Dung Dich Ax Bazo Da Chuc

Nin kho: 2014 - 2016MT S VN TNH TON TRONG DUNG DCH AXIT-BAZ A CHC Ging vin hng dn PGS.TS. NGUYN NH LUYNHc vin thc hinNGUYN TH MINH H (Phn tch)L TH THANH H (V c )HONG TH THNG (V c ) ti*

MC LCA.M UB.NI DUNG I.DUNG DCH AXIT A CHC 1.Mt s vn tnh ton dung dch axit a chc 2. Bi tp vn dng II.DUNG DCH BAZ A CHC 1.Mt s vn tnh ton dung dch baz a chc 2.Bi tp vn dng III.DUNG DCH MUI AXIT 1.Mui axit. 2.Mt s vn trong tnh ton dung dch mui axit HnA- 3.Mt s tnh ton nhanh pH h dung dch mui axit HA- 4.Mt s dng bi tp dung dch mui axit C.KT LUND.TI LIU THAM KHO*

Ha hc phn tch l mn khoa hc v cc phng php xc nh thnh phn nh tnh v nh lng ca cht v hn hp ca chng. Mt trong cc ni dung quan trng ca ho phn tch l hiu v suy on c tnh cht ca axit-baz trong dung dch. Do , chng ti chn ti: Mt s vn tnh ton trong dung dch axit baz a chc. bi tiu lun c hon chnh v y , rt mong nhn c s gp v b sung ca thy v cc bn.

A.M U*

B.NI DUNG*

S phn li ca cc a axit din ra theo tng nc [1]I.AXIT A CHC

1. Mt s vn tnh ton:*

C th coi cc a axit nh mt hn hp gm nhiu n axit. hay* Thnh phn dung dch: *

C 4 trng hpTrng hp 1:Trng hp 2:Trng hp 3:Trng hp 4:*

Trng hp 1:Cn bng (1) quyt nh, ta c:V d 1: [2] Tnh [H+], [OH-], [SO32-] trong dung dch H2SO3 0,01 M.Cc cn bng xy raKa1 Ka2 v Ka1.C W, do c th coi cn bng (1) chim u th:*

C0,01[]0,01-xx xT cn bng (2):iu chng t vic tnh gn ng theo (1) l hp l.*

Trng hp 2:* Cch 1:Gii pt bc cao (theo phng php tip tuynNewton) * Cch 2:Gii theo phng php gn ng lin tcV*

V d 2: [2] Tnh pH ca dung dch H4P2O7 (H4A) 4.10-2 M. Bit pK1 = 1,52; pK2 = 2,36; pK3 = 6,6; pK4 = 9,25.Cc cn bng xy ra:Ta thy:Nn cn bng (1), (2) l ch yu p dng cng thc (1.3) ta c *

*Cch 1: Gii pt bc cao Th (b) vo (a) v bin i (a), ta c:Thay cc gi tr vo ta c:*

* Cch 2:Gii theo phng php gn ng lin tcT cng thc (a), ta c phng trnh khuyt bc 2 i vi h:Chp nhn:+ Bc 1: Thay [H4A]0 v [H+]0 vo (c) ta c: h1 = 0,0388. sau th h1 vo (b) c: [H4A]1 = 0,0216+ Bc 2: Thay [H4A]1 v [H+]1 vo (c), tnh c h2 = 0,0283, th h2 vo (b) c: [H4A]2 = 0,018+ Bc 3: Thay [H4A]2 v [H+]2 vo (c), tnh c h3 = 0,027, th h3 vo (b) c: [H4A]3 = 0,018*

Vy h = 0,027, suy ra pH = 1,57 Trng hp 3:* Cch 1:Gii pt bc cao (theo phng php tip tuynNewton) * Cch 2:Gii theo phng php gn ng lin tcV*

V d 3: [3] Tnh pH trong dung dch axit tactric H2C4H4O6(H2A) 0,03M. Bit pK1 = 3,04; pK2 = 4,07.Cc cn bng xy ra:Ta thy: Nn b qua cn bng (3) T (1.3) v*

* Cch 2:Gii theo phng php gn ng lin tcT (a), ta c pt bc 2 khuyt:Chp nhn [H2A]0= Ca = 0,03M;* Cch 1: Gii pt bc cao Th (b) vo (a) v bin i (a), ta c:Thay cc gi tr vo ta c:*

+ Bc 1: Thay [H2A]0 v h0 vo (c) ta c: h1 = 10-2,28. sau th h1 vo (b) c: [H4A]1 = 2,57.10-2+ Bc 2: Thay [H4A]1 v h1 vo (c), tnh c h2 = 10-2,31, th h2 vo (b) c: [H4A]2 = 2,57.10-2Vy h = 10-2,31, suy ra pH = 2,31 *

Trng hp 4:* Cch 1:Gii pt bc cao (theo phng php tip tuynNewton) * Cch 2:Gii theo phng php gn ng lin tcV*

V d 4: Tnh pH trong dung dch axit H3A 10-4M. Bit pK1 = 7,5; pK2 = 8; pK2 = 8,9.Cc cn bng xy ra:Ta thy: p dng cng thc (1.4) ta c *

* Cch 1: Gii pt bc cao Th (b) vo (a) v bin i (a), ta c:Thay cc gi tr vo ta c:Chn h0:Nghim thc h phi tha mn: 3,55.10-12 < h < 1,74.10-6Chn h0=1,75.10-6

*

Vy h= 8,324.10-7. Suy ra: pH = 6,08*

* Cch 2:Gii theo phng php gn ng lin tcT (a), ta c pt bc 2 khuyt:Chp nhn [H2A]0= Ca = 10-4M;+ Bc 1: Thay [H3A]0 v h0 vo (c) ta c: h1 = 1,79.10-6. sau th h1 vo (b) c: [H4A]1 = 9,83.10-5+ Bc 2: Thay [H3A]1 v h1 vo (c), tnh c h2 = 1,78.10-6, th h2 vo (b) c: [H3A]2 = 9,83.10-5Vy h = 1,78.10-6 , suy ra pH = 5,75 *

2. Bi tp vn dng:Bi 1: [3] Tnh nng cc cu t trong dung dch H3PO4 0,1M trng thi cn bng. Bit H3PO4 c Ka1 = 10-2,15 Ka2 = 10-7,21 Ka3 = 10-12,3 Cc cn bng xy ra:Ta thy: Ka1 >> Ka2, Ka3 v Ka1Ca >> W*

Nn cn bng sau chim u th: C0,1 [ ] 0,1-x x x x2/(0,1-x) = 10-2,15 gii c x = 10-1,64 pH = 1,64.[PO43-] = CaKa1Ka2Ka3 ( [H+]3 + Ka1[H+]2 + Ka1Ka2 [H+] + Ka1Ka2Ka3)-1 Th s vo ta c : [PO43-] = 10-17,8M [HPO42-] = CaKa1Ka2[H+]. ( [H+]3 + Ka1[H+]2 + Ka1Ka2 [H+] + Ka1Ka2Ka3)-1*

Thay s ta c: [HPO42-] = 10-7,2 M [H2PO4-] = CaKa1 [H+]2 ( [H+]3 + Ka1[H+]2 + Ka1Ka2 [H+] + Ka1Ka2Ka3)-1Th s ta c: [H2PO4-] = 10-1,63 M [H3PO4] = 0,1 - 10-1,63 = 0,076 M Bi 2: [1]Tnh pH v cn bng trong h gm HCl 0,010M v H2S 0,10M. Gii:*

Do Ka1 >> Ka2 nn cn bng (2) l ch yu: C 0,10 0,010[ ] 0,10 x 0,010+ x x x

Bi 3: [3]a, Tnh pH ca dung dch gm H3AsO4 0,10M v CH3COOH 0,050M. b, Tnh in li ca axit axetic trong hn hp .Cc cn bng xy ra:Nn c th coi (2) l ch yu*

II.DUNG DCH BAZ A CHC. * Cc qu trnh xy ra trong h 1.Mt s vn tnh ton trong dung dch baz tnh tonS phn li ca cc a baz din ra theo tng nc

*

* Thnh phn dung dch: B, Cb, BH+, BH22+,, BHnn+, Kb1, Kb2Kbn; H2O

*

* C 4 trng hpTrng hp 1:Trng hp 2:Trng hp 3:Trng hp 4:*

*Trng hp 1:Cn bng (1) quyt nh, ta c:*V d 1: Tnh pH ca dung dch Na2S 0,010 M. Bit H2S c Ka1= 10-7 Ka2=10-12,92 .GiiCc qu trnh xy ra trong dung dch: Na2S 2Na+ + S2- S2- + H2O HS- + OH- Kb1 = 10-14/10-12,92= 10-1,08 (1) HS- + H2O H2S + OH- Kb2 = 10-14/10-7= 10-7 (2) H2O H+ + OH- Kw= 10-14 (3)*

V Kb1>>Kb2>>Kw nn trong dung dch cn bng (1) xy ra l ch yu. S2- + H2O HS- + OH- Kb1= 10-1,08 C 10-2 [ ] 10-2 x x x

p dng LTDKL ta c:

x= 9.10-3 = [HS-] = [OH-]

[H+] = 1,1.10-12 pH = 11,95.

*

* Trng hp 2: Cch 1:Gii pt bc cao (theo phng php tip tuynNewton) Cch 2:Gii theo phng php gn ng lin tcV*

Trng hp 3:Cch 1:Gii pt bc cao (theo phng php tip tuynNewton) Cch 2:Gii theo phng php gn ng lin tcV*

V d 2: Tnh pH ca dung dch Natri sucxinat (Na2X) 0,02M.Gii p dng cng thc (2.3), ta cCch 1: Gii phng trnh bc 3Thay cc gi tr Kb1, Kb2, CX2- vo ta c

Bin i t cng thc 2.3, ta c.*

Gii phng trnh, ta c x=9,299.10-6Vy pH=8,97Cch 2: Gii theo phng php gn ng lin tcChp nhn Thay vo phng trnh (1), ta tnh c:*

Thay vo phng trnh (2)Thay vo (1), ta c x2=9,3.10-6Vy x=9,3.10-6pH=8,97*

Trng hp 4:Cch 1:Gii pt bc cao (theo phng php tip tuynNewton) Cch 2:Gii theo phng php gn ng lin tcV*

V d 3: Tnh pH ca hn hp gm 1ml Na3Xit 4,0.10-4M v 99,00 ml H2O.

Cn bng:t Na3Xit thnh Na3X*

Vp dng cng thc 2.4, ta c:(9)Cch 1: Gii phng trnh bc 4T (a) ta c *

Thay cc gi tr Cb, Kb1, Kb2, Kb3 vo ta c Chn x0Chn x0=3,324.10-7*

Vy x=3,329.10-7 pH=7,5

*

Cch 2:Gii theo phng php gn ng lin tcChp nhn Thay vo phng trnh (1), tnh x1 *

Thay vo (2), ta cThay vo (1), ta cVy x=3,329.10-7 pH=7,5

*

1. Tnh nng cc cu t trng thi cn bng ca dung dch Na2CO3 0,1M Bit H2CO3 c Ka1=10-6,34 , Ka2 = 10-10,23

Cc cn bng: Na2CO3 2Na+ + CO32- CO32- + H2O HCO3- + OH- HCO3- + H2O H2CO3 + OH- H2O H+ + OH- V Kb1.Cb1 >> Kb2.Cb2 >> WCn bng (*) quyt nh CO32- + H2O HCO3- + OH- Kb1 = 10-3,77 C 0,1 [ ] 0,1 x x x

2. Bi tp vn dng Dng 1: Dung dch cha mt a baz *

HCO3- + H20 H2CO3 + OH- Kb2 = KH20.Ka1-1 C 10-2,39[ ] 10-2,39 - x x 10-2,39

iu kin x

2. Tnh khi lng mui Na2S phi cho vo 1 lt nc c dung dch c pH = 11,5. Bit H2S c Ka1= 10-7; Ka2=10-12,92. Gii : Cc cn bng xy ra trong dung dch l: Na2S 2Na+ + S2- a a S2-+ H2O HS-+ OH- Kb1 = 10-14/10-12,92= 10-1,08 (1) HS-+ H2O H2S + OH- Kb2 = 10-14/10-7= 10-7 (2) H2O H+ + OH- W= 10-14 (3)V Kb1>>Kb2>>W nn trong dung dch cn bng (1) xy ra l ch yu. S2- + H2O HS-+ OH- Kb1= 10-1,08C0 a[ ] a-x x x Ta c trong x= [OH-] =10-14/ 10-11,5 = 10-2,5

a = 3,28.10-3 = CNa2S mNa2S= 3,28.10-3 .78 = 0,256g

*

Dng 2: Hn hp a baz v baz mnhTrn 10,00 ml dung dch NaOH 8,00.10-3 M vi 30,00 ml dung dch H2S 1,00.10-3M. Tnh pH ca dung dch thu c. Bit H2S c Ka1= 10-7; Ka2=10-12,92.

Gii:y l bi ton pha trn gia a axit (H2S) vi baz mnh nn c phn ng xy ra, chng ta cn xc nh thnh phn gii hn, t m t cc cn bng xy ra trong dung dch v tnh pH.- Nng ban u ca cc cht trong dung dch: NaOH + H2S NaHS + H2O 0,75.10-3 0,75.10-3 0,75.10-3 NaOH + NaHS Na2S + H20 0,75.10-3 0,75.10-3 0,75.10-3 .

*

TPGH gm: Na2S 0,75.10-3M; NaOH d 0,5.10-3 M Vy dung dch l hn hp a baz ( S2-) v baz mnh, nn ta c cc cn bng xy ra:

C0 0,75.10-3 0,5.10-3[ ] 0,75.10-3-x x 0,5.10-3+x

p dng LTDKL ta c: Gii PT bc 2 ta c: x = 7,4.10-4 [OH-]=1,24.10-3[H+]=8,06.10-12 pH= 11,09.*

1.Tnh pH ca hn hp thu c khi trn 40,00 ml NH3 0,25M vi 60,00 ml Na2CO3 0,15M. Kb1>>Kb2, nhng ch c th b qua (3).KP vi mc khng l CO32-, NH3Dng 3: Hn hp a baz v baz yu*

Bc 1: Chp nhnv thay vo (4) c h1=10-11,67Tnh li:Bc 2: Thay [NH3]1=9,96.10-2M v [CO32-]1=0,086M vo (4) tnh h2=2,23.10-12 . Kt qu lp. Vy pH= 11,65 T hp ta c:(4)*

III. DUNG DCH MUI AXIT

1. Mui axit [1] Mui axit l nhng mui m gc axit cn nguyn t H c kh nng phn li cho ion H+. 2.Mt s vn trong tnh ton dung dch mui axit HnA-

Thnh phn dung dch: *

Cc qu trnh xy ra trong h+Nu th ta c mi trng axit.+Nu th ta c mi trng baz.*

Chn mc khng: Theo KP:HayTRNG HP 1:*

TRNG HP 2:TRNG HP 3:TRNG HP 4:*

Nng cc thnh phn cn bng trong h dung dch mui axit*

TRNG HP 1:Ta cViThay vo (1.1) tm h. T tnh(i = 1 n) Ta da vo cn bng th (i+2) vi lu : nng ban u ca ion H+ cn bng (i+2) chnh bng nng cn bng ca ion H+ cn bng (2) *

TRNG HP 2:Ta cGii PT bc cao (theo phng php tip tuyn Newton)Gii theo phng php gn ng lin tcCch 1Cch 2*

V d 1 Tnh pH dung dch NaH2Xit 0,010 M. Cho Ta c: p dng cng thc (2.2) ta c *

Cch 1: Gii phng trnh bc 3Thay cc gi tr vo ta cGii PT c Bin i t cng thc (a) ta c phng trnh:*

Cch 2:Gii theo phng php gn ng lin tcThay vo (b) ta c *

Ta c Vy TRNG HP 3:Ta cGii PT bc cao (theo phng php tip tuyn Newton)Cch 1*

Gii theo phng php gn ng lin tcCch 2Hay*

V d 2: Tnh pH dung dch KH3A 0,010 M. Cho p dng cng thc (2.3) ta c *

Cch 1: Gii phng trnh bc 3Thay cc gi tr vo ta cGii PT c Bin i t cng thc (a) ta c phng trnh:*

Cch 2:Gii theo phng php gn ng lin tcThay vo (b) ta c *

Ta c Vy TRNG HP 4:Ta cGii PT bc cao (theo phng php tip tuyn Newton)Cch 1*

Gii theo phng php gn ng lin tcCch 2Hay*

V d 3: Tnh pH trong dung dch KH2A 5.10-3M bit :

Ta c: *

p dng cng thc (2.4) ta c Cch 1: Gii phng trnh bc 3Bin i t cng thc (a) ta c phng trnh:Thay cc gi tr vo ta c*

Gii PT c Cch 2:Gii theo phng php gn ng lin tcThay vo (b) ta c *

Ta c Vy 3.Mt s tnh ton nhanh pH h dung dch mui axit HA- [2]

T cng thc tng qut (2), p dng n = 1, ta c: Cc qu trnh xy ra: *

V d 1:[2] Tnh pH dung dch NaHA 0,100 M.Cho Dng 1:Gii theo phng php gn ng lin tc*

Bc 1: Theo phng php gn ng lin tc: Thay vo (3.1)T gi tr h1,tnh li Bc 2 :Thay vo (3.1) tnh c h2 =4,97.10-5=10-4,3M kt qu lp li.Vy pH=4,3Vy trong dng 1 ny, nuKhichp nhn*

Dng 2:V d 2: [2] Tnh pH trong dung dch KHA 10-2M bit Ka1 = 10-7 ,Ka2 = 10-12,6.Bc 1: Theo phng php gn ng lin tc: Chp nhnThay vo (3.2)hay Ka1-1Cm >> 1*

Bc 2 :Thay vo (3.2) tnh c h2 =3,543.10-10 = 10-9,45M kt qu lp li.Vy pH=9,45KhiT gi tr h1,tnh li hay Ka1-1C >> 1*

Dng 3:Thay vo (4)hay Ka1-1Cm >> 1 pH = (pKa1 + pKa2) / 2 = (6,34 + 10,33)/2 = 8,34. Vy trong trng hp 3 ny, nng dung dch mui axit khng nh hng n pH ca h.Khi*

Dng 4:hay Ka1-1Cm 1V [HA- ]l cht in li yu nn chp nhn4.2. H c dung dch mui axit

4.1. pH ca dung dch mui axit

4.Mt s dng bi tp dung dch mui axit

*

Gi a l khi lng Na2HPO4. 12H2O phi em ha tan.H3PO4 c pKa1= 2,15; pKa2 = 7,21 ; pKa3 = 12,36.Nhn xt: pH = ( pKa1 + pKa2) / 2 = (2,15+ 7,21)/2 = 4,68.Nn thnh phn chnh ca h l: 0,05.0,1 a/358 Ta c: 0,005 = a/358 a = 1,79g. Bi 1: [3] Tnh s gam Na2HPO4. 12H2O phi ha tan trong 100mL dung dch H3PO4 0,05M sao cho pH ca dung dch thu c bng 4,68. Cho H3PO4 c: pKa1= 2,15; pKa2 = 7,21 ; pKa3 = 12,36.Gii:4.2.1.Tnh khi lng cht *

Bi 2: [2] Tnh s gam axit tactric H2C4H4O6 cn ly khi ha tan vo 50,00ml dung dch NaOH 1,00M th pH ca dung dch thu c l 3,71 (b qua s thay i ca th tchGii: Gi x l s gam axit cn ly Nhn xt Thnh phn chnh trong dung dch l mui axit NaHC4H4O6,xy ra theo phng trnh :H2C4H4O6 + NaOH NaHC4H4O6 + H2O*

Bi 3: [3] a, Tnh th tch NaOH 0,025M cn trung ha hon ton 25,mL dung dch H3AsO4 0,02M. Tnh pH ti thi im . b, Tnh th tch NaOH 0,025M trung ha 25mL dung dch H3AsO4 trn n pH1 = 6,94 v n pH2 = 9,22.4.2.1.Tnh pH, th tcha, Phn ng: H3AsO4 + 3NaOH Na3AsO4 + 3H2O trung ha hon ton 25mL H3AsO4 0,02M cn th tch NaOH l: V = 0,02.25.3/0,025 = 60mLTi thi im trung ha hon ton, thnh phn ca h: AsO43- c C = 0,02.25/(60+25) = 5,88.10-3MGii:*

Do Kb1 >> Kb2 >> Kb3 nn tnh theo (1): *

b, Ta thy: pH1 = 6,94 = pKa2 nn t cn bng: Thnh phn chnh ca h m: H2AsO4- v HAsO42- Ngha l lng NaOH cho vo trung ha ht nc 1 v nc 2 nn Phn ng trung ha: 2H3AsO4 + 3NaOH Na2HAsO4 + NaH2AsO4 + 3H2O*

Nu trung ha n 3.0,02.25 = 2.0,025.V1 V1 = 30mL Nn thnh phn ca h l mui axit HAsO42- Vy phn ng trung ha n ht nc 2: H3AsO4 + 2 NaOH Na2HAsO4 + 2H2O Th tch NaOH cn dng l: V2 = 0,02.25.2/0,025 = 40(mL) Bi 4: [2] Trn 100ml dung dch NaOH 0,102M vi 100ml dung dch NaHCO3 0,100M.Tnh pH v cn bng trong dung dch thu c.Gii:*

Phn ngTPGH: Mi trng baz nn b qua s phn li ca ncCc qu trnh:

*

Bi 5: [3] Tnh s ml dung dch HCl 0,010M phi thm vo 50,00ml dung dch Na2HPO4 0,020 M pH ca dung dch thu c bng 7,00.

Gii: Gi V ml l th tch HCl cn ly.Chn mc khng :KP:*

Bi 6: [2] Tnh s mL dung dch (NH4)2SO4 0,1M cn phi thm vo 100mL dung dch Na2S 0,1M pH ca h gim 0,76 n v.Gii:Khi cha thm (NH4)2SO4 , trong dung dch Na2S c cc cn bng:*

So snh 3 cn bng trn ta thy: cn bng (1) l ch yu: h = 10-12,76 v [H2S] = = 106,98 = 1,04.107 M.*

Go V l s mL dung dch (NH4)2SO4 0,1M cn thm vo 100mL dung dch Na2S 0,1M pH = 12,76 - 0,76 = 12V pH = 12 nn sau phn ng vi NH4+ cn d S2-: *

Cc qu trnh xy ra trong dung dch: C th b qua cn bng (4),(5) Chn mc khng: HS-, NH3 v H2O: Ta c: h = [OH-] - [H2S] + [S2-] - (10-0,2V)/(100+V) - [NH4+]*

V: [H2S]

C.KT LUN

Thng qua bi tiu lun chng em tm hiu, la chn v trnh by mt s vn tnh ton quan trng v thng gp trong dung dch axit, baz a chc v mui axit nh tnh pH, nng cc cu t, khi lng, th tch, c th l:+ Xy dng cng thc tng qut tnh ton trong dung dch axit, baz a chc v mui axit cho tng trng hp.+ Xy dng cng thc tnh nhanh cho tng trng hp c th.+ a ra cc v d c th cho tng trng hp vi nhiu cch gii khc nhau.+ a ra v gii cc bi tp vn dng lin quan n tnh ton trong dung dch axit, baz a chc v mui axit.*

D.TI LIU THAM KHO[1].Nguyn Tinh Dung,(1981), Ha hc phn tch, phn I. L thuyt c s (Cn bng ion), NXBGD, H Ni.[2].Nguyn Tinh Dung o Th Phng Dip, (2007), Ha hc phn tch, Cu hi v bi tp (Cn bng ion trong dung dch), NXB HSP.[3].Nguyn nh Luyn, Ng Vn T, (2011), Ha hc phn tch, NXB H Hu.*

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Dung Dich Ax Bazo Da Chuc