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CHEMISTRY CET - 2014 C-1 ANSWERS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 , 1.78 g of an optically active L-amino acid (A) is treated with NaNO2/HCl at 0 °C. 448 cm3 of nitrogen was at STP is evolved. A sample of protein has 0.25% of this amino acid by mass. The molar mass of the protein is 1) 34,500 g mol–1 2) 35,600 g mol–1 3) 36,500 g mol–1 4) 35,400 g mol–1 Ans (2) Solution : Mass of L amino acid = 178g R CH COOH NH2 + HNO2 1 mol. mass ? 448 cm3 1.78g 22400 ? N2 1 mole of N2 = 22400 cm3 at STP 448 22400 1.78 102 178 89 448 2 Sample of proteins contains 0.25% amino acid Protein amino acid 100 0.25 ? 89 100 89 35600g / mol 0.25 2. 10 g of a mixture of BaO and CaO requires 100 cm3 of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately (Given : molar mass of BaO = 153) 1) 55.1 2) 47.4 3) 52.6 4) 44.9 Ans (3) Solution : Consider the mass % of CaO is 55.1% as given in option 1 Then mass of CaO = 5.51 and BaO = 4.49 5.51 28 4.49 No. of g and of BaO = 76.5 No. of g. eq of CaO = The sum of gram equivalents should be equal to 100 2.5 0.25 1000 For the option (3) which is correct. Mass of CaO = 5.26g and mass of BaO= 4.74g i.e. No. of gram equivalent of CaO = No. of gram equivalent BaO = 5.26 0.1878 28 4.74 0.0619 76.5 Total no. of gram equivalent = 0.1878 + 0.0619 = 0.25 1 3. The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from kg of coke is (Assume coke to be 100% carbon.) (Given enthalpies of combustion of CO2, CO and H2 as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K.) 1) 0.69: 1 2) 0.96: 1 3) 0.79 : 1 4) 0.86 : 1 Ans ( ) Solution : Information given is wrong hence could be consider for grace mark. 4. Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g. Percentage of iron in impure copper is (Given molar mass Fe = 55.5 g mol–1 , molar mass Cu = 63.54 g mol–1 ) 1) 0.85 2) 0.90 3) 0.95 4) 0.97 Ans (2) Solution : No. of gram equivalents of copper deposited 22.011 0.6928 31.77 Same should be the no. of gram equivalents from the anode. No. of gram equivalents from the current = 140 482.5 Q = I t = 140 482.5 = 67550C No. of gram equivalents = 675500 0.7 96500 Since only Cu and Fe are dissolved from the anode no. of g. equivalents of Fe = 0.7 – 0.6928 = 0.0072 Mass of Fe = 0.0072 27.75 = 0.1998 g % Fe = 5. Mass of Fe 0.1998 100 0.89 0.90% Mass of impurities 22.26 25 cm3 of oxalic acid completely neutralised 0.054 g of sodium hydroxide. Molarity of the oxalic acid solution is 1) 0.045 2) 0.032 3) 0.064 4) 0.015 Ans (2) Solution : V N 0.064 25 N 0.064 1000 40 1000 40 N = 0.064 But we know for oxalic acid = Normality 1 = Molarity = 0.032 2 6. The statement that is NOT correct is 1) Energies of stationary states in hydrogen like atoms is inversely proportional to the square of the principal quantum number. 2) The radius of the first orbit of He is half that of the first orbit of hydrogen atom. 3) Angular quantum number signifies the shape of the orbital. 4) Total number of nodes for 3s orbital is three. Ans (4) Solution : Total no. of nodes is (n – 1) = 3 – 1 = 2 7. For the equilibrium : CaCO3(s) CaO(s) + CO2(g) ; Kp = 1.64 atm at 1000 K 50 g of CaCO3 in a 10 litre closed vessel is heated to 1000 K. Percentage of CaCO3 that remains unreacted at equilibrium is (Given R = 0.082 L atm K–1 mol–1) 1) 50 2) 20 3) 40 (4) 60 Ans (4) Solution : CaCO3 CaO + CO2 Kp = 1.64 50g = 0.5 moles K P PCO PV = NRT 1.64 10 = n 0.082 1000 n = 0.2 no. of moles of CO2 formed is 0.2 No. of moles of unreacted CaCO3 = 0.5 – 0.2 = 0.3 2 Percentage of unreacted CaCO3 = 8. 0.3 100 60% 0.5 Conversion of oxygen into ozone is non-spontaneous at 1) high temperature 2) low temperature 3) all temperatures 4) room temperature Ans (3) Solution : 302 203 Here H = + ve S = – ve G = H – TS = + ve at all temperatures. 2 9. Density of carbon monoxide is maximum at 1) 0.5 atm and 273 K 2) 4 atm and 500 K 3) 2 atm and 600 K 4) 6 atm and 1092 K Solution : PV = RT V= Ans (2) RT P Since no. of moles remain constant. V T P Lower the value of T T higher is the density. Option (2) has lowest value. P P 10. The acid strength of active methylene group in (a) CH3COCH2COOC2H5 (b) CH3COCH2COCH3 (c) C2H5OOCCH2COOC2H5 decreases as 1) a > b > c 2) c > a > b 3) a > c > b 4) b > a > c Solution : Electron donating group make the hydrogen active. Ester group is a lesser electron withdrawing group than Carbonyl group. Ans (4) 11. A metallic oxide reacts with water to form its hydroxide, hydrogen peroxide and also liberates oxygen. The metallic oxide could be 1) KO2 (2) Na2O2 3) CaO 4) Li20 Ans (1) 12. Ozonolysis X Y Z (Re ductive) Y can be obtained by Etard' s reaction, Z undergoes disproportionation reaction with concentrated alkali. X could be Ans (1) 3 Solution : CH O3 CH2 CHO + HCHO Obtained from Etard’s reaction undergoes disproportionation reaction (Cannizzaro’s reaction) 13. Gold Sol is not 1) a lyophobic colloid 2) negatively charged colloid 3) a macro molecular 4) a multimolecular colloid Solution : Glucose, protein are macromolecular colloid. Carbocation as an intermediate is likely to be formed in the reaction : OH acetonecyanohydrin 1) Acetone + HCN Anhy. AlCl / HCl 2) Hexane 2 methyl pentane hv 2 chloropropane 3) Propene + Cl2 ethyl alcohol 4) Ethylbromide + Aq KOH Solution : Pyrolysis Ans (3) 14. 3 15. 16. For an ideal binary liquid mixture 1) H(mix) = 0 ; S(mix) < 0 3) S(mix) =0 ; G(mix) = 0 Ans (2) 2) S(mix) > 0 ; G(mix) < 0 4) V(mix) = 0 ; G(mix) > 0 Ans (2) For hydrogen — oxygen fuel cell at one atm and 298 K H2(g) + 1 O2(g) H2O(l) ; G° = —240 kJ 2 E° for the cell is approximately, (Given F = 96,500 C) 1) 1.24 V 2) 1.26 V 3) 2.48 V o o Solution : G = – nFE – 240 = – 2 96500 Eo Eo = + 1.24 V 4) 2.5 V Which one of these is not known ? 1) CuI2 2) CuBr2 3) CuC12 4) CuF2 – +2 + Solution : Here I readily transfers electron to Cu and hence Cu and I2 are formed. Ans (1) 17. 18. 19. The correct statement is 1) The extent of actinoid contraction is almost the same as lanthanoid contraction. 2) Ce+4 in aqueous solution is not known. 3) The earlier members of lanthanoid series resemble calcium in their chemical properties. 4) In general, lanthanoids and actinoids do not show variable oxidation states. 1.CH 3 MgBr P R 2.H O 3 1) ethanamine Ans (1) Ans (1) 1. dil . NaOH 4—methylpent-3—en-2—one P is 2. 2) ethanal 3) propanone 4) ethanenitrile Ans (4) 20. When CH2 = CH — O — CH2 — CH3 reacts with one mole of HI, one of the products formed is 1) ethanol 2) ethanal 3) ethane 4) iodoethene Ans (2) Solution : Vinyl alcohol formed undergoes tautomerisation to form ethanal 4 21. 0.44 g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at S.T.P., 112 cm3 of methane. With PCC the same alcohol forms a carbonyl compound that answers silver mirror test. The monohydric alcohol is 1) (CH3)3C — CH2OH 2) (CH3)2CH — CH2OH 3) (4) Ans (1) Solution : 0.44g of Monohydric alcohol liberates 112 cm3 of CH4 I R OH + CH3MgI CH4 + MG 1 mole OR 1 mole Mass of Monohydric alcohol which given 22400 cm3 of CH4 22400 0.44 88 112 Among the alcohols given option 3 and option 4 cannot be the answer as with PCC they give ketoses which cannot give silver mirror test. Among option (1) and (2) only option (1) has molecular mass 88. 22. The IUPAC name of '13' is 1) 2—methylbutan-3—ol 3) 3—methylbutan-2—ol 23. 2) Pentan-2—ol 4) 2—methylbutan-2—ol For Freundlich isotherm a graph of log Ans (3) x is plotted against log P. The slope of the line and its y-axis m intercept, respectively corresponds to 1) log 1 ,k n 2) log Solution : Freundlich isotherm 1 , log k n 3) 1 ,k n 1 x Pn m 1 x kP n m x 1 log log k log P m n 24. A plot of 4) 4 x m 1 , log k n Ans (4) 1 n } log k log P 1 Vs. k for a reaction gives the slope —1 x 104 K. The energy of activation for the reaction is T (Given R = 8.314 J K–1 mol–1) 1) 1.202 kJ mol–1 2) 83.14 kJ mol–1 3) 8314 J mol–1 4) 12.02 J mol–1 Ans () Solution : Wrong question since the plot is of logk vs 1/T In the the question the plot given is of k not log k Slope = Ea 2.303R 1 104 2.303 8.314 = Ea 25. 26. Ea = 1.9kJ The IUPAC name of the complex ion formed when gold dissolves in aquaregia is 1) tetrachloridoaurate(I) 2) dichloridoaurate(III) 3) tetrachloridoaurate(III) 4) tetrachloridoaurate(II) Solution : Product formed is [AuCl4]The correct sequence of reactions to be performed to convert benzene into m-bromoaniline is 1) bromination, nitration, reduction 2) reduction, nitration, bromination 3) nitration, reduction, bromination 4) nitration, bromination, reduction Ans (3) Ans (4) 5 Solution : NO 2 NO 2 Br2 + HNO 3 NH2 Sn/HCl Br Br 27. 1) 2) 3) 4) Ans (3) Solution : O OH O C + C 6 H5 COCl Activated Benzene nitration Deactivated benzene O O C NO 2 28. A (g) P(g) Q (g) R (g) follows first order kinetics with a half life of 69.3 s at 500 °C. Starting from the gas 'A' enclosed in a container at 500 °C and at a pressure of 0.4 atm, the total pressure of the system after 230 s will be 1) 1.32 atm 2) 1.12 atm 3) 1.15 atm 4) 1.22 atm Ans (2) Solution : A (g ) P(g) +Q (g) + R (g) Half life = 69.3s k= 0.693 0.693 102 s 1 t1 693 2 [R]o 2.303 log t [R]t 2.303 0.4 102 log 230 Pt 0.4 0.4 1 = log = 10 Pt Pt 0.4 Pt 0.04 10 k= A P(g) + Q(g) + R(g) 0.4 – x = 0.04 0.36 0.36 0.36 x = 0.36 Pressure = 0.04 + 0.36 + 0.36 + 0.36 = 1.12 6 29. MnO2 + HCl A (g) 573 A (g) F2(excess) B(g) B(l) + U(S) C(g) + D(g) The gases A, B, C and D are respectively 1) Cl2, ClF3, UF6, ClF 2) O2, O2F2, U2O3, OF2 3) Cl2, ClF, UF6, ClF3 4) O2, OF2, U2O3, O2F2 Ans (1) Solution : MnO2 + HCl MnCl2 + Cl2 + H2O 30. 31. Acetophenone cannot be prepared easily starting from 1) C6H5CH3 2) C6H6 3) C6H5CH(OH)CH3 4) C6H5C CH Ans (1) One mole of ammonia was completely absorbed in one litre solution each of (a) 1M HCl, (b) 1M CH3COOH and (c) 1M H2SO4 at 298 K. The decreasing order for the pH of the resulting solutions is (Given Kb(NH3) = 4.74) 1) a > b > c 2) c > b > a 3) b > c > a 4) b > a > c Ans (4) Solution : NH3 + HCl NH4Cl 1M NH3 completely reacts with 1M HCl. However the salt formed is acidic salt pH is less than 7 NH3 + CH3COOH CH3COONH4 1 mole of NH3 completely reacts with 1 mole of CH3COOH. The salt formed in neutral salt and the pH is very close to 7 2 NH3 + H2SO4 (NH4)2SO4 1 mole of ammonia requires 0.5 mole of H2SO4 and hence H2SO4 remain in the solution in the solution and hence the pH is very less decreasing order of pH is b > a > c 32. 5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately 1) 1 10–5 2) 1 10–4 3) 1 10–6 4) 1 10–3 Ans (2) 5.5 103 Solution : 5.5 mg N2 = moles = 2.36 10–4 moles 28 p N2 k H p N2 n2 n1 2.36 104 10 10 104 kH 4.2 104 2.36 1 kH n2 n1 2.36 104 (1) 10 n 5 2 (2) 10 n 2 / 10 (2) 5 (1) 1 2.36 104 10 n2 5 10 10 104 10 1 50 cm3 of 0.04 M K2Cr2O7 in acidic medium oxidizes a sample of H2S gas to sulphur. Volume of 0.03 M KMnO4 required to oxidize the same amount of H2S gas to sulphur, in acidic medium is 1) 80 cm3 2) 120 cm3 3) 60 cm3 4) 90 cm3 Solution : K2Cl2O4 KMnO4 33. Ans (1) V N V N 1000 1000 50 0.24 V 0.15 50 0.24 80 cm 3 1000 1000 0.15 34. The compound that reacts the fastest with sodium methoxide is 1) 2) 3) 4) Ans (3) 7 Solution : Higher the no. of electron withdrawing groups higher is the reactions • Increasing the number of electron-withdrawing groups increases the reactivity of the aryl halide. Electron-withdrawing groups stabilize the intermediate carbanion, and by the Hammond postulate, lower the energy of the transition state that forms it. 35. The pair of compounds having identical shapes for their molecules is 1) BCl2, ClF3 2) SO2, CO, 3) CH4, SF4 4) XeF2, ZnCl2 Ans (4) Solution : ZnCl2 linear shape in gaseous state Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85 10–5 m–1 Solubility product of the salt AB at 298 K is Given om (AB) = 140 10– 4 S m2 mo1–1 (1) 1.32 10–12 2) 1.74 10–12 3) 5.7 10–12 4) 7.5 10–12 o 4 2 Solution : A m 140 10 Sm K = 1.85 10–5 36. m Ans (2) K 1.85 105 1.85 105 1.85 10 5 140 104 0.132 10 5 1.32 10 6 1000C 1000 2s 1000 2 140 10 4 28 Ks = s2 = (1.32 10–6)2 = 1.74 10–12 37. An incorrect statement with respect to SN1 and SN2 mechanisms for alkyl halide is (1) Competing reaction for an SN2 reaction is rearrangement. (2) A weak nucleophile and a protic solvent increases the rate or favours SN1 reaction. (3) A strong nucleophile in an aprotic solvent increases the rate or favours SN2 reaction. (4) SN1 reactions can be catalysed by some Lewis acids. Butylated hydroxy toluene as a food additive acts as 1) flavouring agent 2) emulsifier 3) antioxidant Solution : BHT Ans (1) 38. Terylene is NOT a 1) polyester fibre 3) copolymer Solution : Chain growth polymer 4) colouring agent Ans (3) 39. 40. 2) step growth polymer 4) chain growth polymer Ans (4) The correct statement is 1) One mole each of benzene and hydrogen when reacted gives 1/3 mole of cyclohexane and 2/3 mole unreacted hydrogen. 2) It is easier to hydrogenate benzene when compared to cyclohexene. 3) Cyclohexadiene and cyclohexene cannot be isolated with ease during controlled hydrogenation of benzene. 4) Hydrogenation of benzene to cyclohexane is an endothermic process. Ans (3) 41. Among the elements from atomic number 1 to 36, the number of elements which have an unpaired electron in their s subshell is 1) 7 2) 9 3) 4 4) 6 Ans (4) Solution : H, Li, Na, K, Cr and Cu 42. The statement that is NOT correct is 1) Van der Waals constant 'a' measures extent of intermolecular attractive forces for real gases. (2) Boyle point depends on the nature of real gas. (3) Compressibility factor measures the deviation of real gas from ideal behaviour. (4) Critical temperature is the lowest temperature at which liquefaction of a gas first occurs. Ans (4) 43. The correct arrangement for the ions in the increasing order of their radii is 1) Ca+2 , K+, S-2 2) Cl–, F–, S–2 3) Na+, Cl–, Ca+2 4) Na+, Al+3, Be+2 Ans (1) 8 44. The correct arrangement of the species in the decreasing order of the bond length between carbon and oxygen in them is 1) CO2 ,HCO2 ,CO,CO32 2) CO,CO32 ,CO2 ,HCO2 3) CO,CO2 ,HCO2 ,CO32 4) CO32 ,HCO2 ,CO2 ,CO Ans (4) Solution : Bond order 45. 1 Bond length The species that is not hydrolysed in water is 1) BaO2 2) CaC2 3) P4 O10 (4) Mg3N2 Ans (1) 46. For the properties mentioned, the correct trend for the differen: ies is in 1) inert pair effect - Al > Ga > In 2) first ionization enthalpy - B > Al > Tl 3) strength as Lewis acid - BCl3 > AlCl3 > GaCl3 4) oxidising property – Al+3 > In+3 > Tl+3 Ans (3) Solution : Lewis acid strength decreases down the group First ionization enthalpy is in the order B > Tl > Al 801 589 577 This is due to poor shielding by f orbitals in the case of T Tl Due to the inert pair effect Tl+3 is unstable and hence easily gets reduced to Tl+1. Therefore Tl+3is a good oxidizing agent 47. A correct statement is 1) [MnBr4]-2 is tetrahedral. 3) [Co(NH3)6]+2 is paramagnetic. 2) [Ni(NH3)6]+2 is an inner orbital complex. 4) [CoBr2(en)2] – exhibits linkage isomerism. Ans (1, 3) Solution : [MnBr4]– 2 is tetrahedral Br – weak ligand does not allow spin pairing. [Mn Br4]–2; x – 4 = – 2 x = + 2 48. Iodoform reaction is answered by all, except 1) CH3CHO 3) 2) CH3 — CH2 — CH2OH 4) CH3 — CH2 — OH Ans (2) Solution : O || Does not have CH3 – C – group 49. A crystalline solid XY3 has ccp arrangement for its element Y. X occupies 1) 33% of tetrahedral voids 2) 33% of octahedral voids 3) 66% of tetrahedral voids 4) 66% of octahedral voids Ans (2) 50. ‘R’ is 1) sulphanilamide 3) o-bromo bromo sulphanilic acid 51. 52. 2) p – bromo sullphanilamide 4) sulphanilic acid Ans (4) The statement that is NOT correct is 1) Carbohydrates are optically active. 2) Lactosee has glycosidic linkage between C4 of glucose lucose and Cl of galactose unit 3) Aldose or ketose sugars in alkaline medium do not isomerise. 4) Penta acetate of glucose does not react with hydroxylamine. Match the reactant in Column - I with the reaction in Column - II : I II (i) Acetic acid (a) Stephen (ii) Sodium phenate (b) Friedel-Crafts (iii) Methyl cyanide (c) HVZ (iv) Toluene (d) Kolbe's (1) i d, b, - c, iv - a (2) i - c, - d, 7 a, iv – b (3) i - c, - a, iii - d, iv – b Ans (3) (4) i - b, c, - a, iv – d Ans (2) 9 53. The statement that is NOT correct is 1) In solid state PCl5 exists as [PCl4]+ [PCl6]– 2) Phosphorous acid on heating disproportionates to give metaphosphoric acid and phosphine. 3) Hypophosphorous acid reduces silver nitrate to silver. 4) Pure phosphine is non-inflammable. Ans (4) 54. In which one of the pairs of ion given, there is an ion that forms a co-ordination compound with both aqueous sodium hydroxide and ammonia and another ion that forms a co-ordination compound only with aqueous sodium hydroxide ? 1) Zn+2, Al+3 2) Al+3, Cu+2 3) Pb+2, Cu+2 4) Cu+2, Zn+2 Ans (1) 55. A crystalline solid X reacts with dil. HCl to liberate a gas Y. Y decolourises acidified KMnO4. When a gas 'Z' is slowly passed into an aqueous solution of Y, colloidal sulphur is obtained. X and Z could be, respectively 1) Na2SO4, H2S 2) Na2SO4, SO2 3) Na2S, SO3 4) Na2SO3, H2S Ans (4) 56. An aromatic compound 'A' (C7H9N) on reacting with NaNO2/HCl/ at 0 °C forms benzyl alcohol and nitrogen gas. The number of isomers possible for the compound 'A' is 1) 7 2) 6 3) 5 4) 3 Ans (3) 57. The statement that is NOT correct is 1) Collectors enhance the wettability of mineral particles during froth flotation. 2) Copper from its low grade ores is extracted by hydrometallurgy. 3) A furnace lined with Haematite is used to convert cast iron to wrought iron. 4) In vapour phase refining, metal should form a volatile compound. Ans (1) 58. A solution of 1.25 g of `P' in 50 g of water lowers freezing point by 0.3 °C. Molar mass of ‘P’ is 94.Kf(water) = 1.86 K kg mol–1 . The degree of association of `P' in water is 1) 60 % 2) 75 % 3) 80 % 4) 65 % Ans (1) 59. Volume occupied by single CsCl ion pair in a crystal is 7.014 10-23 cm3 . The smallest Cs - Cs internuclear distance is equal to length of the side of the cube corresponding to volume of one CsCl ion pair. The smallest Cs to Cs internuclear distance is nearly 0 0 1) 4.3 A 2) 4.5 A 0 3) 4.4 A 0 4) 4 A Ans (*) Solution : Since the answer is 4.125 none of the options can be given as answer For Cr2O72 + 14H+ + 6e– 2Cr+3 + 7H2O ; E° = 1.33 V At Cr2 O 72 = 4.5 millimole, [Cr+3] = 15 millimole, E is 1.067 V. The pH of the solution is nearly equal to 1) 3 2) 4 3) 2 4) 5 Ans (3) Solution : Cr2 O 72 14H 6e 2Cr 3 7H 2 O 60. E = E + 0.0591 [Cr2 O72 ][H ]14 log 6 [Cr 3 ]2 1.067 = 1.33 + 0.0591 [Cr2 O42 ] log[H ]14 log 3 2 6 [Cr ] 0.0591 [Cr2 O 22 log log[H ]14 1.067 – 1.33 = 3 2 6 [Cr ] – 0.263 = 26.3 = log 1 4.5 103 log 14 pH 3 2 100 (15 10 ) 1 + 14 pH = –1.3010 + 14 pH 20 28 = 14 pH pH = 28 2 14 Boscoss, Gopal Nivas Compound, Vriddashram Road, Near Sharada Vidyalaya, Mangalore-3, Ph: 0824-4272728, 9972458537/CET-2014 10