DPP-49 to 50 With Answer Physical Chemistry

DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date DPP No. 49 to 50 Class : XIII Course : 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] ANSWER KEY DPP No.-49 1. 1 2. 3 3. 3 4. 3 5. 9.08. 6. 2.76  10 5 mol Lt1. 7. (1) 4.92 (2) 7.56 8. 0.3 mole of C5H5NHCl should be added to 500 ml solution of C5H5N. 9. 4 10. 4 11. 1 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 ) jDr CO2rFkkHCO –ds lkFkcQj foy;ugSAjDrdhpH7.4ijO;ofLFkr j[kusds fy,{kkj dhlkanzrkovEy dh lkanzrk (i.e.CO2(tyh;)+H2CO3)dkvuqikr D;kgSA H2CO3dkizFke fo;kstufLFkjkad (H2CO3 H++HCO –) 4.2×10–7gSAtgkaH2CO3dksCO2(aq.)dslkFkekuktkrkgSAvFk Zrfoys;CO2 gSaA ( 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 Sol. [1] [B] [B] [B] 7.4 = (7 – log 4.2) + log [A]  log [A] = 1.06  [A] = 11.5 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]? 1M csatksbdvEy(pKa =4.20)rFkk1MC6H5COONa ds foy;udksfeykusijpH4.5çkIrgksrkgS]rks300mlcQj cukusads fy,vko';d csatksbdvEy dkvk;ru D;kgksxkA [log2=0.3] (1) 200 ml (2) 150 ml (3*) 100 ml (4) 50 ml Sol. pH = pKa + log V 300  V V  4.5 = 4.2 + log V V 300  V  0.3 = log V 300  V  log 2 = log 300  V  2 = 300  V  V = 200  Volume of the acid = 300 – 200 = 100. 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 0.02M csUtksbd vEy (pKa=4.2)ds 100mLdk vuqekiu 0.02M NaOH}kjk fd;k tkrk gSA Øe'k% 50mLrFkk 100 mLNaOHfeykusijfoy;udspHdkekugksaxsaA (1) 3.50, 7 (2) 4.2, 7 (3*) 4.2, 8.1 (4) 4.2, 8.25 Sol. When 50 ml of 0.02 M NaOH is added then we will have a buffer solution of pH [salt] pH = pKa + log [acid] = 4.2 + log 1 = 4.2 1 when 100 ml of NaOH is added we will have equivalence point so pH will be calculated according to hydrolysis of salt of WASB. pH = 1 {pK + PK + log C} 2 W a 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) ukbVªsVdsifj{k.kesa]nksijrksadsfeykufcUnq(atthejunction)ij]ladqyvk;u[Fe(H O) NO]2+dsfuekZ.kdsdkj.k] 2 5 ,d**Hkwjhoy;* dkfuekZ.kgksrkgSAblladqy[=3.87B.M.] dsfy,fuEuesalsdkSulkdFkulR;gS (i) FedhvkWDlhdj.kvoLFkk+1rFk NO,NO+ds:iesavfLrRoj[krkgSA (ii) ladqyvk;u sp3d2 ladj.k}kjk] v"VQydh;T;kfefrj[krkgSA (ii)ladqyvuqpqEcdh;gSrFk NOlsFe2+ dksbysDVªkWuLFk ukUrj.kdsdkj.krhuv;qfXerbysDVªkWuj[krkgSA (iv) ladqyvk;ud2sp3 ladj.k}kjk]v"VQydh;T;kfefr j[krkgSA (v) ladqydsHkwjsjaxdsfy,bysDVªkWudkd-dladj.kmÙkjnk;hgSA (1) (i), (ii) rFkk (v)(2) (iii), (iv) rFkk (v) (3*) (i), (ii) rFkk (iii) (4) (ii), (iii) rFkk (v) Sol.   [Fe(H2O)5 NO] Number of unpaired electrons = 3 ; So,  = = 3.87 B.M. Brown colour is due to charge transfer from ligand to metal. 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 . 0.15 eksyNH OHrFkk0.25eksyNH Cl ;qDr cQj foy;u dhpHdh x.kuk djksANH OHds fy,K =2105 gSA 4 4 4 b Ans. 9.08. 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 . 25ºCij0.06MCH COOHrFkk0.04MCH COONa;qDrfeJ.kesaH O+vk;udhlkUnzrkdhx.kukdjksACH COOH 3 3 3 3 dkfo;kstufLFkjkad=1.84×10–5.gSA Ans. 2.76  10 5 mol Lt1. 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 . fn;s x;s feJ.k foy;u dh pH Kkr djksA[log(1.8)= 0.26] (1) 2 gm CH COOH + 100 ml feJ.k esa 4.1 g CH COONa K = 1.8  105 . 3 3 a (2) 0.1 M NH OH ds 5 ml + 0.1 M NH Cl ds 250 ml, K = 1.8  10 5 . 4 4 b Ans. (1) 4.92 (2) 7.56 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 pH=5dscQjfoy;udksçkIrdjusdsfy,0.4MfijhMhuds500mlfoy;uesafeyk;sx;sfijhMhfu;eDyksjkbMds eksy dh x.kuk djksA fijhMhu dkK =1.5109 . Ans. 0.3 mole of C5H5NHCl should be added to 500 ml solution of C5H5N. 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. vuqPNsn # (9 to 11) ikWfyax}kjklaØe.k/k rqladqyksaesajklk;fudca/kudsfy,la;kstdrkca/kfl)kUr(Valencebondtheory) fodflrfd;k x;kAla;kstdrkca/kfl)kUrdhn`f"VlsladqyfuekZ.kesayqbZl{kj¼fyxs.M½rFk yqbZlvEy¼/k rqvFkok/krqvk;u½ds e/;vfHkfØ;kgksrhgSArFk lkFkghbudse/;milgla;kstdca/k(dativebond) dkfuekZ.kgksrkgSA;gfl)kUrladqy dhT;kfefrlajpukrFk pqEcdh;xq.k sadksle>kusdsfy,/k rqdsla;kstdrkd{kds,prFk ddsladj.kdkmi;ksxdjrk gSA la;kstdrkca/kfl)kUrvf/kdka'kf=kfoejklk;fud(stereochemical) rFk pqEcdh;xq.k sadhO;k[;kdjrkgSAijUrq;g Øksfe;e (III)dhjklk;fud cyxfrdhvfØ;f'kyrkdkdkj.k rFkkdksckYV (III) ds U;wupØ.k v"VQydh;ladqy (low- spinoctahedralcomplex)cu sdsckjsesadksbZtkudkjhughansrkgSAbldksle>usdsfy,gevU;fl)kUrtSlsfØLVy {ks=kfl)kUrdhlgk;rkysrsgSaA 'kq)fØLVy{ks=kfl)kUrdsvuqlkjekuktkrkgSfdfyxsa.M,d_.k RedfcUnqvkos'kgSA /k rqvk;urFk fyxs.Mds chpoS|qr(electrostatic)vkd"kZ.kcygSA;gfl)kUrladqyksadsvf/kdka'kegRoiw.kZxq.k/keksZadhlQyrkiwoZdO;k[;kdjrk gSA 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. fØLVy{ks=kfl)kUrdslanHkZesalghdFkudkp;udhft;sA (1)milgla;kstd;kSfxdksaesa/k rqfyxs.Mca/k'kq):ils/k rqvk;urFk fyxs.Mdse/;fLFkjoS|qrvUr%fØ;k (electrostatic interaction) }kjk curs gSaA (2)milgla;kstd;kSfxdksaesa/krqfyxs.Mca/k'kq)lgla;kstdca/kcukrsgaSA (3)d-d{kdksadhfoikVuO;oLFkk(splittingpattern)]fØLVy{kS=kdhizd`frijfuHkZjdjrkgSA (4*)(1)rFk (3)nksuksa 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). [Ti(H O) ]3+ ladqyds fy,fuEu esalsdkSulkdFku lghgS? 2 6 (1)ladqydsdsUnzh;/k rqvk;udkladj.ksp3d2 gSA. (2)ladqy,dv;qfXerbysDVªkWuj[kusdslkFkvuqpqEcdh;gSA (3)ladqydsdsUnzh;/k rqvk;udkladj.kd2sp3gSA (4*)(2)rFk (3)nksuksaA Sol. Ti3+ have same electronic configuration i.e., d1 with low as well as with high crystal field ligand. Ti3+ with high crystal field stabilization energy on account of higher oxidation state favours the formation of inner orbital complex i.e., d2sp3 hybridisation. Ti3+ U;wufØLVy{ks=kfyxs.MdslkFklkFkmPpfØLVy{ks=kfyxs.MdslkFkHkhlekubysDVªkWfudfoU;klj[krkgSAtSls d1 mPpfØLVy{ks=kfoikVuÅtkZdslkFkTi3+mPpvkWDlhdj.kvoLFk j[krkgSAtksdhvkarfjdd{kdfl)kUrfuekZ.kdks izsfjrdjrkgSAtSlsd2sp3ladj.k 11. The hybridisation of [NiCl2 (PPh3)2] and [NiCI2 (PMe3)2] are respectively (consider PPh3 a bulkier ligand than PMe3) : [NiCl2 (PPh3)2]rFkk[NiCI2 (PMe3)2]dhladj.kvoLFkk;sØe'k%D;kgaS\(PPh3dks]PMe3dhrqyukesacM+svkdkjdk fyxs.MekursgSa): (1*) sp3 and dsp2 (2) sp3 and sp3 (3) dsp2 and dsp2 (4) dsp2 and sp3 (1*) sp3 rFkk dsp2 (2) sp3 rFkk sp3 (3) dsp2 rFkk dsp2 (4) dsp2 rFkk sp3 Hint : Both are stronger field ligands but complex with PPh3 is tetrahedral due to its bulkier nature. nksuksamPp{ks=kfyxs.MgSijUrqPPh3dscMs+vkdkjdsdkj.kPPh3 dslkFkladqyleprq"Qydh;gSA DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 50 Class : XIII Course : 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] ANSWER KEY DPP No.-50 1. 2 2. 3 3. 3 4. 3 5. 1 6. 3 7. 3 8. 3 9. 4 10. 2,3 11. 1,2,4 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 1.5feyheksyNaOHdks1feyheksy;qDrH3PO4ds,dtyh;foy;uesafeykusijfoy;umnklhuik;k tkrkgSbl rF;lsfdulwpukvksadksO;qRifÙkdjldrsgSA (1) K of H PO = 10–3 (2*) K of H PO = 10–7 (3) K of H PO = 10–12 (4)mijksDresalsdksbZugha Sol. H3PO4 + NaOH  NaH2PO4 + H2O the reaction leaves 0.5 mmol NaOH and forms 1 mmol NaH2PO4 and then will form 0.5 mmol of Na2HPO4 and 0.5 mmol of NaH2PO4 will remain so we have a buffer with the 'log term' vanishing. pH = pK2  K = 10–7 for H PO . 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 0.1M H3PO4ds50mLesa 0.2MNaOHds 50mlfeyk;s tkrs gSA ifj.kkeh foy;u ds pHdhijkl gksxh& (1) 4 ls5 (2) 6 ls 7 (3*) 9 ls10 (4) 13 ls 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 0.1 M,fufy;eDyksjkbM ds50mLds vuqekiuesa 0.1MNaOHds 40mLds lkFkfeykus dsi'pkr~pH D;k gSA (,fufy;e vk;u ds fy, K = 2.5 x 10-5 )gSA (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] {k jh;tyh;foy;uesaDyksjksesFksuesaçfrLFk iuvfHkfØ;kik;htkrhgS]ftlesaCl– dksOH–ds}kjkçfrLFkfirdjrsgSA CH Cl (aq.) + OH– (aq.) CH OH (aq.) + Cl– (aq.) DyksjksesFksu esFksukWy lkE;koLFkkfLFkjkadK =1×1016gSA0.1MCH ClrFkk0.2MNaOHdslekuvk;rudksfefJrdjdscuk;sx;sfoy;u C 3 dh ifj.kkehpHdhx.kuk djksA[log2=0.30] (1) 1.3 (2) 1 (3*) 12.7 (4) 13 Sol. Let ‘v’ volumes of two solutions are mixed, then CH Cl + OH– (aq) CH OH (aq) + Cl– (aq) 3 3 t = 0 0.1 V 0.2 V 0 0 teq = 0 0 0.1 V 0.1 V 0.1 V 0.1V so, concentration of OH– = [OH–] = 2V = 0.05 so, pOH = 2 – log 5 = 1.3 so, pH = 14 – 1.3 = 12.7. 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) ,d nqcZy vEy rFkk ,d çcy {kkj vuqekiu esa dkSulk lwpd mi;qDr gSA (1*)fQuk¶Fksyhu (8.310.0) (2)esfFky vkjsUt(3.14.4) (3) esfFky jsM(4.26.3) (4) fyVel (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) ,dnqcZy{kkjrFkk,dçcyvEyvuqekiuesadkSulklwpdmi;qDrgSA (1)fQuk¶Fksyhu(8.310.0) (2) fQukWy jsM(6.88.4) (3*)esfFky vkjsUt (3.14.4) (4) fyVel (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 ) H XvEy ds fy,pK =4rFkkpK =10gS] vuqekiu H X+OH- HX- +H O ds fy, fuEu esa ls dkSulklwpd ¼mudh 2 1 2 2 2 pHijklnhxbZgS½mi;qDrgSA (1)esfFkyvkWjsUt(3.1ls4.4) (2)czkseksfØlkWyxzhu(3.8ls5.4) (3*)p-ukbVªksfQukWy(5.6ls7.6) (4)fQukWY¶Fksyhu(8ls9.6) 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 [Co(ox) ]3–, [CoF ]3– rFkk[Co(NH ) ]3+ esa ls 3 6 3 6 (1) [Co(ox) ]3– rFkk[CoF ]3– vuqpqEcdh; vkSj [Co(NH ) ]3+ izfrpqEcdh;gSA 3 6 3 6 (2) [Co(ox) ]3– rFkk[Co(NH ) ]3+ vuqpqEcdh; gSvkSj[CoF ]3– izfrpqEcdh;gSA 3 3 6 6 (3*) [Co(ox) ]3– rFkk[Co(NH ) ]3+ izfrpqEcdh;vkSj [CoF ]3– vuqpqEcdh;gSA 3 3 6 6 (4) [Co(NH ) ]3+ rFkk[CoF ]3– vuqpqEcdh;vkSj [Co(ox) ]3– izfrpqEcdh;gSA 3 6 6 3 Sol. [Co(ox) ]3– , 3d6 All electrons are paired, so diamagnetic. [Co(NH ) ]3+, 3d6 All electrons are paired, so diamagnetic. F– is a weak ligand [CoF ]3–,3d6 It contains four unpaired electrons, so it is paramagnetic. 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 fuEufyf[krladqyksadkstcpqEcdh;rqykesaj[k tkrkgSrksosviusHkjksaesadehiznf'kZrdjrsgSaArcladqyksadk og lewgftldhprq"Qydh;T;kferhgksrhgS]fuEuesalsgSA (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*)buesalsdksbZugha Sol. Ni(CO)4 3 – sp3 (CO strong field ligand) ; [ Ag F ]– – dsp2 (4d8 electron configuration) ; [Zn(CN) ]2 – sp3 (3d10 electron configuration) ; [PtCl ]2– – dsp2 (5d8 electron configuration) ; 4 4 [RhCl(PPh ) ] – dsp2 (4d8 electron configuration) all complexes are diamagnetic in nature which show decrease in their weights in magnetic field. 10. 0.1 M CH3COOH is diluted at 25°C (Ka = 1.8 × 10–5), then which of the following will be found correct tc 0.1MCH3COOH;(Ka =1.8×10–5) foy;u dks 25ºCij ruq fd;k tk;s rks fuEu esa ls dkSu ls dFku lR; gSA (1) [H+] will increase [H+]c<+tk;sxkA (2*) pH will increase pHc<+tk;sxhA (3*) number of H+ will increase H+vk;uksadhla[;kc<+tk;sxh (4) all the above are correct mijksDrlHkhdFkulR;gSA Sol. On the basic of ostwald dilution law, number of H+ ions will increase but increase in volume will be more. Therefore, [H+] decreases, pH increases. 11. Which of the following pairs will not form a buffer solution fuEuesadkSulk;qXecQjfoy;uughacukrkgSA (1*) NaCl + HCl (2*) CH3COOH + NH4Cl (3) Na2HPO4 + Na3PO4 (4*) KNO3 + HNO3 Sol. Buffer will be due to weak acid + conjugate base.