UNIT # 07 (PART - II)

9. ph ph [H + 0. [H + 0.0 V 50 V 50 [H + of miture i[h + N V N V V V 50(0. 0.0) 00 [H + 0. 0.055 ph.6. ph 7 [H + 0 7, [H 0 7 ne ph fter ddition of be ph [H + 0 [H 0 [H + concentrtion incree 0 5 time.. Reltive trength 0.8 0 5 : 6. HCH + H 0mL,0.5(M) 50mL,0.M HCH + H fter rection i form Buffer olution [HCH 0 90 ph p + log [lt [cid [HC 0 90 ph p ph log (.8) ph.75 9. Let e cid i HA it odium lt i NA H H CH 0. (0.0) 0 5 9 0 00. CH CH + NH CH CN + H 00 ml, 0. M 00 ml, 0.M 0 0 After rection [CH CH 0 50, [CH 0 CN 00 ph p + log [lt [cid ph p [H +.8 0 5. CH CN + HCl CH CH + NCl t equivlence the [CH CH ph p log C ph [5 log log0 ph [6 log ph log UNIT # 07 (PART - II) INIC EQUILIBRIUM EXERCISE # 0 00 0. 8. ph p + log HC H C 7 7 log + log HC H C % HC 00 5 80 % HC (H C ) 0. NH + H N P 50mL,0.M 60mL,0.5 0 7. H H + H [m H + obtin from firt ioniztion of H ph p log C p log C.5 +.7.67. Mg(N ) + NF MgF + NN.5 5 0 0.5 (MgF ).5 5 MgF Mg + + F Ionic product. AgCNS Ag + + CNS ( ) (+ ) AgCl Ag + + Cl ( ) (+ ) ( ) ( )...(i) p AgCNS ( ) ( )...(ii) p AgCl dde eqution (i) nd (ii) ( + ) p p AgCNS ( + ) p p AgCNS AgCl (AgT) ( + ). 0 5 AgCl 0.0 0.7 0 divide eq. ii from i 0 [Cl.7 0.7 0 [CNS.0 0 7. IP > p I.P. (C + ) (F) I.P. (0 ) (0 ) 0 8 IP > p 5. ph p HIn In + log HIA In 0 HIn 5. At Hlf y [HIn In 6 5 + log In HIn [lt ph 5.5 + log 5.5 p [cid + [lt [lt log 0.75 [cid [cid 5.6 [lt log [cid

INIC EQUILIBRIUM EXERCISE #. HCl + NCl ml, 0. 99 ml [H + 0. 0.00 ph 00 6. HF + H F + H + ph. C 6 H 5 CH + NH C 6 H 5 CN + H 0 0 ph 7 + p + log C 9. h b p + p b p p 0.8 p.7 6.75 0 h h C 0 8.0 9. 0 h C h.8 %. p p 5 6. 0 8 0 8 mol/m + 5. AgN + NH [Ag(NH ).6 5 0 8 (.6) 0.8 [Ag(NH ) (AgN )(NH ) 0 8 0.8 8 (5 0 )(.6) (.6) 0.6 M mole of NH 8. C 6 H 5 CH + NH C 6 H 5 CN + H 0 ph p + [lt log [cid ph. + log ph 7 +. + log 00 ph 9. ph 8. 5. CH CH + NH CH CN + H 0. 0. At / neutrliztion / / ph p + log (lt) (cid) ph p + log / / t / neutrliztion ph p + log / / 7. HS ph ph log log log log () 0 H + + S 0.09 INIC EQUILIBRIUM EXERCISE #...()...() ph.0 CMPREHENSIN BASED QUESTINS Compreheni on #. Suppoe volume of HC V ml millimole of HC 5V millimole of H C 0 [HC ph p + log [H C 7.0 6. + log V, V 78 ml. If C ecpe, [H + decree, hence ph incree. (C, [H + ph ) Compreheni on #. Phophoric cid ith three ionible hydrogen ion i tribic cid. H-tom re ttched to -tom, ( H- ). If firt tep i only ten ( ) ph [p log c c [H 0.05 % 0.05 0 mol L (M) 98 5. 0 M log c., p. ph. [H [P. [H log [H + + log[p log + log + log log [H phlog[p log[h p p p log[p. + 7. +. log[p.65 [P. 0 M. Zn ( ) Zn + + p [Zn + [ 9. 0 [Zn + (. 0 ) [Zn +.88 0 5 [Zn + 5.7 0 9 M

INIC EQUILIBRIUM. ( i ) H H + + H [H [H [H ( i i ) b 0 0 7 0 7 7 7 0 0 000 / 8. C C C. C.().8 0 6. 0 ph p log C 5 0 0.8 06 55.5 / 00 0.50 p log (0.) 9 p p 8 0 8 5.(c) [H + C 6.8 0 log.8.87 (e) [H + 0 8 + 0 7 0 7 [0. + ph 7 log. 6.95 (f) [H 0 0 + 0 7 0 7 [.00 PH 7 log.00 6.99 ph 7.000 (g) [H + C 5 6.8 0 0 [H +.8 0 8 0. 0 6 ph 6- log. 5.7 6. p - log.56.59.6 p ph 6.795 0. ph.5 [H + 0.5 [H 0.5 NH H 5 0.8 05 C. C 0, [H + 0 b NH + + H 0.5 0.5 C.8 0.556 M 0 0 0 0 0 9. 0. CHCl CH H + + CHCl C 0.0 0.0 0.0+ (0.0 ).55 0 0.0 0.0 +.55 0.55 0 + 0.55 0.00055 0 0.055 0.0775. 0 CHCl C 6.6 0. For e cid EXERCISE # [A [H + C C....8 0 0.0 6. 0.0 0 8 00 0 5 5 [H [AC [H + 0 [ACH [AC.6 0 me [C H 5 6. 0 5. HCN i e cid o H + due to it cn neglect A compriion to HF [H + C 6.7 0 0. 6 67 0 8.8 0 ph log [8.8 0 log [8.8 p H.087 6. H S H + + S [H + 0, [H S 0. M 0 7 0 7 0 0 [ 0 [S [0. 0 0 8 [S 0 [S.5 0 5 [S 7.(i) H H + + H 7.5 0 0.0 M C ( ) C C (ii) H H + H + 6.8 0 8 C ( ) C [C (iii) H + H + 6.8 0 8 C ( ) C ( ) [C 7.5 0 C ( ) RN. (i) ( ) 0.75 0.0 + 0.75 0.75 0 0.56 [H + 0.0 0.56 [H + 5.6 0 [H 5.6 0 RN. (ii) 6.8 0 8 [HP [H [H P from [i rection. [H 6.8 0 8 M RN. (iii).5 0 [P [H [HP.5 0 6.8 0 5.6 0 8 5.6 0 8 [ [

+ 0. NH Cl NH + Cl + NH H NH + H b [NH [H [NH H [NH + i due to lt becue NH H ionie in le mount due to common ion effect 0. [H.8 0 5 0.05 9 0 6 [H. HC H + NH CH C + H 50ml, 0.M 50ml, 0.M 0m mol 5m mol CHCH + H CH C + H 0 5 5 0 5 5 ph p 5 log.8 ph.7. (NH ) S Molrity (NH ) S 00 Molrity of NH + 00 Molrity of NH H INIC EQUILIBRIUM 0. 00 0 / 00 9.6. + log 0. / 00 0 log (0) 0 /0 mole 0.05 mole. Q + + H QH + H + / n V n ' n' + y R + + H RH + H + n V n ' n' y y y+ ( y) ( y)...(i) n ' n ' y( y) y( y)...(ii) n ' y n ' Auming () << n ' & y << n ' from eqution (i) & (ii) ( (i) (ii) ) ( + y) n ' & y ( + y) n ' 0. CH C + H CH CH + H 0 0.08 b 5.8 0 0 0 0.8 0 0 0 0.08.8.8 0. 0 0 0.66 0 5. C 5 H 6 N + + H C 5 H 5 NH + H + ph [p p log C b.699 [ p b + 0.6.98.6 p b p b.6 5.98 9.80 b 0 9.80 p p 8. ph 7 log.5 ph 9 log.5 8.5 0. CH CH+H CH C +H (WASB) ph [p + p + log C [ + 5 log.9 + log 0 [9 log.9 log 0 ph [9 log 0.9 8.78 ph 5.8 [H 0 5.8 [H 0 6 0 0.7 [H 5. 0 6 dding both () ( + y) n ' [H + ( + y) ph log [H + n ' EXERCISE # [B ( n ' n ') ( n n ) V V ph log ( n n ). N Y+H N HY+NH.88 0 0. 0..88 0 5500.55 + 0. 0 0..7 0

N HY+H N H Y+NH.5 0 y y y + ince y << y ~, + y ~.5 0 8 y. y N H Y+H NH Y+NH.7 0 y y z z z + y z ~ y, z + ~.7 0 z. y z.68 0 7 NH Y+H H Y+NH z z t t t + z t ~ z, t + ~ 9.8 0 t. z frction () t 0. 9.8 0 t.8 0 7.8 06 8. [Zn(H) (q)+zn(h) + +Zn + +Zn(H) + Zn(H) + [H + + [H [H + 5 [H 7 0 0 0 6 + + +0 [H +0 [H [H [H () ph 5, ph 9, [H 0 9 0 6 + 0 + 0 + 0 + 0 0 0 M (b) ph 9, ph 5, [H 0 5 0 6 +0 8 +0 7 +0 8 +0. 0 6 M (c) ph, ph, [H 0 0 6 + 0 + 0 5 +0 +0 0 M. Given : Ag+ Cl AgCl G 09.7J/mole Ag Ag + + e G 77. J/mole e + Cl Cl G.J/mole o for rection ( ) AgCl Ag + +Cl G G +G + G G 55.7 J/mole G RT ln p 55.7 0 8. 98 ln p p.7 0 0.7 0 0 [Ag + [Cl 0.05.6 0 9 M 5. Cu (A ) Cu+ +A p 8 0 6 +y 6.() Pb (A ) Pb+ +A p.096 0 6 y y+ Let olubility of Cu (A ) & Pb (A ) i & y repectively. ( Cu (A ) Pb (A ) y ) 08 ( + y) 8 0 6...(i) 08 y ( + y).096 0 6...(ii) (i) (ii) 8.5 y y.096 putting thi in eqution (ii) ((ii)) 08 y (.5 y).096 0 6 y. 0 8.5 y.875 0 8 [Cu + 8.85 0 8 [Pb + y 7. 0 8 Al(H) Al+ + H p Al(H) Al+ + H Al(H) + H Al(H) p 8.6 8.6 Al(H ) [H 0 [H [H.6 0 5 ph.585 ph 9.5 (b) p [Al + [H 5 0 [ 0 [H [H.7 0 0 ph 9.767 ph. 7. HCl 0.09 M CH CH C 0. M,, 0 Cl CHCH C 0.09 M,,? ph, [H + 0. 0. 0.09 + C + C C + C 0.0...(i) CH CH CH C + H + C C C C 0. Cl HCCH Cl HCC + H + C C C C 0. (C )(0.) C ( ) 0 ~ 0. 0 5 5

putting thi in eqution (i)((i) ) 0 0. + 0.09 0.0 0. (C )(0.) C ( ) (0.)(0.) 0. 0 0.95 000 8. C 5.5 M 7 00 C 5.5 0 6 5.5 9. ln 0 [H C 5.5 0 ph.6, ph.7 ln H R 5.7 0.08 0 T T.8 0 H 8. 98 H 595.6 J 5.95 J/mole 0. In begining [H + C [H + 5.8 0 0.00 ph.7 n doubling ph, ne ph(ph, ph).7 CH CH C [H +.8 0 5 CH C + H + C C C C (C ) C( ), C [H+.8 0 5 (.8 0 ).8 0 5 C C C C.8 0 5 C.6 0 5 5 V C.77 0 L.() PV nrt 0.959 n 0.08 98 n 0.099 volume of H ml (per volume of H ) (H ml (H )) C n V 0.099 9.9 M 0 p b.9 b 0 [H bc 0.5M ph 0.90 ph. 097 (b) M 0.5 for NH (NH 0.5). 7.5 0, 6. 0 8, 0 () H + NH NH + H 6 ph p. (b) H + NH NH + H 6 6 6 ph p p.66 (c) H + NH NH + H.8 7...8 NH + NH N H + H.8... ph p 7. (d) H + NH NH + H 0 6 NH + NH N H + H 6 N H + NH N + H ph p. For H C. 0 7,.8 0 H 7.5 0, 6. 0 8, 0 () N C + HCl NHC + NCl p ph p 8.7 (b) N + HCl N H + NCl 0.8.6 0.8 0.8 N H + HCl NH + NCl 0.8 0.8 0.8 ph p p.66 (c) N + NH N H 5 5 0 p p ph 9.6 (d) H + N N H + NH ph p 7.

. BH + HCl BCl + H At end point m mole of BH m mole of HCl ( BH m mole) 0.6 V V 5 ml Totl volume ( ) 0 + 5 65 ml [BCl 65 ince BCl i SAWB ph 7 p b log C 5. 7 p b log 65 p b.75 No on further dding NH NH BCl + NH BH + NCl.8.. 8 ph p b + log..8 5.() ph p + log 0.06 0.05 ph.7 + log.. 8 (b) (c).87 ph 9.68 n diluting olution 0 time (0 ) [HCH 0.005, [HCN 0.006 HCH H + + HC 0.005 0.005 ( ) 0.005 0.005 + 0.006 (0.005 0.006) (0.005 ).8 0 0.005( ) 0.005 0.006.8 0 7.77 +. 0 0.085 [H + 0.005.5 0 ph.86 n further diluting olution by 0 time ( 0 ) [HCH 0.0005, [HCN 0.0006 HCH H + + HC 0.0005 0.0005() 0.0005 0.0005+0.0006 (0.0005 0.0006)(0.0005 ).8 0 0.0005( ) 0.0005 0.0006.8 0.77 +. 0 0.07 [H + 0.0005.05 0 ph.9899 6. Initil () ph p b.7 Let mole of NH h been dded o (NH ) [NH + 0. +, [NH 0. ph 5.7 5.7.7 + log 0. 0. log 0. 0. 0. 0. 0 0.9 0.088 mole 7. N + H N H + NH 0.0 N H + H NH + NH NH + H H + NH.58 07. 0 ince equilibrium contnt of nd & rd rection i very le, [H ill minly come from t rection. ( nd rd, [H t ) N + H N H + NH 0. 0. 0.0 5 + 0. 0 0..7 0 [H.7 0 M N H + NH NH + NH y y y + y ~, y + ~, o.58 0 7 (y ) y y ( y) NH + H H + NH y y z z z + y z ~ y, z + ~. 0 z( z) (y z) z.7 0 7.58 0 z y z 5.9 0 8 [H z 5.9 0 8 M

8. ph 8, [H + 0 8, [H 0 6 HC 0.0005 H + + C 0.0005yz 0 8 y 5 0 HC +H H C +H 08 0.0005 0.0005yz z 0 6 ince equilibrium contnt for firt rection i very le y << z ( y << z) 6 z(0 ) 0 8 0.0005 z 5 z 0.0005, z 9.8 0 6 [H C 9.8 0 6 M [HC 0.0005 9.8 0 6 8 0 y 5 0.9 0 [C y.5 0 8 M.9 0 M 9. Fe + + H Fe(H) + + H + 6.5 0 0.95 0.05 0.05 6.5 0 (0.05) 0.95.7 [H + 0.05 0.5 ph 0.908 0. ph p + log lt cid 6.7 7. + log y 0.005 y.58 0 mole. When indictor i hlf in ionic form ph( ph) p 7. ph 7. + log 5 7.898 no ith thi ph ( ph ) 7.898 p + log p 7.959 gin hen 50% of ne indictor i in ionic form (50% ) ph p 7.959. m mole of H + ion ill produce (H + m mol ) [H + 0 0. 0.0 + H + H 0.0 0.0 0.08 0.0 0. H + H + H 0. 0.0 0.08 0.0 o no they form buffer olution of H & H ( H H ) ph p + log 0.08 ( 0.0 6. 0 8 ) ph 7. + log 7.8. At equivlence point () meq. of HA meq. of NH.6 NA + HCl HA + HCl.6.806.56.806 ph p + log [S [A.9 p + log.56.806 p 5.0 No NH + HA NA [NA 0. 0 ph 7 + p ph 9 + log C 7 + 5.0 log 0.. In begining let m mole of BH re preent ( BH m mole) BH + HCl BCl + H ph p b + log 9. p b log p b 5.7 b 5.8 0 6

no BCl + NH BH 6 6, [BH 0.8 50 [H b ph.77 ph. C.668 0 5. () ph t one fourth neutrliztion ( ph) (ph) p + log / / p + log ph t three fourth neutrliztion ( ph) 0. 9 5 (b).5 p + (c) ph i.e. (ph) p + log / / p + log ph (ph) (ph) log / log / p log p.75 (ph) p +, (ph) p For p + [S 0 [A 0 0 0 0 th 0 i.e. tge For p [S [A 0 i.e. th tge 0 6. Zn(H) () Zn+ (q) + H (q) p Zn(H) () + H [Zn(H) (q) C diolved Zn(H) i preent in form of Zn + & [Zn(H) o olubility [Zn + + [Zn(H) (Zn(H) Zn + [Zn(H) [Zn + + [Zn(H) ) [H p + C [H For min. olubility () d 0 d[h p [H [H [H 9.8 0 5 ph.00869 ph 9.99 7. AgCl () Ag + + C [H 0 p C / + Cl p Ag + + NH Ag(NH ) + AgCl () +NH Ag(NH ) + +Cl, p 0. 0. 0. (0. ) 0.0096 0.058 p 0.0088 Solubility () 9.6 0 8. [Cl 0.0 M Ag(CN) Ag+ +CN int 0 9 0.05 0.05 ~ 0.05 0 9.() 0.05 0.05 09.7 0 7 [Ag [Cl.7 0 7 0.0. 0 9 > p o AgCl ill precipitte. ( AgCl ) M

9. After miing ith equl volume ( ) A + H HA + H [Ag + 0.0 M, HCN 0.0 M HCN H + + CN HA + H H A + H Ag + + CN AgCN () p [HA [H [HA [A [A [H HCN + Ag + H+ + AgCN (), [H A[H [HA [H [HA A [H p.5 06 [H A [A [H 0.0 0.0 From m blnce () 0.0 ince vlue i very high lmot ll of rectnt ill convert into product [A + [H + [A [H ( ) [H [H + 0.0.5 0 6 X 6.6 0 5 [Ag + 6.66 0 5 M [H [H 0. M A M + + A Let olubility i. ( ) p [M + [A. [H [H But ome mount of A ill undergo hydrolyi. Let i the mount of A left in olution. (A A ) [H [H p

INIC EQUILIBRIUM. ph ; H + 0 0. M ph ; H + 0 0.0 M M 0. V From M 0.0 V? M V M V 0. 0.0 V V 0 litre volume of ter dded 0 9 litre.. H + C ; [H C or C 0. 0 0 0 5. Cr(H) () Cr + 7S p (q.) + H (q.) 0 0. 0 6. H C (q) + H () HC (q) + H + (q) 0.0 [HC [H [H C EXERCISE # [B 0.0. 0 7.95 0 0.0 A H C i e cid o the concentrtion of H C ill remin 0.0 0.0 >>. [H + [HC.95 0 No, HC (q) + H () C (q) + H + (q) y y y A HC i gin e cid (eer thn H C ) ith >> y. S p 7 /.6 0 7 0 / [C [H y ( y) [HC ( y). ph 5 men [H + 0 5 HA H + + A t 0 c 0 0 teq c( ) c c [H [A (c ) [H [HA c( ) c [H But, [H + << C (0 5 ) 0 0 5. AgBr Ag + + Br p [Ag + [Br For precipittion to occur Ionic product > Solubility product [Br p 5 0 [Ag 0.05 0 i.e., precipittion jut trt hen 0 mole of Br i dded to AgN olution Number of mole of Br Br 0 needed from M of Br 0 0. 0 9 g Note : [H + H + econd tep(y) ( + y) from firt tep() nd from [A >> y o + y nd y So, y y.8 0 y [C So the concentrtion of [H + [HC concentrtion obtined from the firt tep. A the firt tep. A the diocition ill be very lo in econd tep o there ill beno chnge in thee concentrtion. [H + [HC.95 0 & [C.8 0 7. AgBr Ag + Br p [Ag + [Br For precipittion to occur Ionic product > Solubility product [Br 5 0 [Ag 0.05 p 0 i.e., precipittion jut trt hen 0 mole of Br i dded to AgN olution

Number of mole of Br needed from Br 0 M of Br 0 0. 0 9 g 8. N C N + + C 0 M 0 M 0 M SP (BC ) [B + [C. MX M ++ + X Where i the olubility of MX then p ; () 0 ; 0 [M ++ [M ++ 0 0 [B + 5. 0 0 9 5. 0 5 M. ph log[h + log [H 9. In correpond to choice (c) hich i correct ner. BA + H BH + HA No ph i given by Be Acid ph p + p p b Subtituting given vlue, e get ph ( +.80.78) 7.0 0. Let olubility AgI Ag + + I p [Ag + [I Given p 0 8 p 0 8 5. log [H n olving, [H +.98 0 6. MX M + + X S S SP [ [ 56 5 p 56 / 5. Mg(H) [Mg + + [H p [Mg [H [ [.. 5. AB A + + B [A.0 0 5, [B [.0 0 5, p [B [A [ 0 5 [.0 0 5 0 5.0 0 mol/lit.0 0 8 g/lit ( Moleculr m of Ag I 8).0 0 8 00 000 gm/00ml.8 0 gm/00 ml