DPP-49 to 50 PC Physical Chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 49 to 50 Class : XIII Course : DPP No.49 DPP No.1 Total Marks : 37 Max. Time : 37 min. Single choice Objective ('–1' negative marking) Q.1 to Q.4 (3 marks 3 min.) [12, 12] Subjective Questions ('–1' negative marking) Q.5 to Q.8 (4 marks 4 min.) [16, 16] Comprehension ('–1' negative marking) Q.9 to Q.11 (3 marks 3 min.) [9, 9] 1. Blood is buffered with CO2 and HCO –. What is the ratio of the base concentration to the acid (i.e. CO (aq.) plus H2CO3) concentration to maintain the pH of blood at 7.4 ? The first dissociation constant of H2CO3 (H2CO3 H+ + HCO –) is 4.2 × 10–7 where the H2CO3 is assumed to include CO2(aq.) i.e., dissolved CO2. ( log 2 = 0.3, log 3 = 0.48, log 7 = 0.85, antilog1.06 = 11.5) (1) 11.5 (2) 1.8 (3) 10 (4) 12 2. pH after the mixing of solutions of 1 M benzoic acid (pKa = 4.20) and 1M C6H5 COONa is 4.5, what is the volume of benzoic acid required to prepare a 300 ml buffer [log 2 = 0.3]? (1) 200 ml (2) 150 ml (3) 100 ml (4) 50 ml 3. 100 mL of 0.02 M benzoic acid (pKa = 4.2) is titrated using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are : (1) 3.50, 7 (2) 4.2, 7 (3) 4.2, 8.1 (4) 4.2, 8.25 4. The 'brown ring' formed at the junction of two layers in the test of nitrate is due to the formation of a complex ion, [Fe(H O) NO]2+. Which of the following statements are correct for this complex. [ = 3.87 B.M.] 2 5 (i) Oxidation state of Fe is +1 and NO exists as NO+. (ii) The complex ion is in octahedral geometry as attained by sp3d2 hybrdisation. (iii) The complex is paramagnetic and has three unpaired electrons due to transter of electron from NO to Fe2+. (iv) The complex is in octahedral geometry as attained by d2sp3 hybridisation. (v) The brown colour of the complex is attributed to d-d transition of electron. (1) (i), (ii) and (v) (2) (iii), (iv) and (v) (3) (i), (ii) and (iii) (4) (ii), (iii) and (v) 5. Calculate pH of the buffer solution containing 0.15 moles of NH4OH and 0.25 moles of NH4 Cl . Kb for NH4OH is 2  10 5 . 6. Determine the concentration of H O+ ion in a mixture of 0.06 M CH COOH & 0.04 M CH COONa at 25ºC, dissociation constant of CH COOH = 1.84  10 5 . 7. Calculate the pH of solution of given mixtures. [log (1.8) = 0.26] (1) 2 gm CH3 COOH + 4.1 gm CH3 COONa in 100 ml of mixture , Ka = 1.8  10 5 . (2) 5 ml of 0.1 M NH4 OH + 250 ml of 0.1 M NH4 Cl , Kb = 1.8  10 5 . 8. Calculate the moles of pyridinium chloride (C6H5NHCl) which should be added to 500 ml solution of 0.4 M pyridine (C H N) to obtain a buffer of pH = 5 , K for pyridine is 1.5  10 9 . 5 5 b Comprehension - (9 to 11) Valence bond theory for bonding in transition metal complexes was developed by Pauling. From the valence bond point of view formation of a complex involves reaction between Lewis bases (ligands) and a Lewis acid (metal) or metal ion with the formation of coordinate covalent (or dative) bonds between them. The model utilizes hybridisation of metal s, p and d valence orbitals to account for the observed structures and magnetic properties of complexes. Valence bond theory is able to deal satisfactorily with many stereochemical and magnetic properties but it has nothing to say about electronic spectra or the reason for the kinetic inertness of chromium (III) and low- spin cobalt (III) octahedral complexes. To understand this and more other features of transition metal we must turn to other theories like crystal field theory etc. Pure crystal field theory assumes that the only interaction between the metal ion and the ligands is an electrostatic or ionic one with the ligands being regarded as negative point charges. This theory is quite successful in interpreting many important properties of complexes. 9. Select the correct statement about the crystal filed theory. (1) Metal ligand bond in coordination compounds arises purely from electrostatic interaction between the metal ion and the ligand. (2) Metal-ligand bond in coordination compound is purely a covalent bond. (3) The pattern of splitting of the d-orbitals depends upon the nature of the crystal field. (4) (1) and (3) both. 10. Which of the following is correct for the complex [Ti (H O) ]3+ ? 2 6 (1) Hybridisation of central metal ion of the complex is sp3d2. (2) The complex is paramagnetic containing one unpaired electron. (3) Hybridisation of central metal ion of the complex is d2sp3. (4) Both (2) and (3). 11. The hybridisation of [NiCl2 (PPh3)2] and [NiCI2 (PMe3)2] are respectively (consider PPh3 a bulkier ligand than PMe3) : (1) sp3 and dsp2 (2) sp3 and sp3 (3) dsp2 and dsp2 (4) dsp2 and sp3 DPP No.50 DPP No.2 Total Marks : 35 Max. Time : 35 min. Single choice Objective ('–1' negative marking) Q.1 to Q.9 (3 marks 3 min.) [27, 27] Multiple choice objective ('–1' negative marking) Q.10 to Q.11 (4 marks 4 min.) [8, 8] 1. What information can be derived from the fact that when 1.5 mmol of NaOH are added to an aqueous solution of H3PO4 containing 1.0 mmol, the solution is found to be neutral? (1) K1 of H3PO4 = 10 (2) K of H PO = 10 (3) K of H PO = 10 (4) None of these –3 –7 –12 3 4 3 4 2. To 50 mL of 0.1 M H3PO4 is added 50 mL of 0.2 M NaOH. The pH of the resulting solution would be in the range (take equilibrium data from last question) (1) 4 to 5 (2) 6 to 7 (3) 9 to 10 (4) 13 to 14 3. What is the pH after addition of 40 mL of 0.1 M NaOH in the titration of 50 mL of 0.1 M anilinium chloride? ( Ka for anilinium ion = 2.5 x 10 ) -5 (1) 3.2 (2) 4.2 (3) 5.2 (4) 6.2 4. In a basic aqueous solution, chloromethane undergoes a substitution reaction is which Cl– is replaced by OH– : CH Cl (aq.) + OH– (aq.) CH OH (aq.) + Cl– (aq.) 3 3 Chloromethane Methanol The equilibrium constant KC is 1 × 10 . A solution is prepared by mixing equal volumes of 0.1 M CH Cl and 16 0.2 M NaOH. What will be the pH of the resulting solution. [log2 = 0.30] (1) 1.3 (2) 1 (3) 12.7 (4) 13 5. Which of the following indicators is best suited in the titration of a weak acid versus a strong base ? (1) phenolpthalein(8.310.0) (2) methyl orange (3.14.4) (3) methyl red (4.2  6.3) (4) litmus (4.5  8.3) 6. Which of the following indicators is best suited in the titration of a weak base versus a strong acid ? (1) phenolphthalein (8.3  10.0) (2) phenol red (6.8  8.4) (3) methyl organe (3.1  4.4) (4) litmus (4.5  8.3) 7. For the acid H2X, pK1 = 4 and pK2 = 10. Which of the following indicators( with their ranges provided) is most suitable for the titration H X + OH-  HX- + H O? 2 2 (1) Methyl orange ( 3.1 to 4.4 ) (2) Bromocresol green ( 3.8 to 5.4 ) (3) p-nitrophenol ( 5.6 to 7.6 ) (4) Phenolphthalein ( 8 to 9.6 ) 8. Amongst [Co(ox) ]3–, [CoF ]3– and [Co(NH ) ]3+ : 3 6 3 6 (1) [Co(ox) ]3– and [CoF ]3– are paramagnetic and [Co(NH ) ]3+ is diamagnetic. 3 6 3 6 (2) [Co(ox) ]3– and [Co(NH ) ]3+ are paramagnetic and [CoF ]3– is diamagnetic. 3 3 6 6 (3) [Co(ox) ]3– and [Co(NH ) ]3+ are diamagnetic and [CoF ]3– is paramagnetic. 3 3 6 6 (4) [Co(NH ) ]3+ and [CoF ]3– are paramagnetic and [Co(ox) ]3– is diamagnetic. 3 6 6 3 9. All the following complexes show decrease in their weights when placed in a magnetic balance then the group of complexes having tetrahedral geometry is : (i) Ni(CO)4 (ii) K[AgF4] (iii) Na2 [Zn(CN)4] (iv) K2 [PtCl4] (v) [RhCl(PPh3)3] (1) (ii), (iii), (v) (2) (i), (ii), (iii) (3) (i), (iii), (iv) (4) None of these 10. 0.1 M CH3COOH is diluted at 25°C (Ka = 1.8 × 10–5), then which of the following will be found correct (1) [H+] will increase (2) pH will increase (3) number of H+ will increase (4) all the above are correct 11. Which of the following pairs will not form a buffer solution (1) NaCl + HCl (2) CH3COOH + NH4Cl (3) Na2HPO4 + Na3PO4 (4) KNO3 + HNO3 ANSWER KEY DPP No.-47 1. 1 2. 3 3. 3 4. 1,3,4 5. 900 mL.6. [Sac–] = 4 × 10–12 M 7. [OH–] = [NH+4] = 3 × 10–4 M ; [C5H6N+] = 2.5 × 10–8 M. 8. C1 = 6 M. 9. (i) [OH–] = 9 × 10–5 M, [N H +] = 9 × 10–5 ; (ii) pH = 9.96, [N H 2+] = 9 × 10–16. 2 5 2 6 10. 3 11. 4 12. 3 DPP No.-48 1. (a) 4 ; (b) 4 2. (a) 2 ; (b) 3 3. [H+] = 8 × 10–5, [HCO –] = 5 × 10–5, [CO 2–] = 3 × 10–11. 3 3 4. (a) 1.66 %. ; (b) 8 x 10-5. 5. [ HS– ] = 4 × 10 -8 M ; [ S2– ] = 2.08 × 10 -20 M. 6. [S2] = 2.5 x 1015. 7. pH = 9, [OH¯ ] = 10–5. 8. 2.675 gm. 9. Kh (i) 5 × 10–10 pH 8.7 %Hydrolysis 0.01% (ii) 5 × 10-10 5.7 0.025% (iii) 1 0.0 13.68 95% (iv) 2 .5 × 10–5 11.60 0.625% 10. 8 11. 7