ans. Ionization Constant of Weak Acid (Ka) : In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the equilibrium is obtained as below : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation is α, then following can be written : Reaction : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Initial Concentration(M) C 0 0 Degree of ionisation(α) 1-α Concentration at equilibrium(M) (1 - α)C Equilibrium constant Ke= α αC α αC [H3O+][A-] [HA][H2O] = (αC)(αC) …1 (1 - α)C[H2O] But [H2O] is accepted as constant and so Ke[H2O] = (αC)(αC) (1 - α)C = α2C = Ka …2 (1 - α) Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we consider both the terms same) Hence, for any weak monobasic acid, following can be written [H3O+][A-] Ke= [HA] …3 -1 The unit of Ka will be mollit . As the values of Ka depend on [H3+O] the values of ka will be different for different. Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite temperature. 14. Derive the formula of dissociation constant of weak base Ans. ionisation constant (Kb) of weak base: The ionization of monoacidic weak base MOH will take place in aqueous solution as follows: H2O MOH(aq) M+(aq) + OH-(aq) As base is weak, incomplete ionization will occur and so equilibrium will be obtained and it can be expressed as below: [M+] [OH-] …………4 and Kc = [MOH] [H2O] + [M ] [OH-] =Ke ………. 5 Kex[H2O]= [MOH] Where Kb is the ionization or dissociation constant of monoacidic weak base. If we know the initial concentration of weak base and its degree of ionization, we can calculate the value of Kb as we have studied earlier. [M+] = [OH-]. 15. Obtain relation between ionic equilibrium constant Ka, initial concentration Co and concentration of H3O+ in aqueous solution of weak formic acid. Ans. As formic acid is weak acid, it is partially ionized in water. The equilibrium between the undissociated species of CH3COOH and the ions H+(aq) can be represented as CH3COOH(l) + H2O(l) H3O+(aq) + CHCOO-(aq) The equilibrium constant K for this reaction is as follows : K = [H3O+][CH3COO-] [CH3COOH][H2O] Here the change in concentration of water on dissolving the weak acid will be quite negligible compared to its original concentration, so considering the value of [H2O] as constant and joining it with equilibrium constant K we get a new constant Ka as follows. K[H2O] = Ka = [H3O+][CH3COO-] [CH3COOH] Here formic acid will ionize to much less extent and so the concentration of its unionized part will remain nearly same as its original concentration. [CH3COOH] = Co So, we have Ka = [H3O+][CH3COO-] Co Again, the concentration of the positive ions [H3O+] and that of the negative ions [CH3COO-] will be the same i.e. [H3O+] = [CH3COO-] [CH3COO-] can be replaced by [H3O+], Ka = [H3O+]2 Co + 2 + 1/2 [H3O ] = Ka.Co [H3O ] = (KaCo) 16. Obtain relation between ionic equilibrium constant Kb, initial concentration Co and concentration of OH- in aqueous solution of weak base aniline. Ans. As aniline is weak acid, it is partially ionized in water. The equilibrium between the undissociated species of C6H5NH2 and the ions H+(aq) can be represented as NH3(g) + H2O(l) NH4+(aq) + OH-(aq) The equilibrium constant K for this reaction is as follows : K = [NH4+][OH-] [NH3][H2O] Here the change in concentration of water on dissolving the weak base will be quite negligible compared to its original concentration, so considering the value of [H2O] as constant and joining it with equilibrium constant K we get a new constant Ka as follows. K[H2O] = Kb = [NH4+][OH-] [NH3] Here aniline will ionize to much less extent and so the concentration of its unionized part will remain nearly same as its original concentration. [NH3] = Co So, we have Ka = [NH4+][OH-] Co Again, the concentration of the positive ions [NH4+] and that of the negative ions [OH-] will be the same [OH-] can be replaced by [NH4+], Kb = [OH-]2 Co - 2 1/2 [OH ] = Kb.Co [OH ] = (KbCo) 17. What is meant by ionic product of water? Deduce its equation. Ans. Ionic Product of Water Water is an amphoteric oxide when acid is added to it; it accepts the proton and act as a base and when base is added to it, it donates a proton and acts as acid. When reaction between two molecules of water takes place, one molecule donates proton and other molecule receives proton and shows conjugate acid-base reaction. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Acid-1 Base-2 Conjugate acid-2 Conjugate base-2 If we express the equilibrium constant of above reaction them. [H3O+][OH-] Ka = [H2O][H2O] Where Ka is the dissociation constant of acid. There is no significant change in concentration of water (55.5M) because H2O is a weak acid (possesses about 10-7M H+). Thus, if H2O is considered constant. Ka x [H2O]2 = [H3+O] [OH-] = Kw. Where Kw is ionic product of water. Water is neutral and so [H3+O] and [OH-] in it are 10-7 M. Hence, Kw = [H3+O] [OH-] = (10-7)(10-7) = 10-14 which is constant and equilibrium constant remains constant at constant temperature; so the value of Kw will be constant at constant temperature, viz., the value of Kw is 1 x 10-14 at 298 K. If we find the ratio of concentrations of dissociated and un dissociated water, it will remain on left hand side, i.e. the number of un dissociated molecules or its concentration will be more. 18. Discuss the factors affecting strength of acid. Ans. Factors Affecting strength of acid: From the study of dissociation constant, pH value etc., it is found that their values are different. The reason for this is that [H3+O] available can be more or less. What may be the reason for this? If the acid is strong, its value of Ka will be high and the value of pH will be low. The dissociation of acid will depend on strength of acid and the polarity of H-A bond. As the strength of H-A bond decreases, the energy required for breaking that bond will decrease and HA will be stronger. When difference between electro negativities of A and B will increase, apparently ionization will occur and will be easy to break the structure of the bond. Hence, the acidity will increase. On comparing the elements of the same group of periodic table, the strength of H-A bond will be more important factor than polar nature. As we go down in the group the size of A will increase and so strength of H-A bond will decrease and hence acid strength will increase. Increase in size HF<< HCl <<HBr<<HI Increase in acid strength For this reason, H2O is stronger acid than H2S, but if we discuss the elements in the same period of periodic table, the polarity of H-A bond will determine the strength of acid. As the electronegativity of A increases, the strength of acid will increase. Increase in electro negativity CH4<< NH3<< H2O << HF Increase in acid strength 19. Write a short note on buffer solutions. Ans. Buffer solutions The pH of the fluids like blood in our body and urine is almost constant. If there is change in this pH, it affects biochemical reaction in the body. The pH of chemical and biochemical reactions in our body are constant, biochemical reactions in our body are constant, viz. the pH of human saliva is 6.4. In addition, hydrochloric acid is present in human stomach which helps in digestion. The pH of cosmetics are also kept constant. Hence, the question arises that how pH in any solution can be kept constant. Such solutions are called buffer solutions. Its definition can be given as below: “ The solution which resists the change in pH carried out by addition of acid or base in small proportion to them or are being diluted, and the values of their pH remain constant are called buffer solution.” Buffer solutions can be acidic or basic. If pKa of weak acid and pKb of weak base are known, buffer solutions of known pH can be prepared. Buffer solutions can be of three types as follows: (i) Acidic buffer solution: Acidic buffer solution can be prepared by mixture of weak acid and its salt with strong base. (ii) Basic buffer solution:Basic buffer solution can be prepared by mixture of weak base and its salt with strong acid. (iii) Neutral buffer solution: Neutral buffer solution can be prepared by neutralization of weak acid and weak base. These type of buffer solutions are shown below: Type Substances Value of pH Acidic CH3COOH + CH3COONa <7 Basic NH4OH + NH4Cl >7 Neutral CH3COOH + NH4OH =7 Buffer solution of known pH can be prepared by using the following HendersonHaschelback equation. For acidic solution, [Salt] pH = pKa + log [Acid] …………. 1 where [acid] is concentration a weak acid and its dissociation constant is Ka and [salt] is concentration of the salt of this weak acid with strong base. For a buffer solution, it can be written as pH= pKa + log [CH3COONa] [CH3COOH] similarly, for basic buffer solution e.g., NH4OH + NH4Cl can be written that, [NH4Cl] pH = pKa + log [NH4OH] Such buffer solution can be used in chemical and biochemical reactions and especially in analytical chemistry. In human body buffer solutions containing [HCO-3] and [CO2-3] as well as [H2PO-4] and [HPO2-4] are present. 20. Write a short note on pH scale. Ans. pH Scale. If we express the concentration of hydronium ion [H3+O] in molarity then values like 10-12 to 10-2 are possible. It is difficult to express these values on simple graph paper. Hence, scientist Sorenson found a scale which is called pH scale. According to him pH = -log10[H3+O]. The values 10-12 to 10-2 shown above can be converted to +12 to +2 if calculated on the basis of this relation and plotting of graph can be easy. The definition of pH can be given like this, “pH of a solution is the negative logarithm to the base 10, of the concentration of hydrogen or hydronium ion”. According to thermodynamics, activity is more proper word instead of concentration but in dilute solutions activity and concentration can be considered can be considered to be the same. Now as seen earlier a solution containing 10-7M [H3+O] and [OH-] is neutral. Hence, pH = -log10[H3+O] = -log1010-7M = 7 and for acidic solution [H3+O]> 10-7M, pH < 7 Similarly, for basic solution [H3+O]< 10-7M, Hence, pH > 7. Therefore, it can be written as : pH< 7 Acidic solution pH> 7 Basic solution pH = 7 Neutral solution AS seen above, Kw = [H3+O] [OH-] Putting the values, Kw= (10-7)(10-7) = 10-14and –log Kw = -log(10-14) Therefore, pKw = 14 Therefore, pH + pOH =pKw = 14 Temperature affects the values of pH, pOH, pKw. The above discussion can be shown in the following table : Concentration (M) Acidic Neutral Basic + -7 -7 [H3 O] More than 10 10 Less than 10-7 [OH-] Less than 10-7 10-7 More than 10-7 pH Less than 7 7 More than 7 pOH More than 7 7 Less than 7 pH paper, litmus paper or universal indicator can be used to test whether the solution is acidic or basic but the exact values of pH can be determined with the help of instrument called pH meter. 21. Explain the calculation of pH of a solution from the dissociation constant of weak acid. Ans. In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the equilibrium is obtained as below : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation is α, then following can be written : Reaction : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Initial Concentration(M) C 0 0 Degree of ionisation(α) 1-α α α Concentration at equilibrium(M) (1 - α)C αC αC [H3O+][A-] Equilibrium constant Ke= [HA][H2O] = (αC)(αC) …1 (1 - α)C[H2O] But [H2O] is accepted as constant and so Ke[H2O] = (αC)(αC) (1 - α)C = α2C = Ka …2 (1 - α) Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we consider both the terms same) Hence, for any weak monobasic acid, following can be written [H3O+][A-] Ke= [HA] …3 -1 The unit of Ka will be mollit . As the values of Ka depend on [H3+O] the values of ka will be different for different. Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite temperature. IN table 4.3 the values of ionisation constants of some weak acids are given. According to the relations seen earlier, pH = -log10[H3+O] pOH = -log10[OH-] Similarly, pKa = -log10[Ka] PA- = -log10[A-] can be written. Where [A-] is the concentration of negative ion. It is apparent from above relations that if the values of initial concentration [H3+O] and Ka are known, then at equilibrium, [H3+O] concentration can be determined and pH can be calculated 22. Explain effect of common ion on solubility of sparingly soluble salt. Ans. Effect of Common Ion on Solubility of Sparingly Soluble Salt The sparingly soluble salt that has dissolved in solution that is completely dissociated and so it is ionic in form. Hence, it is a strong electrolyte. Earlier, in chemical equilibrium, effect of concentration, application of Le Chatelier’s principle etc. have been studied. Solubility product is ionic equilibrium and the effect of concentration can be studied. As it is equilibrium constant, it will depend on temperature, but its value will be constant at constant temperature. What will happen if we add soluble ionic substance like KCl in the solution of a sparing soluble salt like AgCl? AgCl(s) Ag+(aq) + Cl-(aq) KCl(s) K+(aq) + Cl-(aq) The Cl- from AgCl in equilibrium and Cl- ion obtained by complete ionization of KCl, the concentration of Cl- will increase. Hence, according to Le Chatelier’s principle, the equilibrium will shift towards left side so as to nullify the effect of Cl- i.e. more AgCl will be formed. In other words, there will decrease in solubility of AgCl. Hence, it can be said that because of the effect of common ion on sparingly soluble alt its solubility decreases and sparingly soluble salt precipitates more. The use of effect of common ion can be made to separate one ion from the other in presence of other ion in qualitative analysis. It can also be used for decrease in solubility of the components in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of second group are less in comparison to solubility products of sulphides of metal of III B group ions, therefore, HCl is added before adding H2S water to test the second group ions. H2S(aq) 2H+(aq) + S2-(aq) HCl(aq) H+(aq) + Cl-(aq) The common ion available from HCl creates common ion effect on the equilibrium and decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group can only be precipitated because their solubility products are less. In the same way, for precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH. The concentration of OH- available from ionization of NH4OH gets decreased due to common ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group only will be precipitated because the values of solubility products of the hydroxides of III A group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions. Its is necessary to note that under certain situations the solubility increases instead of decreasing. The solubility of salt like phosphates increase when acid is added to their solutions or pH of the solution decreases. The reason for this is that, phosphate ion combines with H+ available from acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases. 23. Explain with suitable example the use of common ion effect in qualitative analysis. Ans. The use of effect of common ion can be made to separate one ion from the other in presence of other ion in qualitative analysis. It can also be used for decrease in solubility of the components in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of second group are less in comparison to solubility products of sulphides of metal of III B group ions, therefore, HCl is added before adding H2S water to test the second group ions. H2S(aq) 2H+(aq) + S2-(aq) HCl(aq) H+(aq) + Cl-(aq) The common ion available from HCl creates common ion effect on the equilibrium and decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group can only be precipitated because their solubility products are less. In the same way, for precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH. The concentration of OH- available from ionization of NH4OH gets decreased due to common ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group only will be precipitated because the values of solubility products of the hydroxides of III A group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions. Its is necessary to note that under certain situations the solubility increases instead of decreasing. The solubility of salt like phosphates increase when acid is added to their solutions or pH of the solution decreases. The reason for this is that, phosphate ion combines with H+ available from acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases. 24. What is meant by common ion effect? In qualitative analysis, NH4Cl is added before addition of NH4OH in precopotation of ions of III A group. Explain giving reason. Ans. Let us take the example of weak acid, acetic acid ( ) CH3COOH(aq)+ H2O(l) H3+O(aq) + CH3COO-(aq) OR HAc + H2O(l) H3+O(aq) + Ac-(aq) Where HAc and Ac- are the short forms of CH3COOH and CH3COO- ion. [H3+O][Ac-] Ka = [HAc] ………………..1 Suppose, we add CH3COONa or HCl, to the solution of HAc in equilibrium, then what will happen? As studied earlier in chemical equilibrium if HCl is added, [H3+O] will increase and if CH3COONa is added, [Ac-] will increase. Hence, according to Le Chatelier’s principle, the equilibrium will make negligible change and will keep the same value of equilibrium constant. This means that the equilibrium will be shifted towards left and concentration of HAc will increase i.e. the amount of un dissociated acid will increase and there will be decrease in [H3+O] and hence, there will be increase in pH. By addition of HCl due to increase in [H3+O] similar result will be obtained. This effect is known as effect of common ion effect on dissociation constant of acid. In the same way, in the case of ionization of weak base NH3, if we increase [NH4+] by adding salt like NH4Cl then, according to Le Chatelier’s principle, as [NH4+] increases the equilibrium will shift towards left and hence, un dissociated NH3 will increase, i.e. [OH-] will decrease. As a result, pH will increase. In qualitative analysis, to precipitate the radicals of III A viz Fe2+, Fe3+, Al3+, Cr3+ as their hydroxides, NH4Cl is added to the solution before adding NH4OH. Consequently, the reverse reaction is favoured in the equilibrium, resulting in decrease of OH- ion concentration. Because of this, the hydroxides of III B, IV groups and that of Mg2+ are not precipitated alon with those of group III A. NH4Cl(aq) NH4+(aq) + Cl-(aq) NH4OH(aq) NH4+(aq) + OH-(aq) The hydroxides of the radicals of III A group are extremely less soluble compared to those of succeeding groups. So, in presence of NH4Cl the hydroxides of III A group are precipitated by NH4OH. (25). 2 moles of PCl5 is heated in closed vessel of 4 liter a t definite temperature. At equilibrium state PCl5 remained undecomposed. Calculate Kc of the reaction. Solution: Reaction, PCl5(g) PCl3(g) +Cl2(g) Now, 55% PCl5 remains undecomposed. Therefore, 45 % PCl5 goes under reaction. = 45 x 2 = 0.9 moles PCl5 100 As ratio of PCl5 and (PCl3 and Cl2 ) is 1 : 1 in number of moles, 0.9 moles of PCl5 will yield 0.9 moles of PCl3 and Cl2. PCl5(g) PCl3(g) +Cl2(g) In the beginning: 2 moles 0 0 Used moles at equilibrium: 0.9 moles 0.9 0.9 Unused moles at equilibrium: 1.1 moles Now, Kc = [concentration of products] [concentration of reactants] ……..(i) Kc = equilibrium constant Now, at equilibrium, moles of reactants = 1.1 moles and moles of products = 0.9 moles ( in solution) Now, this vessel consists of 4 liter solution Therefore, concentration = 1.1 = 0.275 M/L (for reactant) 4 = 0.9 = 0.225 M/L (for products) …………(ii) 4 From equation (i) Kc = [PCl3] [Cl2] [PCl5] From equation (ii) Kc = (0.225) (0.225) = 0.050625 (0.275) 0.275 Therefore, Kc = 0.184 mol/lit Kc = 1.84 x 10-1 mol/lit. (26). The value of Kp obtained at 1060 kelvin temperature is 0.033 mollit-1 for the reaction 2NOCl(g) 2NO(g) + Cl2(g). What will be the value of Kc for the reaction. Solution: 2NOCl(g) 2NO(g) + Cl2(g). ∆n(g) = np – nr (for gaseous molecules) =(2+1)–2=1 Data: Kp = 0.033 atm T = 1060K 0.082 atm lit mol-1 K-1 Kp = Kc x (RT)∆n(g) Kc = Kp (RT)∆n(g) Substituting the values from data, Kc = 0.033 = 0.033 (0.082 x 1060) 86.92 -4 = 3.796 x 10 M Therefore, Kc = 3.796 x 10-4 M (27). At definite temperature and 3 atm pressure, 75 % PCl5 is decomposed into PCl3 and Cl2. Calculate Kp of this reaction. PCl5(g) PCl3(g) +Cl2(g). Solution: 75% PCl5 is decomposed into PCl3 and Cl2. At equilibrium, moles PCl5 decomposed 100 75 1 x [ suppose initial moles of PCl5 is 1.] x= 75 x 1 = 0.75 moles 100 0.75 moles of PCl3 and Cl2 are formed and unreacted moles of PCl5 = ( 1 – 0.75) = 0.25 moles PCl5(g) PCl3(g) +Cl2(g). Initial moles: 1 0 0 Used moles at equilibrium: 0.75 0.75 0.75 Unused moles at equilibrium: 0.25 Now, total moles at equilibrium = 0.25 + 0.75 + 0.75 = 1.75 moles Total pressure in system = 3 qtm Now, partial pressure at equilibrium: moles of 1 element x total pressure Total moles Hence, partial pressure of PCl3 ( Ppcl3) = 0.75 x 3 = 3 x 3 1.75 7 = 9 atm 7 ………… (i) Similarly , partial moles of Cl2 Pcl2 = 0.75 x 3 = 3 x3 =9 1.75 7 7 atm …………..(ii) Partial pressure of PCl5 Ppcl5 = 0.25 x 3 = 1 x3 = 3 atm 1.75 7 7 ……………….. (iii) Now, Kp = (Ppcl3) x (Pcl2) (Ppcl5) From equation (i) , (ii) and (iii) Kp = 9/7 x 9/7 3/7 Kp = 3.857 atm. (28). 1 mole N2 and 3 moles H2 are heated at 473 K temperature and 100 atmosphere pressure. If 0.50 mole of NH3 are formed at equilibrium time calculate equilibrium constant Kp for this reaction. N2(g) + 3H2(g) 2NH3(g) Solution: N2(g) + 3H2(g) 2NH3(g) Now, the ratio of production of moles of this reaction is 1 : 3 : 2 Initial moles of N2 = 1mole Data Initial moles of 3H2 = 3 moles Therefore, from the ratio, moles of NH3 formed = 2 moles But from the data, moles of NH3 fromed = 0.50 moles. Therefore, ¼ th of NH3 is formed [ because 2 /0.50 = 4] Hence, 1/4th of reactants must be used. For N2 , ¼ x 1 = ¼ moles For H2 , ¼ x 3 = ¾ moles For NH3 = 0.50 moles ……(i) Above these are the moles formed in solution Therefore, unused moles at equilibrium For N2 , 1 – ¼ = ¾ moles ….(ii) For H2 , 3 – ¾ = 9/4 moles ……(iii) N2(g) + 3H2(g) 2NH3(g) Initial moles: 1 3 2 Used moles at equilibrium ¼ ¾ 0.50 Unused moles at equilibrium ¾ 9/4 Partial pressure = moles of 1 element x total pressure Total number of moles Total pressure = 100 atm Total number of moles = ¾ + 9/4 + 0.50 = 3.50 moles Partial pressure of N2 PN2 = ¾ x 100 3.50 = 21.428 atm …………..(iv) Partial pressure of H2 PH2 = 9/4 x 100 3.50 = 64.285 atm ……….(v) Partial pressure of NH3 PNH3 = 0.50 x 100 3.5 = 14.2857 atm …….(vi) Now, from equation N2(g) + 3H2(g) 2NH3(g) Kp = (PNH3)2 = (14.2857)2 (PN2) x (PH2)3 21. 2428 x (64.285)3 Kp ≈ 3.616 x 10-5 atm -2 (29). On heating 0.5 mole solid calcium carbonate in a vessel of 500ml volume, at 400 K temperature , Kc 0.9 mol lit-1 is obtained at equilibrium state, then calculate mole of CO2 at equilibrium. How many percent the reaction would have been completed? CaCO3(s) CaO(s) + CO2(g) Solution: CaCO3(s) CaO(s) + CO2(g) Initial moles: Used moles at equilibrium: 0.5 - - x x x [as ratio is 1 : 1] unused moles at equilibrium : (0.5 – x) concentration (0.5 – x) x x (mole/ liter) 0.5 0.5 0.5 [ volume in solution is 500ml (from data)] Kc = [CaO][CO2] [CaCO3] = (x/0.5) (x/0.5) = (2x) (2x) ((0.5 –x)/0.5) 2(0.5 – x) 2 Kc = 4x 1 – 2x ……….(i) From data, Kc = 0.9 mol/lit Therefore, from equation (i), 0.9 = 4x2 => 9 = 4x2 1-2x 10 1 – 2n 9 – 18 x=40x2 40x2 + 18x -9 = 0 Comparing with quadrative equation, ax2 + bx + c = 0 a = 40 b = 18 ………(ii) c= -9 Solution, x = - b ±√ b2 – 4ac 2a Substituting values of a, b ,c from equation (ii) x = -18 ± √ (18)2 – 4(40) (-9) 2( 40) x= -18 ± √ 324 + 1440 80 x = -18 ± √ 1764 = -18 + 42 80 80 x = -18 + 42 or x = -18 -42 80 80 x= 24 or x= -60 80 80 (not possible as solution of x is negative value) Therefore 3 => x = 0.3 moles 10 Therefore, moles of Co2 at equilibrium = x = 0.3 moles Therefore, from data , 0.5 ml of CaCO3 0.3 moles of CO2 and 0.3 moles of CaO Therefore, moles of CO2 formed = 0.3 moles CaCO3(s) CaO(s) + CO2(g) 0.5 0.3 100 (?) 0.3 x 100 = 60 0.5 Therefore, % of reaction completed = 60% (30). Mg(HCO3)2(s) is decomposed as follows in a closed vessel. If mole fraction of CO2 is 0.8 and Kp = 64 atm3 then calculate total pressure of mixture at equlibrium. Mg(HCO3)2 MgCO3(s) + CO2(g) + H2O(g) Solution: Mg(HCO3)2 MgCO3(s) + CO2(g) + H2O(g) Now, gaseous mole in system are CO2 and H2O Mole fraction of CO2 = 0.8 Total moles = 1 Therefore, mole fraction of H2O = 1 -0.8 = 0.2 …….(i) Therefore, mole fraction Xco2 = 0.8 Mole fraction XH2O = 0.2 Now, partial pressure = moles of element x total pressure Total moles = mole fraction x total pressure Therefore, partial pressure of CO2 (PCO2) = Xco2 x total pressure = 0.8 x ……….(ii) [let total pressure be x] Partial pressure of H2O (PH2O) = XH2O x total pressure = 0.2 x ………………(iii) Now, Kp = [PCO2] [PH2O] But, from data, Kp = 64 atm2 and from equation (ii) and (iii) 64 = (0.8x) ( 0.2x) Therefore, 64 = 0.16x2 Therefore, x2 = 64 0.16 2 x = 400 therefore, x = 20 atm Therefore, total pressure in system = 20 atm (PCO2 = 16 atm PH2O = 4 atm) (31). If 3.65 x 10 -2 gram HCl is dissolved in 500 ml solution, then find its pH (pH = 2.699) Solution: Data, For HCl, weight (W) = 3.65 x 10-2 gram M = 36.5 gram/ mole V = 500 ml = 0.5L Moles of HCl = W = 3.65 x 10-2 M 36.5 = 1 x 10-3 mole Concentration of [H+] in HCl = 1 x 10-3 0.5 = 2 x 10-3 M/L pH= -log[H+] = -log[2 x 10-3] = - [log 2 + log (10-3)] [ log(axb) = log a + log b] = 3 – log2 = 3 – 0.3010 pH= 2.6990. (32). Find pH of mixture of 50ml 0.03 M HCl and 60 ml 0.02M NaOH . Solution: 0.03 ml HCl 1000ml HCl solution contains 0.03 mol HCl Therefore, 50 ml solution contains = 0.03 x 50 1000 = 1.5 x 10-3 moles HCl ……..(i) Similarly, 0.02 M NaOH 1000ml NaOH contains 0.02 mol NaOH Therefore, 60 ml solution contains = 0.02 x 60 1000 = 1.2 x 10-3 moles NaOH …..(ii) NaOH + HCl neutralisation NaCl + H2O Therefore, 1.2 x 10-3 moles of NaOH will neutralize 1.2 x 10-3 moles of HCl out of 1.5 x 10-3 moles of HCl Therefore, moles of HCl that remains unneutralised are = (1.5 – 1.2) x 10-3 = 0.3 x 10-3 = 3 x 10-4 moles ………(iii) Total volume of the mixture = 110ml Therefore, molarity of HCl after neutralization = moles = moles x 1000 Volume(L) V (ml) -4 = 3 x 10 x 1000 [from equation (iii)] 110 Therefore, molarity of HCl = 2.727 x 10-3 M Now, HCl is strong acid and hence complete ionization takes place Therefore, HCl(aq) + H2O(l) Cl-(aq) + H3O+(aq) 2.727 x 10-3 M 2.727 x 10-3 M pH= - log[H3O+] = - log [ 2.727 x 10-3] Therefore, pH = [ log 2.727 + ( -3) log 10] pH = - log 2.727 + 3log10 therefore, pH = -0.4357 + 3 therefore, pH=2.5643 (33). Find weight of CH3COOH in 100ml aqueous solution of CH3COOH having pH 3.0 Ka for CH3COOH = 1.75 x 10-5 Solution: CH3COOH + H2O H3O+ + CH3COOAs CH3COOH is weak acid, partial ionization takes place Therefore, Ka = 1.75 x 10-5 Therefore, for weak acid, Ka = [H3O+]2 Co Where,Co = initial concentration of weak acid i.e. CH3COOH Co = [H3O+]2 Ka ………….. (i) Now, pH = - log10 [H3O+] = 3 [given] Therefore, -log[H3O+] = 3 Therefore, [H3O+] = 1 x 10-3 M …………………..(ii) Substituting the value of [H3O+] from equation (ii) in equation (i) Therefore, Co = [H3O+]2 = ( 1 x 10-3)2 Ka 1.75 x 10-5 Co = 5.7142 x 10-2 M Now, concentration = mole Liter Therefore, 1 liter solution conatin 5.7142 x 10-2 moles of acetic acid 1L 5.7142 x 10-2 moles 100ml(10-1L) ? -2 = 5.7142 x 10 = 5.714 x 10-3 moles 10-1 Now, mole = weight Molecular weight M.W. of CH3COOH = 60 gram / mole Therefore, moles= W 60 Therefore, W = 60 x moles = 60 x 5.7412 x 10-3 W = 3.428 x 10-1 gram = 0.342 gram. (34).0.02 M CH3NH2 is ionized to 15% at equilibrium .Find its ionization constant. Solution: CH3NH2(g) + H2O(l) CH3NH+3(aq) + OH-(aq) Methyl amine ionizes to 15% at equilibrium If 100 moles of CH3NH2 are taken, then 15 moles of OH- are formed CH3NH2 OH100 : 15 0.02 : (?) = 15 x 0.02 = 0.3 x 10-2 M 100 Therefore, [OH ] = 3 x 10-3 M Now, CH3NH2 is weak base Therefore, weak base , Kb = [OH-]2 Co Where, Kb = ionization constant of weak base Co = initial concentration of base = 0.02 M Therefore, Kb = ( 3 x 10-3)2 0.02 = 9 x 10-6 2x 10-2 Therefore, Kb = 4.5 x 10-4 (35). For NH3 Kb = 1.77 x 10-5 , what will be the pH of 500 ml solution containing 1.7 gram ammonia? Solution: NH3(g) + H2O(l) NH+4(aq) + OH-(g) For weak base, Kb = 1.77 x 10-5 V (ml) = 500 ml = 500 x 10-3 L W = 1.7 gram M.W. of ammonia = 17 gm/ mole Concentration = moles Liter = weight of NH3 (gm) M.W. of NH3 (gm/mole) x volume (L) = 1.7 17 x 500 x 10-3 = 0.2 M Therefore, concentration = 0.2 M ……..(i) For weak base, Kb = [OH-]2 Co Where, Co = initial concentration of NH3 = 0.2 M (from equation (i)) Therefore, [OH-]2 = Kb x Co = 1.77 x 10-5 x 0.2 = 3.54 x 10-6 Therefore, [OH ] = Kb x Co = 1.77 x 10-5 x 0.2 = 3.54 x 10-6 Therefore, [OH-] = √ 3.54 x 10-6 Therefore, [OH-] = 1.881 x 410-3 M Now, pOH = -log10[OH-] = -log10 [ 1.881 x 10-3] Therefore, pOH = 3 – log(1.881) = 3 – 0.2744 Therefore, pOH = 2.7256 Now, pH + pOH = 14 Therefore, pH = 14 – pOH 14 – 2.7256 Therefore, pH = 11.2744 (36). pH of 0.05M solution of a weak acid is 3.68 then find Ka of that acid. Solution: for weak acid, Initial concentration (Co) = 0.05M pH = 3.68 Ka = ? Now ,pH = - log10[H3O+] = 3.68 Therefore, -log[H3O+]=3.68 Shifting minus (-) sign to right side and taking anti –log on both sides Therefore, [H3O+] = - antilog(3.68) anti-log (-3.68) = antilog(-3.68) -1 +1 = antilog( 4 + 0.32) antilog (-4 + 0.32) Therefore, [H3O+] = antilog(0.32) x 10-4 Therefore, [H3O+] = 2.089 x 10-4 M Now, for weak acid, Ka = [H3O+]2 Co Therefore, Ka = (2.089 x 10-4)2 0.05 = 87.278 x 10-8 = 8.727 x 10-7 Therefore, Ka = 8.7278 x 10-7 (37). Find pH of 0.025M CH3COONa solution. Ka of CH3COOH = 1.75 x 10-5. Solution: CH3COONa CH3COO(aq)- + Na(aq)+ 0.025M 0.025M 0.025M Now, CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) . ……….(i) [ A- + H2O HA + OH-(aq)] Weak acid + H2O(l) H(aq) + OH-(aq) Now, hydrolysis constant, Kn = Kw Ka [CH3COONa is salt of strong base (NaOH) and weak acid (CH3COOH) and hence, Kn = Kw Ka ] Therefore, Kn = Kw = [CH3COOH][OH ] and from reaction (i) Ka [CH3COO-] As concentration of [OH ] produced by the self- ionization of water is negligible in comparison with the concentration of OH- produced by hydrolysis reaction. Therefore, [CH3COOH] = [OH-] (At equilibrium) Therefore, Kn = Kw = [OH-]2 Ka [CH3COO-] Kw = ionic product of water = 1x 10-14 Ka = 1.75 x 10-5 [CH3COO-] = initial concentration of CH3COONa = Co = 0.025 M Substituting , these values in above equation Kw = [OH-]2 => 1 x 10-14 = [OH-]2 Ka [CH3COO ] 1.75 x 10-5 0.025 Therefore, [OH-]2 = 1.4285 x 10-11 = 14.285 x 10-12 Therefore, [OH-] = 3.7796 x 10-6 M ≈ 3.78 x 10-6 M Now, pOH = -log10 [OH-] = -log[3.78 x 10-6] = -log 3.78 + 6log10 = 6 –log(3.78) = 6- 0.5775 Therefore, pOH= 5.4225 Now, pH + pOH = 14 pH= 14 –pOH = 14 – 5.4225 = 8.5775 Therefore, pH = 8.5775 (38). Kb of NH3 is 1.8 x 10-5.Calculate pH of 0.20M solution of NH4Cl Solution: NH4Cl NH4+ + ClNow, NH4Cl is salt of strong acid (HCl) and weak base (NH3) Kn = Kw Kb Where, Kn = hydrolysis constant Kw = ionic product = 1 x 10-14 Kb = ionization constant of weak base = 1.8 x 10-5. Now, NH+4 + H2O NH3 + H3O+ Now, Kn = Kw = [NH3] [H3O] = [H3O+]2 Kb [NH+4] [NH+4] + [ because at equilibrium, [NH3] = [H3O ] ] Therefore, Kw = 1x 10-14 = [H3O+]2 Kb 1.8 x 10-5 0.20 + 2 Therefore ,[H3O ] = 1 x 0.20 x 10-9 1.8 = 0.111 x 10-9 = 1.111 x 10-10 + Therefore, [H3O ] = 1.054 x 10-5 M Now, pH = -log10 [H3O+] = -log10[1.054 x 10-5] = -log 1.054 + 5 log10 = 5 – log(1.054) (log10 = 1) = 5 -0.0229 = 4.9771 Therefore, pH = 4.9771 ≈ 4.98 39. Find solubility of PbSO4 in water, having solubility product 1.3 x 10-8 (Molecular mass of PbSO4 = 303 gram mol-) Ans. Let, the solubility of PbSO4 be S mol/litre PbSO4(s) Pb+2(aq) + SO42-(aq) S S +2 2Solubility product, Ksp = [Pb ][SO4 (aq)] = [S][S] = [S2] But, Ksp = 1.3 x 10-8(given) Ksp = S2 = 1.3 x 10-8 S =1.140 x 10-4 mol/litre Now, molecular weight of PbSO4 = 303 gm/litre Solubility (in gram/litre) = mol/litre x M.W. = 1.140 x 10-4 x 303 = 345.47 x 10-4 gm/litre = 3.45 x 10-2 gm/L S = 3.45 x 10-2 gm/litre 40. Ksp of CaF2 is 1.7 x 10-10. What will be the volume of saturated solution of 10 milligram salt? (Atomic mass : Ca = 40 and F = 19) Ans. Let, molar solubility of CaF2 be S mol/litre Solubility Equilibrium, CaF2 Ca2+(aq) + 2F-(aq) S 2S (Because 2 moles of F- are formed) Ksp(given) = 1.7 x 10-10 Ksp = [Ca2+][F-]2 = s[2s]2 = 4S3 From above data, Ksp = 4S3 = 1.7 x 10-10 4S3 = 1.7 x 10-10 S3+ = 0.425 x 10-10 = 42.5 x 10-12 S = 3.489 x 10-4 mol/litre [ S3 = 42.5 x 10-12 S = (42.5 x 10-4)1/3 (42.5)1/3 = (log 42.5)/3 = (1.6284)/3 = 0.5428 = antilog(0.5428) = 3.489 3.49] -4 S 3.49 x 10 mol/L Where S= Solubility of CaF2 Now, molecular weight of CaF2 = Ca + 2(F) = 40 + 2(19) = 78 g/mol Now, molar Solubility = Wt.(gm) x 1 Mol. Wt. (gm/mole) x v(in litre) Solubility of CaF2 = Wt. of CaF2(gm) x m.w of CaF2 (gm/mole) Substituting the values in above reaction W= 10 mg = 10 x 10-3 gm 3.49 x 10-4 = 10 x 10-3 x 1000 78 x V(ml) V(ml) = 10 x 1 78 x 3.49 x 10-4 V(ml) = 367.34 ml 1000 V(ml) 41. Ksp of CaF2 at 298 K temperature is 1.7 x 10-10. If a person drinks 2.5 litre of water saturated with this salt every day, how much CaF2 will enter in his body every day? (Molecular mass of CaCl2 =78 g mol-) Ans. Suppose, Solubility of CaF2 be S mol/L CaF2 Ca2+(aq) + 2F-(aq) S 2S Ksp = [Ca2+][F-]2 = s[2s]2 = 4S3 Now, Ksp(given) = 1.7 x 10-10 = 4S3 S3 = 0.425 x 10-10 = 42.5 x 10-12 S = (42.5)1/3 x 10-4 mol/L S = 3.489 x 10-4 mol/L 3.49 x 10-4 mol/L [ S = (42.5)1/3 = (log(42.5))/3 = (1.6248)/3 = 0.5428 = antilog(0.5428) = 3.489 ] Now, person drinks 2.5 litre of water saturated with CaF2 salt every day V = 2.5 L Molecular mass = 7.8g/mol S = 3.49 x 10-4 mol/litre Solubility Product (s) = Wt (gm) x 1 M.W(gm/mole) x V(L) 3.49 x 10-4 = W(gm) x 1 78 x 2.5 W(gm) = (3.49 x 10-4) x 78 x 2.5 = 680.55 x 10-4 = 0.068055 gm W(gm) = 0.068055 gm = 6.8055 x 10-2 gm 42. pH of saturated aqueous solution of Ca(OH)2 is 12.25. Find Ksp of Ca(OH)2 Ans. Suppose, the solubility product of Ca(OH)2 be S mol/litre Equilibrium state Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) S 2S Now, pH = 12.25 PH + pOH = 14 pOH = 14 - pH = 14 – 12.25 = 1.75 OH Now, p = -log10[OH-] = 1.75 -log10[OH-] = 1.75 Shifting ‘mius sign’ to the right side and taking ‘antilog’ on both sides [OH-] = -antilog(1.75) = antilog(2.25) = 1.778 x 10-2 [OH-] = 1.778 x 10-2 M But, according to reaction [OH-] = 2S = 1.778 x 10-2M ….(i) S = 0.889 x 10-2 M …. (ii) Now, Ksp = [Ca2+][OH-]2 = [S][2S]2 (from reaction) = 4S3 = 4(0.889 x 10-2)3 [from equation (i)] Ksp = 2.810 x 10-6 M3 43. What will be the value of Ksp if the concentration in saturated solution of Mg(OH)2 is 8.2 x 10-4 %W/V. (Molecular Mass of Mg(OH)2 = 58.3 g.mol-). Ans. 8.2 x 10-4% W/V means 100 ml saturated solution contains 8.2 x 10-4 gm Mg(OH)2 100 ml 8.2 x 10-4 gm Mg(OH)2 1000 ml(1L) (?) -4 = 8.2 x 10 x 1000 100 = 8.2 x 10-3 gm/litre M.W. of Mg(OH)2 = 58.3 gm/mol Molarity of saturated solution of Mg(OH)2 = 8.2 x 10-3 gm/litre 58.3 gm/mole -4 S = 1.4065 x 10 mol/litre Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) S 2S 2+ - 2 Now, Ksp = [Mg ][OH ] = [S][2S]2 = 4S3 = 4(1.4065 x 10-4)3 = 11.130 x 10-12 M3 Ksp = 1.113 x 10-11 M3 44. How many grams of Zn(OH)2 can be dissolved in 2 litres of 0.02 M NaOH solution ?ksp for Zn(OH)2 = 4.5 x 10-17. Ans. Suppose solubility of Zn(OH)2 is S mol/litre Zn(OH)2(s) Zn2+(aq) + 2OH-(aq) S 2S NaOH(aq) Na+(aq) + OH-(aq) 0.02M 0.02M 0.02M (given) So, total concentration of [OH ] in the solution = [OH-] from Zn(OH)2 + [OH-] from NaOH = 2S + 0.02 (But here, 0.02 >>> S, hence [OH-] = 0.02 M ) (because Zn(OH)2 is sparingly soluble) TotL [OH ] = 0.02 M Now, Ksp for Zn(OH)2 = [Zn2+][OH-]2 = [S][0.02]2 Ksp = S x 4 x 10-4 But Ksp = 4.5 x 10-17(given) Ksp = 4.5 x 10-17 = S x 4 x 10-4 S = 4.5 x 10-17 4 x 10-4 S = 1.125 x 10-13 mol/litre Mol. Wt of Zn(OH)2 = Zn + 2(O) = 2(H) = 65 + 2(16) + 2(1) = 99 gm/mol Solubility of Zn(OH)2 in o.o2M NaOH solution = (1.125 x 10-13) x 99 gm/litre = 1.11375 x 10-11 grams/litre But, given solution is of 2 litres Solubility = 1.11375 x 10-11(gm/L) x 2 L Solubility of Zn(OH)2 = 2.2275 x 10-11 gm 45. Solubility product of Mg(OH)2 is 1.2 x 10-11. Calculate its solubility in pure water as well as in aqueous solution of 0.05 M NaOH Ans. Suppose, solubility of Mg(OH)2 be S mol/L Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) S 2S NaOH(aq) Na+(aq) + OH-(aq) 0.05M 0.05M 0.05M (a) Solubility in pure water Ksp = [Mg2+][OH-]2 = [S][2S]2 = 4S3 But, Ksp = 1.2 x 10-11 (given) 4S3 = 1.2 x 10-11 S3 = 0.3 x 10-4 = 3 x 10-12 1/3 S = (3) x 10-4 S= 1.442 x 10-4 mol/litre [ (3)1/3 = (log3)/3 = (0.4771)/3 = 0.1590 = antilog(0.1590) = 1,442] S = 1.442 x 10-4 mol/litre……(a) (In pure water) (b) Solubility in 0.05M NaOH Total concentration of [OH-] = 2S + 0.05 [ But, here 0.05 >>> S as Mg(OH)2 is sparingly soluble) [OH-] = 0.05 M Concentration of Mg2+ = [S] = ? - Ksp = [Mg2+][OH-]2 Ksp = 1.2 x 10 = [S][0.05]2 = 1.2 X 10-11 = S x 25 x 10-4 S = 1.2 x 10-11 = 0.048 x 10-7 = 4.8 x 10-9 25 x 10-4 S = 4.8 x 10-9 mol/litre = 4.8 x 10-19 M 46. At 298 K temperature, 2.901 litre saturated solution can be prepared by dissolving 0.08 gram CaF2. Calculate Ksp of salt. (Molecular mass of CaF2 = 78.0 g.mol-) Ans. W = 0.08 gm Molecular wt = 78 gm/mol V(L) = 2.901 litre Now, concentration = 3.5354 x 10-4 M Now, CaF2 Ca2+ + 2F(3.53 x 10-4) (2 x 3.53 x 10-4) 2+ - 2 Now, Ksp =[Ca ][F ] = (3.53 x 10-4)(2 x 3.53 x 10-4)2 = 4 x (3.53 x 10-4)3 = 4 x 44.191 x 10-12 M3 = 176.767 x 10-12 M3 = 1.7676 x 10-10 M3 Ksp = 1.768 x 10-10 M3 47. Equal volumes of 2 x 10-4 M BaCl2 and 5.0 x 10-3 M H2SO4 are mixed. Will precipitation occur or not? (Ksp of BaSO4 = 1.1 x 10-10) Ans, If equal volumes of solutions are mixed, concentration of each salt will be halved [C1V1 = C2V2 But if V2 = 2V1 C1V1 = C2(2V1) C2 = (C1)/2] Concentration of BaCl2 in the mixture = 2 x 10-4 M = 1 x 10-4 M ….(i) 2 Concnetration of H2SO4 in the mixture = (5 x 10-3)/2 M = 2.5 x 10-3 M …(ii) BaCl2 Ba2+ + 2Cl-4 -4 (1 x 10 ) 1 x 10 M (2 x 1 x 10-4) M + H2SO4 2H + SO4-2 (2.5 x 10-3)M (2 x 2.5 x 10-3)M 2.5 x 10-3 M When, they are mixed, salt formed is BaSO4 BaCl2 + H2SO4 BaSO4 + 2HCl 2+ BaSO4(aq) Ba (aq) + SO42-(aq) Ionic Product of BaSO4, Ip = [Ba2+][SO42-] = (1 x 10-4)(2.5 x 10-3) Ip = 2.5 x 10-7 M2 But, Ksp = 1.1 x 10-10(given) Comparing above both values Ip > Ksp -11 Precipitation will occur 48. By addition of how many grams of FeSO4 to 500 ml solution of 0.02 M NaOH, Fe(OH)2 will be precipitated? (Ksp of Fe(OH)2 = 1.5 x 10-15 ) Ans. NaOH Na+ + OH0.02M 0.02M Fe(OH)2 Fe+2(aq) + 2OH-(aq) S S 2S [Suppose, molar solubility of Fe(OH)2 be S mol/L Now, Fe(OH)2 is sparingly soluble salt S <<< 0.02 M [OH-] = 0.02 M Now, Ksp = [Fe2+][OH-]2 …(i) But Ksp = 1.5 x 10-15 (given) …..(ii) For, precipitation to occur, Ip > Ksp …(iii) From equation (i), (ii), (iii) Ip = Ksp = 1.5 x 10-15 = [Fe2+][OH-]2 S[0.02]2 1.5 x 10-15 = S x 4 x 10-4 S = 3.75 x 10-12 M S = [Fe2+] = 3.75 x 10-12 M Now, l mole [Fe2+] = 1 mol FeSO4 [ because FeSO4 Fe2+(aq) + SO42-(aq)] [Fe2+] = [FeSO4] = 3.75 x 10-12 mol/litre For l litre, FeSO4 = 3.75 x 10-12 moles Hence for 500 ml = (3.75 x 10-12)/2 = 1.875 x 10-12 moles/500 ml Mol. Wt of FeSO4 = Fe + S + 4(O) = 56 + 32 + 4(16) = 152 gm/mol Wt of FeSO4 = 1.875 x 10-12 x 152 = 285 x 10-12 = 2.85 x 10-10 gm For precipitation, Ip > Ksp Wt (FeSO4) > 2.85 x 10-10 gm. Thus, if FeSO4 added in slightly greater than 2.85 x 10-10 gm, Ip > Ksp and Fe(OH)2 will precipitate. 49. Will precipitates of PbI2 be obtained by mixing 20 ml 3 x 10-3 M pb(NO3)2 and 80 ml. 2 x 10-3 M NaI Solutions ? (Ksp for PbI2 = 6.0 x 10-9) Ans. Pb(NO3)2(aq) Pb2+(aq) + 2NO3-(aq) -3 3 x 10 M 3 x 10-3M 2(3 x 10-3M) [Pb2+] = 3 x 10-3 M….(i) NaI(aq) Na+(aq) + I-(aq) 2 x 10-3M 2 x 10-3M 2 x 10-3M -3 [I ] = 2 x 10 M….(ii) 20 ml Pb(NO3)2 and 80 ml NaI solution is mixed. Total volume, V2 = (20 + 80) ml = 100 ml V2 = 100 ml Now, concentrations of Pb2+ and I- ions in mixture according to M1V1 = M2V2. For Pb2+ M1 = 3 x 10-3M V1 = 20 ml V2 = 100 ml. M1V1 = M2V2 M2 = M1 x V1 V2 -3 M2 = (3 x 10 ) x 20 = 6 x 10-4 100 2+ [Pb ] 6 x 10-4 M…(iii) For IM1 = 2 X 10-3 M V1 = 80 ml V2 = 100 ml M1V1 = M2V2 M2 = M1 x V1 V2 = (2 x 10-3) x 80 = 1.6 x 10-3 M 100 [I-] = 1.6 x 10-3 M….(iv) Now, Ksp = 6 x 10-4 (given)…(v) For PbI2, Ip = [Pb2+][I-]2 [ PbI2(S) Pb2+(aq) + 2I-(aq)] From equation (iii) and (iv) Ip = (6 x 10-4) x (1.6 x 10-3)2 Ip = 15.36 x 10-10 Ip = 1.536 x 10-9 ….(vi) From equation (v) and (vi) Ip(1.536 x 10-9) < Ksp(6.0 x 10-9) Ip < Ksp PbI2 will not precipitate 50. The concentration of F- in a sample of water is 2 x 10-5 M. How many minimum grams of solid CaCl2 will have to be added for precipitation of F-?( Ksp for CaF2 = 1.7 x 10-10, molecular mass of CaCl2 = 111 grammolAns. [F-] = 2 x 10-5 M CaCl2 Ca2+(aq) + 2Cl-(aq) Suppose, x mole of CaCl2 should be added to 1 litre of water concenetration of Ca2+ in water will be x M [Ca2+] = x M CaF2(aq) Ca2+(aq) + 2F-(aq) xM (3 x 10-5) M -10 Ksp = 1.7 x 10 (given) Now, Ksp = [Ca2+][F-]2 = (X)(3 x 10-5M)2 1.7 x 10-10 = X x 9 x 10-10 X = (1.7)/9 = 0.1888 M X = 0.189 M …..(i) Now, 1 mol of Ca2+ = 1 mole of CaF2 = 1 mole of CaCl2 1 mole of CaF2 = 1 mole of CaCl2. 0.189 moles of CaF2 = 0.189 moles of CaCl2 [from equation (i)] If little more than 0.189 moles of CaCl2 is added i.e 0.20 moles, then CaF2 will precipitate. moles = 0.20 Molar mass of CaCl2 = 111 gm/mol 0.20 mol of CaCl2 will contain = 111 gm/mol x 0.20 = 22.2 gm solid CaCl2 22.2 gms solid CaCl2 is added for precipitation of F- (minimum value) Section –D 1. Explain dynamic nature of chemical and physical equilibrium. Ans. Dynamic Nature of Equilibrium The most important matter in the case of equilibrium is that there is a continuous transformation of reactant to product and product to reactant. This state appears to be steady but it is not so. This type of reaction which takes place in both the directions is called reversible reaction and it is expresses by the symbol of two half arrows(↔). This symbol indicates that such reaction occurs simultaneously in both (forward and reverse) directions. Generally, the change of reactant to product is called forward reaction and the change of product to reactant is called reverse reaction. Thus, in reversible reactions, forward and reverse reactions continuously occur and we find it as equilibrium state. The misture of reactants and products obtained at equilibrium time is called equilibrium misture. The decomposition reaction of solid calcium carbonate in a cold vessel, at high temperature can be shown as below. ∆ CaCO3(s) ↔CaO(s) + CO2(g) The equilibrium is dynamic and not steady or static as the forward and the reverse reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed vessels. In the above reaction obtaining CaO and CO2 by decomposing of CaCO3and obtaining CaCO3 by combination of CaO and CO2 continuously take place. Suppose, if we deposit, some amount in our bank account and withdraw the same amount, then balance in the account appears steady or static. But this can be considered operative or dynamic and not closed or static. It is very difficult to determine the dynamic nature of equilibrium, even then with the help of radioactive isotope, it can be proved, viz. 14CO2 gas containing radioactive isotope, 14C and CaCO3 are taken in two different flasks and CO2 obtained by decomposition is connected with vessel containing 14CO2 gas, after sometime, Ca14CO3 will be formed in the vessel of CaCO3 and CO2 will be obtained in the vessel containing 14CO2. Thus if the equilibrium would have been steady, there must not be exchange of 12C and 14C. With the help of suitable counter, the radioactivity can be measured and the proof for the dynamic nature of equilibrium can be obtained through the proportions of concentrations of reactants and products remain constant. The reaction can be fast or slow depending upon the nature of the reactant and the experimental conditions. Equilibrium reactions can be divided into following three categories: (i) Reactions which are almost at the extent of completion and concentration of reactants may be negligible. It is not possible to detect this experimentally. (ii) Reactions in which the products are formed in very less proportions and most part of the reactant remains unchanged at the equilibrium. (iii) Reactions in which the concentrations of reactants and products are in comparable proportions at equilibrium. Dynamic Nature of Chemical Equilibrium The dynamic nature of chemical equilibrium can be demonstrated by taking example of the production of ammonia. By keeping known quantities of dinitrogen and dihydrogen at high temperature and pressure in a closed vessel, the amount of ammonia formed can be determined at constant intervals by a series of experiments. The quantities of unreacted dinitrogen and dihydrogen also can be determined. From this it is concluded that even if the reactants and products are in different proportions, their concentrations are same at equilibrium. This constancy in composition indicates dynamic nature of equilibrium. For this, in synthesis of ammonia, deuterium (D2) instead of dihydrogen, (H2) is used and ammonia gas is produced by Haber process and it is studied. The results obtained are similar to those obtained above. In the mixture proportions of N2, Dand ND3 instead of N2, H2, NH3 can be determined and equilibrium can be obtained. If D2 is added after the formation of ammonia by reaction of N2 and H2, the reaction may not occur but H in NH3 is displaced by D and ND3 can be determined by mass spectrometer. Thus, it is proved that in the reaction That the reactions from reactants to products and products to reactants that is forward and reverse reactions continuously occur with the same rates and so ND3 instead of NH3is obtained. By the use of radioactive isotope, the dynamic nature of equilibrium can be proved viz. For the reaction, H2(g) + I2(g) 2HI(g) Radioactive isotope 131I of iodine can be used to study the dynamic nature of chemical equilibrium. As the equilibrium is dynamic, certain properties or factors are found similar. E.g. Intensity of colour, constant pressure, constant concentration etc. 2. Explain equilibrium solid in solution and gas in solution. Ans. Equilibrium Involving Dissolution of Solid in Solution For the study of this type of equilibrium, the example of sugar and its solution in water can be taken. At constant temperature, take some water. Add some sugar into it and stir. In the beginning sugar easily dissolves, but as more and more amount of sugar is added, it dissolves according to its solubility and then some amount of sugar remains in solid form without dissolution. We know this state as saturated solution but in this system the equilibrium is also establishes sugar(solid) and liquid(solution of sugar). This can be expressed as below: Sugar(s) Sugar solution(aq) Solid liquid As studies earlier, equilibrium is dynamic because the forward and the reverse reactions continuously occur in each system. In this system, the amount of sugar that dissolves in water, the same amount of sugar separates from solution of sugar. Hence, the number of molecules of sugar and number of aqueous molecules of sugar in the solution remain constant in this system. Equilibrium involving Gas and Solution At constant temperature and pressure, carbon dioxide can be dissolved in water in a closed vessel(system), so that a system containing gas and a solution of carbon dioxide can be formed. As temperature and pressure are constant, carbon dioxide dissolves according to its pressure and temperature and forms solution of carbon dioxide and the excess carbon dioxide gas remains in equilibrium with it. Equilibrium is dynamic and so molecules have carbon dioxide gas that dissolve in water is the same as the number of molecules of the gas that release from the solution in the system. Thus, in this closed system, the total number of molecules(amount) of carbon dioxide in gaseous form and those that have dissolved in water remain constant. This equilibrium can be shown as below: CO2(g) CO2(soln) All the reactions studies above are processes in which only physical change takes place and so they all are examples of physical process equilibrium. 3. Mention the characteristics of chemical equilibrium Ans. Characteristics of Chemical Equilibrium (viii) During chemical equilibrium, the properties like colour, concentration, pressure or temperature of the system remain constant and they are similar in the total area of the system. (ix) When chemical equilibrium is attained, the rates of forward and reverse reactions become equal. (x) If the factors like concentration, pressure, temperature etc. which affect the chemical equilibrium are changed, they produce effect on equilibrium. (xi) Even if the initial concentrations are different, the equilibrium constant remains constant at constant temperature. (xii) The value of equilibrium constant changes if the temperature changes. (xiii) For attainment of equilibrium, the reaction can be carried out from left to right (reactants to products) or from right to left (products to reactants) (xiv) There is no effect of catalyst on the equilibrium constant and so the proportions of products remain same but the rate of reaction to attain equilibrium increases. 4. Give chemical equilibrium (active masses) law and derive its formula Ans. Law of Chemical Equilibrium and Equilibrium Constant The mixture of reactants and products at equilibrium is called equilibrium mixture. We shall study the relation between concentrations of reactants and products at equilibrium state. Let us take a simple reversible reaction as follows: A+B C + D In this reaction A and B are reactants and C and D are products. This means that in this reaction moles of reactants and products are one each but in all reactions this may not happen. Hence, it is necessary that their moles are expressed. Balanced reaction determines their moles. Viz. N2(g) + 3H2(g) 2NH3(g) From the experimental studies of many reversible reactions scientists of Norway, Guldberg and Waage mentioned in 1864 that the concentrations of substances in equilibrium mixture can be expressed by following equilibrium equation. Where Kc is equilibrium constant and [ ] bracket expresses concentration of reactant or product on mol/lit or M. The equilibrium equation is also known as law of active masses because in the early years of chemistry, concentration was said to be ‘active mass’. Now, we shall derive the equation for equilibrium constant of a general reaction. Suppose, if a reaction takes places, as given below in which the reactants and products are shown in balanced form with their proper moles (a, b, c or d) aA + bB cC + dD On the basis of Guldberg and Waage’s law the rate of forward reaction Vf [A]a [B]b or Vf = Kf [A]a [B]b Where Kf is the proportionality constant for forward reaction. The rate of reverse reaction Vr [C]c [D]d Vr= Kr [C]c [D]d Where Vr is the proportionality constant for reverse reaction. At equilibrium the rates of forward and reverse reaction will be equal and so Vf = Vr That is, Kf[A]a [B]b = Kr [C]c[D]d Thus, when equilibrium is attained if we determine the concentration of the reactants and the products in any reaction and their stoichiometric multiples, the equilibrium constant Kc can be obtained. 5. What is meant by Kc and Kp? Deduce relation between Kc and Kp. Ans. Relation between Kp and Kc As seen earlier the equilibrium constant of a gaseous reaction can be written as …. 1 But we know that according to simple gas equation PV = nRT. Hence, it can be written as P = (n/V)RT = CRT (where n/V = C = concentration in mol lit-1) Substituting the values of p in the above equation 4.10, it can be written as ….2 ng …3 = Kc X (RT) …. 4 Where ∆ng = (c+d) – (a+b) Means number of total moles of gaseous products minus number of total moles of gaseous reactants. Hence, it can be written as Kp = Kc X (RT)ng…. 4 6 Deduce the equilibrium constant Kp pCO2 for decomposition of solid CaCO3 in closed vessel Ans. Kc = [CaO][CO2] Now, Density of CaCO3 = Constant [CaCO3] Density of CaO Kc x Density of CaCO3 = [CO2] Density of CaO Kp = PCO2 7. Deduce the formula Kp = (P2)/4 for decomposition of solid NH4HS in closed vessel. Ans. NH4HS(s) NH3(g) + H2S(g) x x Total moles = 2x Total pressure = Patm P/2 P/2 Kp = [PNH3][PH2S] = (P/2)(P/2) = (P2/4) atm2 8. Derive the formula for decomposition constant of solid ammonium carbonate and prove that Kp = (4/27)P3. Ans. NH4COONH2(s) 2NH3(g) + CO2(g) Moles = 2x x Total moles = 3x Kp = [PNH3]2[PCO2] = (2P/3)2(P/3)2 = (4P3/27) atm3 9. Give Le-Chatelier’s principle and explain in detail the effect of concentration on equilibrium. Ans. Equilibrium constant is independent of initial concentration of reactant but, if change is carried out in concentration of pressure, there is an effect on equilibrium, the equilibrium tries to nullify this effect. Reaction can be endothermic or exothermic. Hence, the heat absorbed or evolved, functions like that of the reactant and its effect is on equilibrium which tries to nullify the effect. Le Chatelier studies the effect of concentration and temperature and his presentation is called Le Chatelier’s principle whose statement can be written as below: “ If from the factors determining the equilibrium state, any one factor is changed, there will be such a change in the system that the effect will be nullified or made negligible so that the value of equilibrium constant at that temperature will remain constant.” This principle can be applied to both physical and chemical equilibrium. We shall study in detail the factors like (1) change in concentration (2) change in pressure (3) addition of inert gases (4) change in temperature and (5) use of catalyst affecting the equilibrium (1)Effect of change in concentration: If we add or remove the reactant or the product from the reaction in equilibrium, its effect on equilibrium according to Le Chatelier’s principle will be as follows: (a)If the concentration of reactant or product is increased by addition of reactant or product, the reaction will occur in such a way that the increase in concentration will be taken for use, i.e. the increase in concentration of reactant, the concentration of product will increase and if concentration of product is increased the reaction will result in the direction of increase in concentration of the reactant. (b)If the concentration of reactant or product is decreased by removing reactant or product, the reaction will occur in such a way that product or reactant will be established again. Hence, if any change in concentration of reactant of product is carried out then the equilibrium will try to make this effect minimum and equilibrium will be established accordingly. If we take this as an example, in the reaction If concentration of reactants dinitrogen or dihydrogen is increased, the reaction will proceed towards right hand side and the product ammonia obtained will be more. If the concentration of nitrogen or hydrogen is decreased, the reaction will proceed towards left hand side and reactants of nitrogen or hydrogen will be obtained back i.e. production of ammonia will decrease. Here, it is necessary to remember that 1 mole dinitrogen combines with 3 moles of dihydrogen in this reaction and forms 2 moles of ammonia. Hence, increase in concentration of dihydrogen rather than dinitrogen, will give more product. We take another example of heterogeneous equilibrium. If solid CaCO3(s) is heated in closed vessel, the following decomposition reaction will occur. CaCO3(s) CaO(s) + CO2(g) Hence, if more CaO(s) is to be obtained then, the CO2(g) formed in the reaction should be removed because CO2(g), gas can combine with solid CaO(s) and carry out reverse reaction then proportion of product will decrease. Hence, by removing CO2(g) from the reaction vessel, more CaO(s) can be obtained. 10. Explain Arrhenius acid-base theory and mention it’s drawbacks Ans. Arrhenius concept about Acid and Base: According to Arrhenius concept, substances which dissociate in water and give hydrogen ion (H+) are called acids and substances which dissociate in water and give hydroxyl ions (OH-) are called bases. E.g; (1) HCl(g) +H2O +H2O H+ + Cl- Acid (2) NaOH(s) Na+ + OHBase The limitations of this concept are as follows: (i) proton (H+) is highly unstable. (ii) It can not exist independently. (iii) It immediately combines with molecules of solvent water and gives H+3O ions. Its addition in certain bases OH- is not present even then they show properties of base, viz., NH3. Similarly compounds like BF3 do not possess H+, even then they act as acid. (iv) this concept can only be applied to aqueous solution because salt like NH4Cl reacts as acid in solvent like liquid NH3. Hence, it is difficult to accept Arrhenius concept as the universal one. Because ionization is given importance. (In addition it is necessary to know that ionic radius of H+ is about 10-15 meter and so it is very small in size9. Hence it easily gets combined with molecules of water and forms H3+O ion which is called as hydronium ion. One estimate is that one H+ can be combined with four molecules of water showing H+ + 4H2O H9O4+ ion. 11. Explain with suitable example, Lowry-Bronsted acid-base theory(proton transfer theory) Ans. Concept of Bronsted – Lowry for acid and base: Danish chemist Bronsted, an English chemist Lowry presented the concept of acid and base. They made H+ (proton) as a base. According, to their concept , the substance which gives a proton or donates a proton is called the acid and the substance which receives the proton or accepts a proton is called the base. Thus, acid is a proton donor and base is proton acceptor. Let us take the dissociation reaction of hydrogen chloride in water. Base Acid H3+O (aq) + Cl-(aq) HCl(g) + H2O(l) Acid Similarly, Base Acid NH3(g) + H2O(l) Base NH4+(aq)+ OH-(aq) Base Acid We shall understand in detail the first from the above reactions : HCl H+ + ClAcid-1 proton conjugate base-1 As it gives proton, HCl is an acid H2O + H+ H3O+ Base-2 Proton Conjugate acids-2 As it accepts a proton, H2O is a base. Total reaction: HCl + H2O H3O+(aq) + Cl-(aq). Acid-1 Base-2 Acid-2 Base-1. In the above reaction giving-taking of proton is not shown. Hence, it can be said that only transfer of proton takes place, it is not obtained free. Every acid will lose proton and so its conjugate base will be formed and every base will accept a proton and so its conjugate acid will be formed. Hence, this concept is known as proton transfer or conjugate acid-base concept. We have earlier seen in limitations of Arrhenius concept that OH- is not present in NH3 even then it acts as a base but according to Bronsted-Lowry concept it can be explained. NH3 + H2O NH+4(aq) + OH-(aq) Base-1 Acid -2 Acid-1 Base-2 In the reaction, base NH3 accepts a proton and forms conjugate acid NH4+ ion and acid H2O loses proton and forms conjugate base OH-. Thus the concept of Bronsted - Lowry is found to be more applicable and acceptable than Arrhenius concept. Even then its limitations are also known. This difficulties are observed in the study of reactions in organic chemistry and complex salts. BF3 has no proton even then it acts as an acid. Hence, the third concept has come in to existence, which is known as Lewis acid- base concept. Proton is given importance in Bronsted – Lowry concept. 12. Explain in detail Lewis acid-base theory. Ans. Lewis Concept of Acid and Base : Lewis, in presenting this concept, in 1923, mentioned that acid means a substance which can accept a pair of electrons and base means a substance which can donate a pair of electrons. Thus, instead of the concepts of proton, ionization, conjugate acid or base, he made electrons associated with all reactions and substances and its arrangement as the base of the concept. As seen earlier BF3 can be said acid or not and NH3 can be said a base or not can be solved by this concept. It will be clear from the following reaction: Thus, BF3 accepts the pair of electron and so it is acid and NH3 donates a pair of electron, so it is acid and NH3 donates a pair of electron, so it is base. Electron deficient substances or ions like AlCl3, Co3+, Mg+2, will act as acid and substances like H2O, NH3, OH-, F- will act as base. They are respectively called Lewis acid and Lewis base. 13. Derive the formula of dissociation constant of Weak acid Ans. Ionization Constant of Weak Acid (Ka) : In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the equilibrium is obtained as below : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation is α, then following can be written : Reaction : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Initial Concentration(M) C 0 0 Degree of ionisation(α) 1-α α α Concentration at equilibrium(M) (1 - α)C αC αC Equilibrium constant Ke= [H3O+][A-] [HA][H2O] = (αC)(αC) …1 (1 - α)C[H2O] But [H2O] is accepted as constant and so Ke[H2O] = (αC)(αC) (1 - α)C = α2C = Ka …2 (1 - α) Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we consider both the terms same) Hence, for any weak monobasic acid, following can be written [H3O+][A-] Ke= [HA] …3 The unit of Ka will be mollit-1. As the values of Ka depend on [H3+O] the values of ka will be different for different. Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite temperature. 14. Derive the formula of dissociation constant of weak base Ans. ionisation constant (Kb) of weak base: The ionization of monoacidic weak base MOH will take place in aqueous solution as follows: H2O MOH(aq) M+(aq) + OH-(aq) As base is weak, incomplete ionization will occur and so equilibrium will be obtained and it can be expressed as below: [M+] [OH-] …………4 and Kc = [MOH] [H2O] [M+] [OH-] =Ke ………. 5 Kex[H2O]= [MOH] Where Kb is the ionization or dissociation constant of monoacidic weak base. If we know the initial concentration of weak base and its degree of ionization, we can calculate the value of Kb as we have studied earlier. [M+] = [OH-]. 15. Obtain relation between ionic equilibrium constant Ka, initial concentration Co and concentration of H3O+ in aqueous solution of weak formic acid. Ans. As formic acid is weak acid, it is partially ionized in water. The equilibrium between the undissociated species of CH3COOH and the ions H+(aq) can be represented as CH3COOH(l) + H2O(l) H3O+(aq) + CHCOO-(aq) The equilibrium constant K for this reaction is as follows : K = [H3O+][CH3COO-] [CH3COOH][H2O] Here the change in concentration of water on dissolving the weak acid will be quite negligible compared to its original concentration, so considering the value of [H2O] as constant and joining it with equilibrium constant K we get a new constant Ka as follows. K[H2O] = Ka = [H3O+][CH3COO-] [CH3COOH] Here formic acid will ionize to much less extent and so the concentration of its unionized part will remain nearly same as its original concentration. [CH3COOH] = Co So, we have Ka = [H3O+][CH3COO-] Co Again, the concentration of the positive ions [H3O+] and that of the negative ions [CH3COO-] will be the same i.e. [H3O+] = [CH3COO-] [CH3COO-] can be replaced by [H3O+], Ka = [H3O+]2 Co + 2 + 1/2 [H3O ] = Ka.Co [H3O ] = (KaCo) 16. Obtain relation between ionic equilibrium constant Kb, initial concentration Co and concentration of OH- in aqueous solution of weak base aniline. Ans. As aniline is weak acid, it is partially ionized in water. The equilibrium between the undissociated species of C6H5NH2 and the ions H+(aq) can be represented as NH3(g) + H2O(l) NH4+(aq) + OH-(aq) The equilibrium constant K for this reaction is as follows : K = [NH4+][OH-] [NH3][H2O] Here the change in concentration of water on dissolving the weak base will be quite negligible compared to its original concentration, so considering the value of [H2O] as constant and joining it with equilibrium constant K we get a new constant Ka as follows. K[H2O] = Kb = [NH4+][OH-] [NH3] Here aniline will ionize to much less extent and so the concentration of its unionized part will remain nearly same as its original concentration. [NH3] = Co So, we have Ka = [NH4+][OH-] Co Again, the concentration of the positive ions [NH4+] and that of the negative ions [OH-] will be the same [OH-] can be replaced by [NH4+], Kb = [OH-]2 Co [OH-]2 = Kb.Co [OH-] = (KbCo)1/2 17. What is meant by ionic product of water? Deduce its equation. Ans. Ionic Product of Water Water is an amphoteric oxide when acid is added to it; it accepts the proton and act as a base and when base is added to it, it donates a proton and acts as acid. When reaction between two molecules of water takes place, one molecule donates proton and other molecule receives proton and shows conjugate acid-base reaction. H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Acid-1 Base-2 Conjugate acid-2 Conjugate base-2 If we express the equilibrium constant of above reaction them. [H3O+][OH-] Ka = [H2O][H2O] Where Ka is the dissociation constant of acid. There is no significant change in concentration of water (55.5M) because H2O is a weak acid (possesses about 10-7M H+). Thus, if H2O is considered constant. Ka x [H2O]2 = [H3+O] [OH-] = Kw. Where Kw is ionic product of water. Water is neutral and so [H3+O] and [OH-] in it are 10-7 M. Hence, Kw = [H3+O] [OH-] = (10-7)(10-7) = 10-14 which is constant and equilibrium constant remains constant at constant temperature; so the value of Kw will be constant at constant temperature, viz., the value of Kw is 1 x 10-14 at 298 K. If we find the ratio of concentrations of dissociated and un dissociated water, it will remain on left hand side, i.e. the number of un dissociated molecules or its concentration will be more. 18. Discuss the factors affecting strength of acid. Ans. Factors Affecting strength of acid: From the study of dissociation constant, pH value etc., it is found that their values are different. The reason for this is that [H3+O] available can be more or less. What may be the reason for this? If the acid is strong, its value of Ka will be high and the value of pH will be low. The dissociation of acid will depend on strength of acid and the polarity of H-A bond. As the strength of H-A bond decreases, the energy required for breaking that bond will decrease and HA will be stronger. When difference between electro negativities of A and B will increase, apparently ionization will occur and will be easy to break the structure of the bond. Hence, the acidity will increase. On comparing the elements of the same group of periodic table, the strength of H-A bond will be more important factor than polar nature. As we go down in the group the size of A will increase and so strength of H-A bond will decrease and hence acid strength will increase. Increase in size HF<< HCl <<HBr<<HI Increase in acid strength For this reason, H2O is stronger acid than H2S, but if we discuss the elements in the same period of periodic table, the polarity of H-A bond will determine the strength of acid. As the electronegativity of A increases, the strength of acid will increase. Increase in electro negativity CH4<< NH3<< H2O << HF Increase in acid strength 19. Write a short note on buffer solutions. Ans. Buffer solutions The pH of the fluids like blood in our body and urine is almost constant. If there is change in this pH, it affects biochemical reaction in the body. The pH of chemical and biochemical reactions in our body are constant, biochemical reactions in our body are constant, viz. the pH of human saliva is 6.4. In addition, hydrochloric acid is present in human stomach which helps in digestion. The pH of cosmetics are also kept constant. Hence, the question arises that how pH in any solution can be kept constant. Such solutions are called buffer solutions. Its definition can be given as below: “ The solution which resists the change in pH carried out by addition of acid or base in small proportion to them or are being diluted, and the values of their pH remain constant are called buffer solution.” Buffer solutions can be acidic or basic. If pKa of weak acid and pKb of weak base are known, buffer solutions of known pH can be prepared. Buffer solutions can be of three types as follows: (i) Acidic buffer solution: Acidic buffer solution can be prepared by mixture of weak acid and its salt with strong base. (ii) Basic buffer solution:Basic buffer solution can be prepared by mixture of weak base and its salt with strong acid. (iii) Neutral buffer solution: Neutral buffer solution can be prepared by neutralization of weak acid and weak base. These type of buffer solutions are shown below: Type Substances Value of pH Acidic CH3COOH + CH3COONa <7 Basic NH4OH + NH4Cl >7 Neutral CH3COOH + NH4OH =7 Buffer solution of known pH can be prepared by using the following HendersonHaschelback equation. For acidic solution, [Salt] pH = pKa + log [Acid] …………. 1 where [acid] is concentration a weak acid and its dissociation constant is Ka and [salt] is concentration of the salt of this weak acid with strong base. For a buffer solution, it can be written as pH= pKa + log [CH3COONa] [CH3COOH] similarly, for basic buffer solution e.g., NH4OH + NH4Cl can be written that, [NH4Cl] pH = pKa + log [NH4OH] Such buffer solution can be used in chemical and biochemical reactions and especially in analytical chemistry. In human body buffer solutions containing [HCO-3] and [CO2-3] as well as [H2PO-4] and [HPO2-4] are present. 20. Write a short note on pH scale. Ans. pH Scale. If we express the concentration of hydronium ion [H3+O] in molarity then values like 10-12 to 10-2 are possible. It is difficult to express these values on simple graph paper. Hence, scientist Sorenson found a scale which is called pH scale. According to him pH = -log10[H3+O]. The values 10-12 to 10-2 shown above can be converted to +12 to +2 if calculated on the basis of this relation and plotting of graph can be easy. The definition of pH can be given like this, “pH of a solution is the negative logarithm to the base 10, of the concentration of hydrogen or hydronium ion”. According to thermodynamics, activity is more proper word instead of concentration but in dilute solutions activity and concentration can be considered can be considered to be the same. Now as seen earlier a solution containing 10-7M [H3+O] and [OH-] is neutral. Hence, pH = -log10[H3+O] = -log1010-7M = 7 and for acidic solution [H3+O]> 10-7M, pH < 7 Similarly, for basic solution [H3+O]< 10-7M, Hence, pH > 7. Therefore, it can be written as : pH< 7 Acidic solution pH> 7 Basic solution pH = 7 Neutral solution AS seen above, Kw = [H3+O] [OH-] Putting the values, Kw= (10-7)(10-7) = 10-14and –log Kw = -log(10-14) Therefore, pKw = 14 Therefore, pH + pOH =pKw = 14 Temperature affects the values of pH, pOH, pKw. The above discussion can be shown in the following table : Concentration (M) Acidic Neutral Basic [H3+O] More than 10-7 10-7 Less than 10-7 -7 -7 [OH ] Less than 10 10 More than 10-7 pH Less than 7 7 More than 7 pOH More than 7 7 Less than 7 pH paper, litmus paper or universal indicator can be used to test whether the solution is acidic or basic but the exact values of pH can be determined with the help of instrument called pH meter. 21. Explain the calculation of pH of a solution from the dissociation constant of weak acid. Ans. In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the equilibrium is obtained as below : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation is α, then following can be written : Reaction : HA(aq) + H2O(l) H3+O(aq) + A-(aq). Initial Concentration(M) C 0 0 Degree of ionisation(α) 1-α α α Concentration at equilibrium(M) (1 - α)C αC αC [H3O+][A-] Equilibrium constant Ke= [HA][H2O] = (αC)(αC) …1 (1 - α)C[H2O] But [H2O] is accepted as constant and so Ke[H2O] = (αC)(αC) (1 - α)C = α2C = Ka …2 (1 - α) Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we consider both the terms same) Hence, for any weak monobasic acid, following can be written [H3O+][A-] Ke= [HA] …3 -1 The unit of Ka will be mollit . As the values of Ka depend on [H3+O] the values of ka will be different for different. Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite temperature. IN table 4.3 the values of ionisation constants of some weak acids are given. According to the relations seen earlier, pH = -log10[H3+O] pOH = -log10[OH-] Similarly, pKa = -log10[Ka] PA- = -log10[A-] can be written. Where [A-] is the concentration of negative ion. It is apparent from above relations that if the values of initial concentration [H3+O] and Ka are known, then at equilibrium, [H3+O] concentration can be determined and pH can be calculated 22. Explain effect of common ion on solubility of sparingly soluble salt. Ans. Effect of Common Ion on Solubility of Sparingly Soluble Salt The sparingly soluble salt that has dissolved in solution that is completely dissociated and so it is ionic in form. Hence, it is a strong electrolyte. Earlier, in chemical equilibrium, effect of concentration, application of Le Chatelier’s principle etc. have been studied. Solubility product is ionic equilibrium and the effect of concentration can be studied. As it is equilibrium constant, it will depend on temperature, but its value will be constant at constant temperature. What will happen if we add soluble ionic substance like KCl in the solution of a sparing soluble salt like AgCl? AgCl(s) Ag+(aq) + Cl-(aq) KCl(s) K+(aq) + Cl-(aq) The Cl from AgCl in equilibrium and Cl- ion obtained by complete ionization of KCl, the concentration of Cl- will increase. Hence, according to Le Chatelier’s principle, the equilibrium will shift towards left side so as to nullify the effect of Cl- i.e. more AgCl will be formed. In other words, there will decrease in solubility of AgCl. Hence, it can be said that because of the effect of common ion on sparingly soluble alt its solubility decreases and sparingly soluble salt precipitates more. The use of effect of common ion can be made to separate one ion from the other in presence of other ion in qualitative analysis. It can also be used for decrease in solubility of the components in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of second group are less in comparison to solubility products of sulphides of metal of III B group ions, therefore, HCl is added before adding H2S water to test the second group ions. H2S(aq) 2H+(aq) + S2-(aq) HCl(aq) H+(aq) + Cl-(aq) The common ion available from HCl creates common ion effect on the equilibrium and decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group can only be precipitated because their solubility products are less. In the same way, for precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH. The concentration of OH- available from ionization of NH4OH gets decreased due to common ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group only will be precipitated because the values of solubility products of the hydroxides of III A group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions. Its is necessary to note that under certain situations the solubility increases instead of decreasing. The solubility of salt like phosphates increase when acid is added to their solutions or pH of the solution decreases. The reason for this is that, phosphate ion combines with H+ available from acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases. 23. Explain with suitable example the use of common ion effect in qualitative analysis. Ans. The use of effect of common ion can be made to separate one ion from the other in presence of other ion in qualitative analysis. It can also be used for decrease in solubility of the components in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of second group are less in comparison to solubility products of sulphides of metal of III B group ions, therefore, HCl is added before adding H2S water to test the second group ions. H2S(aq) 2H+(aq) + S2-(aq) HCl(aq) H+(aq) + Cl-(aq) The common ion available from HCl creates common ion effect on the equilibrium and decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group can only be precipitated because their solubility products are less. In the same way, for precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH. The concentration of OH- available from ionization of NH4OH gets decreased due to common ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group only will be precipitated because the values of solubility products of the hydroxides of III A group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions. Its is necessary to note that under certain situations the solubility increases instead of decreasing. The solubility of salt like phosphates increase when acid is added to their solutions or pH of the solution decreases. The reason for this is that, phosphate ion combines with H+ available from acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases. 24. What is meant by common ion effect? In qualitative analysis, NH4Cl is added before addition of NH4OH in precopotation of ions of III A group. Explain giving reason. Ans. Let us take the example of weak acid, acetic acid ( ) CH3COOH(aq)+ H2O(l) H3+O(aq) + CH3COO-(aq) OR HAc + H2O(l) H3+O(aq) + Ac-(aq) Where HAc and Ac- are the short forms of CH3COOH and CH3COO- ion. [H3+O][Ac-] Ka = [HAc] ………………..1 Suppose, we add CH3COONa or HCl, to the solution of HAc in equilibrium, then what will happen? As studied earlier in chemical equilibrium if HCl is added, [H3+O] will increase and if CH3COONa is added, [Ac-] will increase. Hence, according to Le Chatelier’s principle, the equilibrium will make negligible change and will keep the same value of equilibrium constant. This means that the equilibrium will be shifted towards left and concentration of HAc will increase i.e. the amount of un dissociated acid will increase and there will be decrease in [H3+O] and hence, there will be increase in pH. By addition of HCl due to increase in [H3+O] similar result will be obtained. This effect is known as effect of common ion effect on dissociation constant of acid. In the same way, in the case of ionization of weak base NH3, if we increase [NH4+] by adding salt like NH4Cl then, according to Le Chatelier’s principle, as [NH4+] increases the equilibrium will shift towards left and hence, un dissociated NH3 will increase, i.e. [OH-] will decrease. As a result, pH will increase. In qualitative analysis, to precipitate the radicals of III A viz Fe2+, Fe3+, Al3+, Cr3+ as their hydroxides, NH4Cl is added to the solution before adding NH4OH. Consequently, the reverse reaction is favoured in the equilibrium, resulting in decrease of OH- ion concentration. Because of this, the hydroxides of III B, IV groups and that of Mg2+ are not precipitated alon with those of group III A. NH4Cl(aq) NH4+(aq) + Cl-(aq) NH4OH(aq) NH4+(aq) + OH-(aq) The hydroxides of the radicals of III A group are extremely less soluble compared to those of succeeding groups. So, in presence of NH4Cl the hydroxides of III A group are precipitated by NH4OH. (25). 2 moles of PCl5 is heated in closed vessel of 4 liter a t definite temperature. At equilibrium state PCl5 remained undecomposed. Calculate Kc of the reaction. Solution: Reaction, PCl5(g) PCl3(g) +Cl2(g) Now, 55% PCl5 remains undecomposed. Therefore, 45 % PCl5 goes under reaction. = 45 x 2 = 0.9 moles PCl5 100 As ratio of PCl5 and (PCl3 and Cl2 ) is 1 : 1 in number of moles, 0.9 moles of PCl5 will yield 0.9 moles of PCl3 and Cl2. PCl5(g) PCl3(g) +Cl2(g) In the beginning: 2 moles 0 0 Used moles at equilibrium: 0.9 moles 0.9 0.9 Unused moles at equilibrium: 1.1 moles Now, Kc = [concentration of products] [concentration of reactants] ……..(i) Kc = equilibrium constant Now, at equilibrium, moles of reactants = 1.1 moles and moles of products = 0.9 moles ( in solution) Now, this vessel consists of 4 liter solution Therefore, concentration = 1.1 = 0.275 M/L (for reactant) 4 = 0.9 = 0.225 M/L (for products) …………(ii) 4 From equation (i) Kc = [PCl3] [Cl2] [PCl5] From equation (ii) Kc = (0.225) (0.225) = 0.050625 (0.275) 0.275 Therefore, Kc = 0.184 mol/lit Kc = 1.84 x 10-1 mol/lit. (26). The value of Kp obtained at 1060 kelvin temperature is 0.033 mollit-1 for the reaction 2NOCl(g) 2NO(g) + Cl2(g). What will be the value of Kc for the reaction. Solution: 2NOCl(g) 2NO(g) + Cl2(g). ∆n(g) = np – nr (for gaseous molecules) =(2+1)–2=1 Data: Kp = 0.033 atm T = 1060K 0.082 atm lit mol-1 K-1 Kp = Kc x (RT)∆n(g) Kc = Kp (RT)∆n(g) Substituting the values from data, Kc = 0.033 = 0.033 (0.082 x 1060) 86.92 = 3.796 x 10-4 M Therefore, Kc = 3.796 x 10-4 M (27). At definite temperature and 3 atm pressure, 75 % PCl5 is decomposed into PCl3 and Cl2. Calculate Kp of this reaction. PCl5(g) PCl3(g) +Cl2(g). Solution: 75% PCl5 is decomposed into PCl3 and Cl2. At equilibrium, moles PCl5 decomposed 100 75 1 x [ suppose initial moles of PCl5 is 1.] x= 75 x 1 = 0.75 moles 100 0.75 moles of PCl3 and Cl2 are formed and unreacted moles of PCl5 = ( 1 – 0.75) = 0.25 moles PCl5(g) PCl3(g) +Cl2(g). Initial moles: 1 0 0 Used moles at equilibrium: 0.75 0.75 0.75 Unused moles at equilibrium: 0.25 Now, total moles at equilibrium = 0.25 + 0.75 + 0.75 = 1.75 moles Total pressure in system = 3 qtm Now, partial pressure at equilibrium: moles of 1 element x total pressure Total moles Hence, partial pressure of PCl3 ( Ppcl3) = 0.75 x 3 = 3 x 3 1.75 7 = 9 atm 7 ………… (i) Similarly , partial moles of Cl2 Pcl2 = 0.75 x 3 = 3 x3 =9 1.75 7 7 atm …………..(ii) Partial pressure of PCl5 Ppcl5 = 0.25 x 3 = 1 x3 = 3 atm 1.75 7 7 ……………….. (iii) Now, Kp = (Ppcl3) x (Pcl2) (Ppcl5) From equation (i) , (ii) and (iii) Kp = 9/7 x 9/7 3/7 Kp = 3.857 atm. (28). 1 mole N2 and 3 moles H2 are heated at 473 K temperature and 100 atmosphere pressure. If 0.50 mole of NH3 are formed at equilibrium time calculate equilibrium constant Kp for this reaction. N2(g) + 3H2(g) 2NH3(g) Solution: N2(g) + 3H2(g) 2NH3(g) Now, the ratio of production of moles of this reaction is 1 : 3 : 2 Initial moles of N2 = 1mole Data Initial moles of 3H2 = 3 moles Therefore, from the ratio, moles of NH3 formed = 2 moles But from the data, moles of NH3 fromed = 0.50 moles. Therefore, ¼ th of NH3 is formed [ because 2 /0.50 = 4] Hence, 1/4th of reactants must be used. For N2 , ¼ x 1 = ¼ moles For H2 , ¼ x 3 = ¾ moles For NH3 = 0.50 moles ……(i) Above these are the moles formed in solution Therefore, unused moles at equilibrium For N2 , 1 – ¼ = ¾ moles ….(ii) For H2 , 3 – ¾ = 9/4 moles ……(iii) N2(g) + 3H2(g) 2NH3(g) Initial moles: 1 3 2 Used moles at equilibrium ¼ ¾ 0.50 Unused moles at equilibrium ¾ 9/4 - Partial pressure = moles of 1 element x total pressure Total number of moles Total pressure = 100 atm Total number of moles = ¾ + 9/4 + 0.50 = 3.50 moles Partial pressure of N2 PN2 = ¾ x 100 3.50 = 21.428 atm …………..(iv) Partial pressure of H2 PH2 = 9/4 x 100 3.50 = 64.285 atm ……….(v) Partial pressure of NH3 PNH3 = 0.50 x 100 3.5 = 14.2857 atm …….(vi) Now, from equation N2(g) + 3H2(g) 2NH3(g) Kp = (PNH3)2 = (14.2857)2 (PN2) x (PH2)3 21. 2428 x (64.285)3 Kp ≈ 3.616 x 10-5 atm -2 (29). On heating 0.5 mole solid calcium carbonate in a vessel of 500ml volume, at 400 K temperature , Kc 0.9 mol lit-1 is obtained at equilibrium state, then calculate mole of CO2 at equilibrium. How many percent the reaction would have been completed? CaCO3(s) CaO(s) + CO2(g) Solution: CaCO3(s) CaO(s) + CO2(g) Initial moles: 0.5 Used moles at equilibrium: x x x [as ratio is 1 : 1] unused moles at equilibrium : (0.5 – x) concentration (0.5 – x) x (mole/ liter) 0.5 0.5 Kc = [CaO][CO2] [CaCO3] = (x/0.5) (x/0.5) = (2x) (2x) ((0.5 –x)/0.5) 2(0.5 – x) Kc = 4x2 1 – 2x ……….(i) From data, Kc = 0.9 mol/lit Therefore, from equation (i), 0.9 = 4x2 => 9 = 4x2 1-2x 10 1 – 2n 9 – 18 x=40x2 40x2 + 18x -9 = 0 x 0.5 [ volume in solution is 500ml (from data)] Comparing with quadrative equation, ax2 + bx + c = 0 a = 40 b = 18 ………(ii) c= -9 Solution, x = - b ±√ b2 – 4ac 2a Substituting values of a, b ,c from equation (ii) x = -18 ± √ (18)2 – 4(40) (-9) 2( 40) x= -18 ± √ 324 + 1440 80 x = -18 ± √ 1764 = -18 + 42 80 80 x = -18 + 42 or x = -18 -42 80 80 x= 24 or x= -60 80 80 (not possible as solution of x is negative value) Therefore 3 => x = 0.3 moles 10 Therefore, moles of Co2 at equilibrium = x = 0.3 moles Therefore, from data , 0.5 ml of CaCO3 0.3 moles of CO2 and 0.3 moles of CaO Therefore, moles of CO2 formed = 0.3 moles CaCO3(s) CaO(s) + CO2(g) 0.5 0.3 100 (?) 0.3 x 100 = 60 0.5 Therefore, % of reaction completed = 60% (30). Mg(HCO3)2(s) is decomposed as follows in a closed vessel. If mole fraction of CO2 is 0.8 and Kp = 64 atm3 then calculate total pressure of mixture at equlibrium. Mg(HCO3)2 MgCO3(s) + CO2(g) + H2O(g) Solution: Mg(HCO3)2 MgCO3(s) + CO2(g) + H2O(g) Now, gaseous mole in system are CO2 and H2O Mole fraction of CO2 = 0.8 Total moles = 1 Therefore, mole fraction of H2O = 1 -0.8 = 0.2 …….(i) Therefore, mole fraction Xco2 = 0.8 Mole fraction XH2O = 0.2 Now, partial pressure = moles of element x total pressure Total moles = mole fraction x total pressure Therefore, partial pressure of CO2 (PCO2) = Xco2 x total pressure = 0.8 x ……….(ii) [let total pressure be x] Partial pressure of H2O (PH2O) = XH2O x total pressure = 0.2 x ………………(iii) Now, Kp = [PCO2] [PH2O] But, from data, Kp = 64 atm2 and from equation (ii) and (iii) 64 = (0.8x) ( 0.2x) Therefore, 64 = 0.16x2 Therefore, x2 = 64 0.16 2 x = 400 therefore, x = 20 atm Therefore, total pressure in system = 20 atm (PCO2 = 16 atm PH2O = 4 atm) (31). If 3.65 x 10 -2 gram HCl is dissolved in 500 ml solution, then find its pH (pH = 2.699) Solution: Data, For HCl, weight (W) = 3.65 x 10-2 gram M = 36.5 gram/ mole V = 500 ml = 0.5L Moles of HCl = W = 3.65 x 10-2 M 36.5 = 1 x 10-3 mole Concentration of [H+] in HCl = 1 x 10-3 0.5 = 2 x 10-3 M/L pH= -log[H+] = -log[2 x 10-3] = - [log 2 + log (10-3)] [ log(axb) = log a + log b] = 3 – log2 = 3 – 0.3010 pH= 2.6990. (32). Find pH of mixture of 50ml 0.03 M HCl and 60 ml 0.02M NaOH . Solution: 0.03 ml HCl 1000ml HCl solution contains 0.03 mol HCl Therefore, 50 ml solution contains = 0.03 x 50 1000 = 1.5 x 10-3 moles HCl ……..(i) Similarly, 0.02 M NaOH 1000ml NaOH contains 0.02 mol NaOH Therefore, 60 ml solution contains = 0.02 x 60 1000 = 1.2 x 10-3 moles NaOH …..(ii) NaOH + HCl neutralisation NaCl + H2O -3 Therefore, 1.2 x 10 moles of NaOH will neutralize 1.2 x 10-3 moles of HCl out of 1.5 x 10-3 moles of HCl Therefore, moles of HCl that remains unneutralised are = (1.5 – 1.2) x 10-3 = 0.3 x 10-3 = 3 x 10-4 moles ………(iii) Total volume of the mixture = 110ml Therefore, molarity of HCl after neutralization = moles = moles x 1000 Volume(L) V (ml) -4 = 3 x 10 x 1000 [from equation (iii)] 110 Therefore, molarity of HCl = 2.727 x 10-3 M Now, HCl is strong acid and hence complete ionization takes place Therefore, HCl(aq) + H2O(l) Cl-(aq) + H3O+(aq) 2.727 x 10-3 M 2.727 x 10-3 M pH= - log[H3O+] = - log [ 2.727 x 10-3] Therefore, pH = [ log 2.727 + ( -3) log 10] pH = - log 2.727 + 3log10 therefore, pH = -0.4357 + 3 therefore, pH=2.5643 (33). Find weight of CH3COOH in 100ml aqueous solution of CH3COOH having pH 3.0 Ka for CH3COOH = 1.75 x 10-5 Solution: CH3COOH + H2O H3O+ + CH3COOAs CH3COOH is weak acid, partial ionization takes place Therefore, Ka = 1.75 x 10-5 Therefore, for weak acid, Ka = [H3O+]2 Co Where,Co = initial concentration of weak acid i.e. CH3COOH Co = [H3O+]2 Ka ………….. (i) Now, pH = - log10 [H3O+] = 3 [given] Therefore, -log[H3O+] = 3 Therefore, [H3O+] = 1 x 10-3 M …………………..(ii) Substituting the value of [H3O+] from equation (ii) in equation (i) Therefore, Co = [H3O+]2 = ( 1 x 10-3)2 Ka 1.75 x 10-5 -2 Co = 5.7142 x 10 M Now, concentration = mole Liter Therefore, 1 liter solution conatin 5.7142 x 10-2 moles of acetic acid 1L 5.7142 x 10-2 moles -1 100ml(10 L) ? = 5.7142 x 10-2 = 5.714 x 10-3 moles -1 10 Now, mole = weight Molecular weight M.W. of CH3COOH = 60 gram / mole Therefore, moles= W 60 Therefore, W = 60 x moles = 60 x 5.7412 x 10-3 W = 3.428 x 10-1 gram = 0.342 gram. (34).0.02 M CH3NH2 is ionized to 15% at equilibrium .Find its ionization constant. Solution: CH3NH2(g) + H2O(l) CH3NH+3(aq) + OH-(aq) Methyl amine ionizes to 15% at equilibrium If 100 moles of CH3NH2 are taken, then 15 moles of OH- are formed CH3NH2 OH100 : 15 0.02 : (?) = 15 x 0.02 = 0.3 x 10-2 M 100 Therefore, [OH ] = 3 x 10-3 M Now, CH3NH2 is weak base Therefore, weak base , Kb = [OH-]2 Co Where, Kb = ionization constant of weak base Co = initial concentration of base = 0.02 M Therefore, Kb = ( 3 x 10-3)2 0.02 = 9 x 10-6 2x 10-2 Therefore, Kb = 4.5 x 10-4 (35). For NH3 Kb = 1.77 x 10-5 , what will be the pH of 500 ml solution containing 1.7 gram ammonia? Solution: NH3(g) + H2O(l) NH+4(aq) + OH-(g) For weak base, Kb = 1.77 x 10-5 V (ml) = 500 ml = 500 x 10-3 L W = 1.7 gram M.W. of ammonia = 17 gm/ mole Concentration = moles Liter = weight of NH3 (gm) M.W. of NH3 (gm/mole) x volume (L) = 1.7 17 x 500 x 10-3 = 0.2 M Therefore, concentration = 0.2 M ……..(i) For weak base, Kb = [OH-]2 Co Where, Co = initial concentration of NH3 = 0.2 M (from equation (i)) Therefore, [OH-]2 = Kb x Co = 1.77 x 10-5 x 0.2 = 3.54 x 10-6 Therefore, [OH ] = Kb x Co = 1.77 x 10-5 x 0.2 = 3.54 x 10-6 Therefore, [OH-] = √ 3.54 x 10-6 Therefore, [OH-] = 1.881 x 410-3 M Now, pOH = -log10[OH-] = -log10 [ 1.881 x 10-3] Therefore, pOH = 3 – log(1.881) = 3 – 0.2744 Therefore, pOH = 2.7256 Now, pH + pOH = 14 Therefore, pH = 14 – pOH 14 – 2.7256 Therefore, pH = 11.2744 (36). pH of 0.05M solution of a weak acid is 3.68 then find Ka of that acid. Solution: for weak acid, Initial concentration (Co) = 0.05M pH = 3.68 Ka = ? Now ,pH = - log10[H3O+] = 3.68 Therefore, -log[H3O+]=3.68 Shifting minus (-) sign to right side and taking anti –log on both sides Therefore, [H3O+] = - antilog(3.68) anti-log (-3.68) = antilog(-3.68) -1 +1 = antilog( 4 + 0.32) antilog (-4 + 0.32) Therefore, [H3O+] = antilog(0.32) x 10-4 Therefore, [H3O+] = 2.089 x 10-4 M Now, for weak acid, Ka = [H3O+]2 Co Therefore, Ka = (2.089 x 10-4)2 0.05 = 87.278 x 10-8 = 8.727 x 10-7 Therefore, Ka = 8.7278 x 10-7 (37). Find pH of 0.025M CH3COONa solution. Ka of CH3COOH = 1.75 x 10-5. Solution: CH3COONa CH3COO(aq)- + Na(aq)+ 0.025M 0.025M 0.025M Now, CH3COO (aq) + H2O(l) CH3COOH(aq) + OH-(aq) . ……….(i) [ A- + H2O HA + OH-(aq)] Weak acid H2O(l) H(aq)+ + OH-(aq) Now, hydrolysis constant, Kn = Kw Ka [CH3COONa is salt of strong base (NaOH) and weak acid (CH3COOH) and hence, Kn = Kw Ka Therefore, Kn = Kw = [CH3COOH][OH-] and from reaction (i) ] Ka [CH3COO-] As concentration of [OH ] produced by the self- ionization of water is negligible in comparison with the concentration of OH- produced by hydrolysis reaction. Therefore, [CH3COOH] = [OH-] (At equilibrium) Therefore, Kn = Kw = [OH-]2 Ka [CH3COO-] Kw = ionic product of water = 1x 10-14 Ka = 1.75 x 10-5 [CH3COO-] = initial concentration of CH3COONa = Co = 0.025 M Substituting , these values in above equation Kw = [OH-]2 => 1 x 10-14 = [OH-]2 Ka [CH3COO ] 1.75 x 10-5 0.025 Therefore, [OH-]2 = 1.4285 x 10-11 = 14.285 x 10-12 Therefore, [OH-] = 3.7796 x 10-6 M ≈ 3.78 x 10-6 M Now, pOH = -log10 [OH-] = -log[3.78 x 10-6] = -log 3.78 + 6log10 = 6 –log(3.78) = 6- 0.5775 Therefore, pOH= 5.4225 Now, pH + pOH = 14 pH= 14 –pOH = 14 – 5.4225 = 8.5775 Therefore, pH = 8.5775 (38). Kb of NH3 is 1.8 x 10-5.Calculate pH of 0.20M solution of NH4Cl Solution: NH4Cl NH4+ + ClNow, NH4Cl is salt of strong acid (HCl) and weak base (NH3) Kn = Kw Kb Where, Kn = hydrolysis constant Kw = ionic product = 1 x 10-14 Kb = ionization constant of weak base = 1.8 x 10-5. Now, NH+4 + H2O NH3 + H3O+ Now, Kn = Kw = [NH3] [H3O] = [H3O+]2 Kb [NH+4] [NH+4] [ because at equilibrium, [NH3] = [H3O+] ] Therefore, Kw = 1x 10-14 = [H3O+]2 Kb 1.8 x 10-5 0.20 + 2 Therefore ,[H3O ] = 1 x 0.20 x 10-9 1.8 = 0.111 x 10-9 = 1.111 x 10-10 + Therefore, [H3O ] = 1.054 x 10-5 M Now, pH = -log10 [H3O+] = -log10[1.054 x 10-5] = -log 1.054 + 5 log10 = 5 – log(1.054) (log10 = 1) = 5 -0.0229 = 4.9771 Therefore, pH = 4.9771 ≈ 4.98 39. Find solubility of PbSO4 in water, having solubility product 1.3 x 10-8 (Molecular mass of PbSO4 = 303 gram mol-) Ans. Let, the solubility of PbSO4 be S mol/litre PbSO4(s) Pb+2(aq) + SO42-(aq) S S +2 2Solubility product, Ksp = [Pb ][SO4 (aq)] = [S][S] = [S2] But, Ksp = 1.3 x 10-8(given) Ksp = S2 = 1.3 x 10-8 S =1.140 x 10-4 mol/litre Now, molecular weight of PbSO4 = 303 gm/litre Solubility (in gram/litre) = mol/litre x M.W. = 1.140 x 10-4 x 303 = 345.47 x 10-4 gm/litre = 3.45 x 10-2 gm/L -2 S = 3.45 x 10 gm/litre 40. Ksp of CaF2 is 1.7 x 10-10. What will be the volume of saturated solution of 10 milligram salt? (Atomic mass : Ca = 40 and F = 19) Ans. Let, molar solubility of CaF2 be S mol/litre Solubility Equilibrium, CaF2 Ca2+(aq) + 2F-(aq) S 2S (Because 2 moles of F- are formed) Ksp(given) = 1.7 x 10-10 Ksp = [Ca2+][F-]2 = s[2s]2 = 4S3 From above data, Ksp = 4S3 = 1.7 x 10-10 4S3 = 1.7 x 10-10 S3+ = 0.425 x 10-10 = 42.5 x 10-12 S = 3.489 x 10-4 mol/litre [ S3 = 42.5 x 10-12 S = (42.5 x 10-4)1/3 (42.5)1/3 = (log 42.5)/3 = (1.6284)/3 = 0.5428 = antilog(0.5428) = 3.489 3.49] -4 S 3.49 x 10 mol/L Where S= Solubility of CaF2 Now, molecular weight of CaF2 = Ca + 2(F) = 40 + 2(19) = 78 g/mol Now, molar Solubility = Wt.(gm) x 1 Mol. Wt. (gm/mole) x v(in litre) Solubility of CaF2 = Wt. of CaF2(gm) x m.w of CaF2 (gm/mole) Substituting the values in above reaction W= 10 mg = 10 x 10-3 gm 3.49 x 10-4 = 10 x 10-3 x 1000 78 x V(ml) V(ml) = 10 x 1 78 x 3.49 x 10-4 V(ml) = 367.34 ml 1000 V(ml) 41. Ksp of CaF2 at 298 K temperature is 1.7 x 10-10. If a person drinks 2.5 litre of water saturated with this salt every day, how much CaF2 will enter in his body every day? (Molecular mass of CaCl2 =78 g mol-) Ans. Suppose, Solubility of CaF2 be S mol/L CaF2 Ca2+(aq) + 2F-(aq) S 2S Ksp = [Ca2+][F-]2 = s[2s]2 = 4S3 Now, Ksp(given) = 1.7 x 10-10 = 4S3 S3 = 0.425 x 10-10 = 42.5 x 10-12 S = (42.5)1/3 x 10-4 mol/L S = 3.489 x 10-4 mol/L 3.49 x 10-4 mol/L [ S = (42.5)1/3 = (log(42.5))/3 = (1.6248)/3 = 0.5428 = antilog(0.5428) = 3.489 ] Now, person drinks 2.5 litre of water saturated with CaF2 salt every day V = 2.5 L Molecular mass = 7.8g/mol S = 3.49 x 10-4 mol/litre Solubility Product (s) = Wt (gm) x 1 M.W(gm/mole) x V(L) -4 3.49 x 10 = W(gm) x 1 78 x 2.5 W(gm) = (3.49 x 10-4) x 78 x 2.5 = 680.55 x 10-4 = 0.068055 gm W(gm) = 0.068055 gm = 6.8055 x 10-2 gm 42. pH of saturated aqueous solution of Ca(OH)2 is 12.25. Find Ksp of Ca(OH)2 Ans. Suppose, the solubility product of Ca(OH)2 be S mol/litre Equilibrium state Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) S 2S Now, pH = 12.25 PH + pOH = 14 pOH = 14 - pH = 14 – 12.25 = 1.75 OH Now, p = -log10[OH-] = 1.75 -log10[OH-] = 1.75 Shifting ‘mius sign’ to the right side and taking ‘antilog’ on both sides [OH-] = -antilog(1.75) = antilog(2.25) = 1.778 x 10-2 [OH ] = 1.778 x 10-2 M But, according to reaction [OH-] = 2S = 1.778 x 10-2M ….(i) S = 0.889 x 10-2 M …. (ii) Now, Ksp = [Ca2+][OH-]2 = [S][2S]2 (from reaction) = 4S3 = 4(0.889 x 10-2)3 [from equation (i)] Ksp = 2.810 x 10-6 M3 43. What will be the value of Ksp if the concentration in saturated solution of Mg(OH)2 is 8.2 x 10-4 %W/V. (Molecular Mass of Mg(OH)2 = 58.3 g.mol-). Ans. 8.2 x 10-4% W/V means 100 ml saturated solution contains 8.2 x 10-4 gm Mg(OH)2 100 ml 8.2 x 10-4 gm Mg(OH)2 1000 ml(1L) (?) = 8.2 x 10-4 x 1000 100 = 8.2 x 10-3 gm/litre M.W. of Mg(OH)2 = 58.3 gm/mol Molarity of saturated solution of Mg(OH)2 = 8.2 x 10-3 gm/litre 58.3 gm/mole S = 1.4065 x 10-4 mol/litre Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) S 2S 2+ - 2 Now, Ksp = [Mg ][OH ] = [S][2S]2 = 4S3 = 4(1.4065 x 10-4)3 = 11.130 x 10-12 M3 Ksp = 1.113 x 10-11 M3 44. How many grams of Zn(OH)2 can be dissolved in 2 litres of 0.02 M NaOH solution ?ksp for Zn(OH)2 = 4.5 x 10-17. Ans. Suppose solubility of Zn(OH)2 is S mol/litre Zn(OH)2(s) Zn2+(aq) + 2OH-(aq) S 2S + NaOH(aq) Na (aq) + OH-(aq) 0.02M 0.02M 0.02M (given) So, total concentration of [OH ] in the solution = [OH-] from Zn(OH)2 + [OH-] from NaOH = 2S + 0.02 (But here, 0.02 >>> S, hence [OH-] = 0.02 M ) (because Zn(OH)2 is sparingly soluble) TotL [OH-] = 0.02 M Now, Ksp for Zn(OH)2 = [Zn2+][OH-]2 = [S][0.02]2 -4 Ksp = S x 4 x 10 But Ksp = 4.5 x 10-17(given) Ksp = 4.5 x 10-17 = S x 4 x 10-4 S = 4.5 x 10-17 4 x 10-4 S = 1.125 x 10-13 mol/litre Mol. Wt of Zn(OH)2 = Zn + 2(O) = 2(H) = 65 + 2(16) + 2(1) = 99 gm/mol Solubility of Zn(OH)2 in o.o2M NaOH solution = (1.125 x 10-13) x 99 gm/litre = 1.11375 x 10-11 grams/litre But, given solution is of 2 litres Solubility = 1.11375 x 10-11(gm/L) x 2 L Solubility of Zn(OH)2 = 2.2275 x 10-11 gm 45. Solubility product of Mg(OH)2 is 1.2 x 10-11. Calculate its solubility in pure water as well as in aqueous solution of 0.05 M NaOH Ans. Suppose, solubility of Mg(OH)2 be S mol/L Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) S 2S NaOH(aq) Na+(aq) + OH-(aq) 0.05M 0.05M 0.05M (a) Solubility in pure water Ksp = [Mg2+][OH-]2 = [S][2S]2 = 4S3 But, Ksp = 1.2 x 10-11 (given) 4S3 = 1.2 x 10-11 S3 = 0.3 x 10-4 = 3 x 10-12 S = (3)1/3 x 10-4 S= 1.442 x 10-4 mol/litre [ (3)1/3 = (log3)/3 = (0.4771)/3 = 0.1590 = antilog(0.1590) = 1,442] S = 1.442 x 10-4 mol/litre……(a) (In pure water) (b) Solubility in 0.05M NaOH Total concentration of [OH-] = 2S + 0.05 [ But, here 0.05 >>> S as Mg(OH)2 is sparingly soluble) [OH-] = 0.05 M Concentration of Mg2+ = [S] = ? Ksp = [Mg2+][OH-]2 -11 Ksp = 1.2 x 10 = [S][0.05]2 = 1.2 X 10-11 = S x 25 x 10-4 S = 1.2 x 10-11 = 0.048 x 10-7 = 4.8 x 10-9 25 x 10-4 S = 4.8 x 10-9 mol/litre = 4.8 x 10-19 M 46. At 298 K temperature, 2.901 litre saturated solution can be prepared by dissolving 0.08 gram CaF2. Calculate Ksp of salt. (Molecular mass of CaF2 = 78.0 g.mol-) Ans. W = 0.08 gm Molecular wt = 78 gm/mol V(L) = 2.901 litre Now, concentration = 3.5354 x 10-4 M Now, CaF2 Ca2+ + 2F-4 (3.53 x 10 ) (2 x 3.53 x 10-4) Now, Ksp =[Ca2+][F-]2 = (3.53 x 10-4)(2 x 3.53 x 10-4)2 = 4 x (3.53 x 10-4)3 = 4 x 44.191 x 10-12 M3 = 176.767 x 10-12 M3 = 1.7676 x 10-10 M3 Ksp = 1.768 x 10-10 M3 47. Equal volumes of 2 x 10-4 M BaCl2 and 5.0 x 10-3 M H2SO4 are mixed. Will precipitation occur or not? (Ksp of BaSO4 = 1.1 x 10-10) Ans, If equal volumes of solutions are mixed, concentration of each salt will be halved [C1V1 = C2V2 But if V2 = 2V1 C1V1 = C2(2V1) C2 = (C1)/2] Concentration of BaCl2 in the mixture = 2 x 10-4 M = 1 x 10-4 M ….(i) 2 Concnetration of H2SO4 in the mixture = (5 x 10-3)/2 M = 2.5 x 10-3 M …(ii) BaCl2 Ba2+ + 2Cl-4 -4 (1 x 10 ) 1 x 10 M (2 x 1 x 10-4) M H2SO4 2H+ + SO4-2 -3 -3 (2.5 x 10 )M (2 x 2.5 x 10 )M 2.5 x 10-3 M When, they are mixed, salt formed is BaSO4 BaCl2 + H2SO4 BaSO4 + 2HCl BaSO4(aq) Ba2+(aq) + SO42-(aq) Ionic Product of BaSO4, Ip = [Ba2+][SO42-] = (1 x 10-4)(2.5 x 10-3) -7 2 Ip = 2.5 x 10 M But, Ksp = 1.1 x 10-10(given) Comparing above both values Ip > Ksp Precipitation will occur 48. By addition of how many grams of FeSO4 to 500 ml solution of 0.02 M NaOH, Fe(OH)2 will be precipitated? (Ksp of Fe(OH)2 = 1.5 x 10-15 ) Ans. NaOH Na+ + OH0.02M 0.02M Fe(OH)2 Fe+2(aq) + 2OH-(aq) S S 2S [Suppose, molar solubility of Fe(OH)2 be S mol/L Now, Fe(OH)2 is sparingly soluble salt S <<< 0.02 M [OH-] = 0.02 M Now, Ksp = [Fe2+][OH-]2 …(i) But Ksp = 1.5 x 10-15 (given) …..(ii) For, precipitation to occur, Ip > Ksp …(iii) From equation (i), (ii), (iii) Ip = Ksp = 1.5 x 10-15 = [Fe2+][OH-]2 S[0.02]2 -15 1.5 x 10 = S x 4 x 10-4 S = 3.75 x 10-12 M S = [Fe2+] = 3.75 x 10-12 M Now, l mole [Fe2+] = 1 mol FeSO4 [ because FeSO4 Fe2+(aq) + SO42-(aq)] 2+ -12 [Fe ] = [FeSO4] = 3.75 x 10 mol/litre For l litre, FeSO4 = 3.75 x 10-12 moles Hence for 500 ml = (3.75 x 10-12)/2 = 1.875 x 10-12 moles/500 ml Mol. Wt of FeSO4 = Fe + S + 4(O) = 56 + 32 + 4(16) = 152 gm/mol Wt of FeSO4 = 1.875 x 10-12 x 152 = 285 x 10-12 = 2.85 x 10-10 gm For precipitation, Ip > Ksp Wt (FeSO4) > 2.85 x 10-10 gm. Thus, if FeSO4 added in slightly greater than 2.85 x 10-10 gm, Ip > Ksp and Fe(OH)2 will precipitate. 49. Will precipitates of PbI2 be obtained by mixing 20 ml 3 x 10-3 M pb(NO3)2 and 80 ml. 2 x 10-3 M NaI Solutions ? (Ksp for PbI2 = 6.0 x 10-9) Ans. Pb(NO3)2(aq) Pb2+(aq) + 2NO3-(aq) -3 3 x 10 M 3 x 10-3M 2(3 x 10-3M) [Pb2+] = 3 x 10-3 M….(i) NaI(aq) Na+(aq) + I-(aq) -3 -3 2 x 10 M 2 x 10 M 2 x 10-3M [I-] = 2 x 10-3 M….(ii) 20 ml Pb(NO3)2 and 80 ml NaI solution is mixed. Total volume, V2 = (20 + 80) ml = 100 ml V2 = 100 ml Now, concentrations of Pb2+ and I- ions in mixture according to M1V1 = M2V2. For Pb2+ M1 = 3 x 10-3M V1 = 20 ml V2 = 100 ml. M1V1 = M2V2 M2 = M1 x V1 V2 M2 = (3 x 10-3) x 20 = 6 x 10-4 100 [Pb2+] 6 x 10-4 M…(iii) For IM1 = 2 X 10-3 M V1 = 80 ml V2 = 100 ml M1V1 = M2V2 M2 = M1 x V1 V2 = (2 x 10-3) x 80 = 1.6 x 10-3 M 100 [I-] = 1.6 x 10-3 M….(iv) Now, Ksp = 6 x 10-4 (given)…(v) For PbI2, Ip = [Pb2+][I-]2 [ PbI2(S) Pb2+(aq) + 2I-(aq)] From equation (iii) and (iv) Ip = (6 x 10-4) x (1.6 x 10-3)2 Ip = 15.36 x 10-10 Ip = 1.536 x 10-9 ….(vi) From equation (v) and (vi) Ip(1.536 x 10-9) < Ksp(6.0 x 10-9) Ip < Ksp PbI2 will not precipitate 50. The concentration of F- in a sample of water is 2 x 10-5 M. How many minimum grams of solid CaCl2 will have to be added for precipitation of F-?( Ksp for CaF2 = 1.7 x 10-10, molecular mass of CaCl2 = 111 grammolAns. [F-] = 2 x 10-5 M CaCl2 Ca2+(aq) + 2Cl-(aq) Suppose, x mole of CaCl2 should be added to 1 litre of water concenetration of Ca2+ in water will be x M [Ca2+] = x M CaF2(aq) Ca2+(aq) + 2F-(aq) xM (3 x 10-5) M -10 Ksp = 1.7 x 10 (given) Now, Ksp = [Ca2+][F-]2 = (X)(3 x 10-5M)2 1.7 x 10-10 = X x 9 x 10-10 X = (1.7)/9 = 0.1888 M X = 0.189 M …..(i) Now, 1 mol of Ca2+ = 1 mole of CaF2 = 1 mole of CaCl2 1 mole of CaF2 = 1 mole of CaCl2. 0.189 moles of CaF2 = 0.189 moles of CaCl2 [from equation (i)] If little more than 0.189 moles of CaCl2 is added i.e 0.20 moles, then CaF2 will precipitate. moles = 0.20 Molar mass of CaCl2 = 111 gm/mol 0.20 mol of CaCl2 will contain = 111 gm/mol x 0.20 = 22.2 gm solid CaCl2 22.2 gms solid CaCl2 is added for precipitation of F- (minimum value)