·ÖÎö»¯Ñ§×Ôѧָµ¼·½°¸

x??CH2x?CH2c???N???OH??OH??K?NH??1.42.4?64??64b1.0?10?91.0??1.0?10?9?1.4?10?92.42.30x??CH2?6N4??x?CH2?6N4H?¦Â??? 0.28?0.50?mol/L?1.01.42.30??2.42.4m??CH2?6N4??0.50?V?HCl??0.50?200?140.19?14?g?10001.4?2002.4?4.9?mL?12Àý12£ºÄ³Ò»ÈÜÒºÓÉHCl,KH2PO4ºÍHAc»ìºÏ¶ø³É,ÆäŨ¶È·Ö±ðΪc(HCl) = 0.10 mol/L, c(KH2PO4) = 1.0¡Á10-3mol/L, c(HAc) = 2.0¡Á10-6mol/L ¡£¼ÆËã¸ÃÈÜÒºµÄpH¼°[Ac-],[PO43-]¸÷Ϊ¶àÉÙ? (ÒÑÖªH3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20, 12.36, HAc µÄpKa = 4.74) ½â

c(HCl)>>c(H2PO4-),ÈÜÒº[H+]ÓÉHCl¾ö¶¨ pH = 1.00 10-4.74

[Ac-]= cx0=10-5.7©¤©¤©¤©¤©¤©¤©¤©¤ =10-9.44 10-4.74+10-1.0 = 3.6¡Á10-10 (mol/L) [PO43-]= cx0

10-21.68

= 10-3.0©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤

10-3.0+10-4.12+10-10.32+10-21.68

Àý13£ºÄ³·ÖÎö¹¤×÷ÕßÓûÅäÖÆpH = 0.64µÄ»º³åÈÜÒº¡£³ÆÈ¡´¿ÈýÂÈÒÒËá(CCl3COOH) 16.3 g,ÈÜÓÚË®ºó,¼ÓÈë2.0 g¹ÌÌåNaOH,ÈܽâºóÒÔˮϡÖÁ1 L.ÊÔÎÊ:

(1) ʵ¼ÊÉÏËùÅ仺³åÈÜÒºµÄpHΪ¶àÉÙ? (2) ÈôÒªÅäÖÆpH = 0.64µÄÈýÂÈÒÒËỺ³åÈÜÒº,Ðè¼ÓÈë¶àÉÙĦ¶ûÇ¿Ëá»òÇ¿¼î? [ÒÑÖªMr(CCl3COOH) = 163.4, Ka(CCl3COOH) = 0.23, Mr(NaOH) = 40.0] ½â(1)½â·¨¢Ù Éè¼ÓÈëNaOHµÄŨ¶ÈΪcb(mol/L),Ôòcb = 2.00/40.0 = 0.050 (mol/L) 16

c(CCl3COOH) = ©¤©¤©¤©¤ = 0.10 (mol/L) 163.4

ÈÜÒºµÄÖÊ×ÓÆ½ºâʽΪ: [H+] = [CCl3COO-]-cb 0.23

[H+] = ©¤©¤©¤©¤©¤¡Á0.10-0.050 [H+]+0.23

½â·½³Ì,[H+] = 0.036 mol/L pH = 1.44

½â·¨¢Ú ÒòCl3CCOOH¼ÓÈëNaOHºó,¹¹³É»º³åÈÜÒº,ca = cb = 0.050 mol/L ca-[H+]

¿ÉÖ±½Ó´úÈë [H+] = Ka ©¤©¤©¤©¤©¤ ¼ÆËã, cb+[H+] 0.050-[H+]

[H+] = 0.23¡Á©¤©¤©¤©¤©¤©¤©¤©¤, ½âµÃ[H+] = 0.037 mol/l, pH = 1.43 0.050+[H+]

16

(2) Éè¼ÓÈëHClµÄŨ¶ÈΪca(mol/L), ¸ù¾ÝÖÊ×ÓÆ½ºâʽ Ka

[H+] = ©¤©¤©¤©¤©¤¡Ác(CCl3COOH)+ca [H+]+Ka 0.23

0.23 = ©¤©¤©¤©¤©¤©¤¡Á0.10+ca 0.23+0.23

ca = 0.18 mol/L ¼ÓÈë n(HCl) = 0.18¡Á1 £«0.05= 0.23 (mol)

Àý14£º³ÆÈ¡0.5000 gij´¿Ò»ÔªÈõËáHB, ÈÜÓÚÊÊÁ¿Ë®ÖÐ, ÒÔ0.1000 mol/L NaOHÈÜÒºµÎ¶¨, ´ÓµçλµÎ¶¨ÇúÏߵõ½ÏÂÁÐÊý¾Ý:

V (NaOH)/mL 0.00 20.47 40.94(»¯Ñ§¼ÆÁ¿µã) pH 2.65 4.21 8.43 ÊÔ¼ÆËã¸ÃÒ»ÔªÈõËáHBµÄĦ¶ûÖÊÁ¿ºÍpKaÖµ¡£ ½âÉèHBµÄĦ¶ûÖÊÁ¿ÎªM,¸ù¾Ý»¯Ñ§¼ÆÁ¿¹ØÏµ,¿ÉµÃ 0.500

©¤©¤©¤©¤©¤¡Á1000 = 0.100¡Á40.94 M(HB)

0.500¡Á1000

M(HB) = ©¤©¤©¤©¤©¤©¤©¤ = 122.1 (g/mol) 0.100¡Á40.94

ÓÉÉϱí¿ÉÖª,»¯Ñ§¼ÆÁ¿µãʱÏûºÄNaOHÈÜҺΪ40.94 mL,ÖкÍÒ»°ëʱµÄV(NaOH) = 20.47 mL,´Ëʱ[HB]=[B-] pKa= 4.21

Àý15£ºÈç¹û½«25 mL 0.40 mol/L H3PO4Óë30 mL 0.50 mol/L Na3PO4»ìºÏ²¢ÓÃˮϡÊÍÖÁ100 mL, ÒÆÈ¡25 mL½øÐеζ¨,ÊÔÎÊ: (1) ÈôÓü׻ù³ÈΪָʾ¼Á,ÐèºÄÓÃ0.10 mol/L HCl¶àÉÙºÁÉý? (2) ÈôÓ÷Ó̪Ϊָʾ¼Á,ÐèºÄÓÃ0.10 mol/L NaOH¶àÉÙºÁÉý?

½âÈÜÒºÖУºn(H) = 25¡Á0.40¡Á3 = 30 (mmol) n(PO4) = 30¡Á0.50+25¡Á0.40 = 25 (mmol) ËùÒÔ»ìºÏºóÉú³É HPO42- 20 mmol, H2PO4- 5 mmol 20/4

1. V(HCl) = ©¤©¤©¤©¤ = 50 (mL) 0.10 5/4

2. V(NaOH) = ©¤©¤©¤©¤ = 12.5 (mL)

Àý16£ºÊÔÉè¼ÆÒ»ÖֲⶨÑÎËáºÍÅðËá»ìºÏÈÜÒºÁ½×é·ÖŨ¶ÈµÄ¼òµ¥·½°¸(°üÀ¨µÎ¶¨¼Á¡¢±ØÒªÊÔ¼Á¡¢Ö¸Ê¾¼Á¼°ÖÕµãʱÑÕÉ«µÄ¸Ä±ä)¡£

´ðHCl NaOH±ê×¼ÈÜÒº,V1(mL) Cl- NaOH±ê×¼ÈÜÒº,V2(mL)

©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤¡ú ©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤¡ú H2BO3- H3BO3 ¼×»ù³Èºì±ä»Æ H3BO3 ¸ÊÓÍ,·Ó̪Óɺì¡ú»Æ¡ú·Ûºì

²âHCl ²âH3BO3

Àý17£ºÉè¼Æ²â¶¨HCl-NH4Cl»ìºÏÒºÖÐÁ½×é·ÖŨ¶ÈµÄ·ÖÎö·½°¸¡£(Ö¸³öµÎ¶¨¼Á¡¢±ØÒªÌõ¼þ¡¢Ö¸Ê¾¼Á) ´ðµÚÒ»²½

©°©¤©¤©¤©¤©¤©¤ ÏÈÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁ»Æ (²âHCl) ©¦ ¼×»ùºì ¡ý

©°©¤©¤©¤©¤©´ ©¦ HCl ©¦

17

©¦ ©¦ ©¦ NH4Cl ©¦ ©¸©¤©¤©¤©¤©¼ ¡ü

©¦µÚ ¶þ ²½ K2CrO4

©¸©¤©¤©¤©¤©¤©¤©¤©¤ÔÙÓÃAgNO3±ê×¼ÈÜÒºµÎ¶¨µÚÒ»²½µÎ¶¨ºóµÄÈÜÒºÖÁשºìÉ« ³Áµí³öÏÖ(²âCl-×ÜÁ¿) »òÖ±½ÓÓü×È©·¨²â NH4Cl

Îå¡¢×Ô²âÁ·Ï°Ìâ

Ò»£®Ñ¡ÔñÌâ

1.HPO42-µÄ¹²éî¼îÊÇ-----------------------------------( ) (A) H2PO4- (B) H3PO4 (C) PO43- (D) OH- 2. OH-µÄ¹²éîËáÊÇ--------------------------------( ) (A) H+ (B) H2O (C) H3O+ (D) O2-

3.ÔÚÏÂÁи÷×éËá¼î×é·ÖÖÐ,ÊôÓÚ¹²éîËá¼î¶ÔµÄÊÇ-----------( )

(A) HCN-NaCN (B) H3PO4-Na2HPO4 (C)+NH3CH2COOH-NH2CH2COO- (D)H3O+-OH- 4.Ë®ÈÜÒº³ÊÖÐÐÔÊÇÖ¸--------------------------( )

(A) pH = 7 (B) [H+] = [OH-] (C) pH+pOH = 14 (D) pOH = 7 5.ÔÚH2C2O4ÈÜÒºÖÐ,ÏÂÁлî¶ÈϵÊýµÄÅÅÁÐ˳ÐòÕýÈ·µÄÊÇ---------------( ) (A) ?(HC2O4-) > ?(H+) > ?(C2O42-) (B) ?(H+) > ?(HC2O4-) > ?(C2O42-) (C) ?(C2O42-) > ?(HC2O4-) > ?(H+) (D) ?(HC2O4-) > ?(C2O42-) > ?(H+)

6.ijMA2ÐÍ(M2+¡¢A-)µç½âÖÊÈÜÒº,ÆäŨ¶Èc(MA2) = 0.10mol/L, Ôò¸ÃÈÜÒºµÄÀë×ÓÇ¿¶ÈΪ-----( ) (A) 0.10 mol/L (B) 0.30 mol/L (C) 0.40 mol/L (D) 0.60 mol/L 7.ÔÚÒ»¶¨Î¶ÈÏÂ,Àë×ÓÇ¿¶ÈÔö´ó,´×ËáµÄŨ¶È³£ÊýKa½«---------( )

(A) Ôö´ó (B) ¼õС (C) ²»±ä (D) ¼õСÖÁÒ»¶¨³Ì¶ÈºóÇ÷ÓÚÎȶ¨ 8.ÔÚÁ×ËáÑÎÈÜÒºÖÐ,HPO42-Ũ¶È×î´óʱµÄpHÊÇ------( ) (ÒÑÖªH3PO4µÄ½âÀë³£ÊýpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36) (A) 4.66 (B) 7.20 (C) 9.78 (D) 12.36

9.ÒÑÖªH3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36¡£½ñÓÐÒ»Á×ËáÑÎÈÜÒºµÄpH = 4.66, ÔòÆäÖ÷Òª´æÔÚÐÎʽÊÇ-------( ) (A) HPO42- (B) H2PO4- (C) HPO42- + H2PO4 (D) H2PO4-+ H3PO4 10.¶þÒÒÈý°±ÎåÒÒËá( ETPA, ÓÃHL±íʾ)µÄpKa1~pKa5·Ö±ðÊÇ1.94¡¢2.87¡¢4.37¡¢8.69ºÍ10.56, ÈÜÒºÖеÄH3L2-×é·ÖŨ¶È×î´óʱµÄpHÊÇ-----------------( )

(A) 2.87 (B) 3.62 (C) 5.00 (D) 9.62

11.pH = 7.00µÄH3AsO4ÈÜÒºÓйØ×é·ÖƽºâŨ¶ÈµÄ¹ØÏµÓ¦¸ÃÊÇ--------( ) (ÒÑÖª H3AsO4 pKa1 = 2.20, pKa2 = 7.00, pKa3 = 11.50)

(A) [H3AsO4] = [H2AsO4-] (B) [H2AsO4-] = [HAsO42-] (C) [HAsO42-] >[H2AsO4-] (D) [H3AsO4] >[HAsO42-]

12.ÒÑÖªEDTAµÄ¸÷¼¶½âÀë³£Êý·Ö±ðΪ10-0.9¡¢10-1.6¡¢10-2.0¡¢10-2.67¡¢10-6.16ºÍ10-10.26, ÔÚpH = 2.67~6.16µÄÈÜÒºÖÐ, EDTA×îÖ÷ÒªµÄ´æÔÚÐÎʽÊÇ----------------( )

(A) H3Y- (B) H2Y2- (C) HY3- (D) Y4-

13. c(Na2CO3) = 0.1mol/LµÄNa2CO3ÈÜÒºµÄÎïÁÏÆ½ºâʽÊÇ-------------------( )

(A) 0.1 mol/L = [CO32-] = 2[Na+] (B) 0.1 mol/L = [CO32-]+[Na+] (C) 0.1 mol/L = [H2CO3]+[HCO3-]+[CO32-] = [Na+]/2 (D) 0.1 mol/L = [H2CO3]+[HCO3-]+2[CO32-] = [Na+]

18

C14.ÓÃNaOHÈÜÒºµÎ¶¨H3PO4ÈÜÒºÖÁpH = 4.7ʱ,ÈÜÒºµÄ¼ò»¯ÖÊ×ÓÌõ¼þΪ ( ) (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12¡¢7.20¡¢12.36)

(A) [H3PO4] = [H2PO4-] (B) [H2PO4-] = [HPO42-] (C) [H3PO4] = [HPO42-] (D) [H3PO4] = 2[PO43-] 15. NH4H2PO4Ë®ÈÜÒºÖÐ×îÖ÷ÒªµÄÖÊ×Ó×ªÒÆ·´Ó¦ÊÇ-------------------( ) (ÒÑÖªH3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36;NH3µÄpKbΪ4.74)

(A) NH4++H2PO4- = NH3 +H3PO4 (B) H2PO4-+H2PO4- = H3PO4 +HPO42-

(C) H2PO4-+H2O = H3PO4 +OH- (D) H2PO4-+H2O = HPO42- +H3O+

16.½«1.0 mol/L NaAcÓë0.10 mol/L H3BO3µÈÌå»ý»ìºÏ,ËùµÃÈÜÒºpHÊÇ-( )

[pKa(HAc) = 4.74, pKa(H3BO3) = 9.24] (A) 6.49 (B) 6.99 (C) 7.49 (D) 9.22 17.0.10 mol/L HClºÍ1.0 mol/L HAc(pKa = 4.74)»ìºÏÒºµÄpHΪ---( ) (A) 2.37 (B) 1.30 (C) 1.00 (D) 3.37

18.pH=1.00µÄHClÈÜÒººÍpH=13.00 µÄ NaOH ÈÜÒºµÈÌå»ý»ìºÏ,ËùµÃÈÜÒºµÄpHÊÇ- ( ) (A) 14 (B) 12 (C) 7 (D) 6

19.ÒÑÖªH3PO4µÄKa1 = 7.6¡Á10-3, Ka2 = 6.3¡Á10-8, Ka3 = 4.4¡Á10-13,ÈôÒÔNaOHÈÜÒºµÎ¶¨H3PO4ÈÜÒº,ÔòµÚ¶þ»¯Ñ§¼ÆÁ¿µãµÄpHԼΪ-------( )

(A) 10.7 (B) 9.7 (C) 7.7 (D) 4.9 20.ÒÔÏÂÈÜҺϡÊÍ10±¶Ê±pH¸Ä±ä×îСµÄÊÇ------------------( )

(A) 0.1 mol/L NH4AcÈÜÒº (B) 0.1 mol/L NaAcÈÜÒº (C) 0.1 mol/L HacÈÜÒº (D) 0.1 mol/L HClÈÜÒº 21. 1.0 mol/L NH4HF2ÈÜÒºµÄpHÊÇ-------------( )

[pKa(HF) = 3.18, pKb(NH3) = 4.74] (A) 1.59 (B) 3.18 (C) 6.22 (D)9.26 22.Áù´Î¼×»ùËİ·[(CH2)6N4]»º³åÈÜÒºµÄ»º³åpH·¶Î§ÊÇ---------( )

?pKb[(CH2)6N4] = 8.85? (A) 4~6 (B) 6~8 (C) 8~10 (D) 9~11 23.ÏÂÁÐÑεÄË®ÈÜÒº»º³å×÷ÓÃ×îÇ¿µÄÊÇ-----------------( ) (A) NaAc (B) Na2CO3 (C) Na2B4O7¡¤10H2O (D) Na2HPO4

24.½ñÓûÓÃNa3PO4ÓëHClÀ´ÅäÖÆpH = 7.20µÄ»º³åÈÜÒº,ÔòNa3PO4ÓëHClÎïÖʵÄÁ¿Ö®±Èn(Na3PO4)¡Ãn(HCl)Ó¦µ±ÊÇ- ( ) (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) (A) 1:1 (B) 1:2 (C) 2:3 (D) 3:2 25.ÓûÅäÖÆpH=5.1µÄ»º³åÈÜÒº,×îºÃÑ¡Ôñ--------( )

(A) Ò»ÂÈÒÒËá(pKa = 2.86) (B) °±Ë®(pKb = 4.74) (C) Áù´Î¼×»ùËİ·(pKb = 8.85) (D) ¼×Ëá(pKa = 3.74)

26.ÔÚÒÔÏ´¿Á½ÐÔÎïÈÜÒºÖÐ,ÄÜ×÷Ϊ±ê×¼»º³åÈÜÒºµÄÊÇ-----------( )

(A) NaH2PO4 (B) Na2HPO4 (C) °±»ùÒÒËáż¼«Àë×Ó (D) ¾ÆÊ¯ËáÇâ¼Ø ÒÑÖª: H3PO4 °±»ùÒÒËá ¾ÆÊ¯Ëá pKa1 2.12 2.35 3.04 pKa2 7.20 9.60 4.37 pKa3 12.36 27.½«Å¨¶ÈÏàͬµÄÏÂÁÐÈÜÒºµÈÌå»ý»ìºÏºó,ÄÜʹ·Óָ̪ʾ¼ÁÏÔºìÉ«µÄÈÜÒºÊÇ----------( )

(A) °±Ë®+´×Ëá (B) ÇâÑõ»¯ÄÆ+´×Ëá (C) ÇâÑõ»¯ÄÆ+ÑÎËá (D) Áù´Î¼×»ùËİ·+ÑÎËá 28.Ç¿ËáµÎ¶¨Èõ¼î,ÒÔÏÂָʾ¼ÁÖв»ÊÊÓõÄÊÇ------( )

(A) ¼×»ù³È (B) ¼×»ùºì (C) ·Ó̪ (D) äå·ÓÀ¶ (pT = 4.0)

29.ÓÃ0.1 mol/L HClµÎ¶¨0.1 mol/L NH3Ë®(pKb = 4.7)µÄpHͻԾ·¶Î§Îª6.3~4.3, ÈôÓÃ0.1 mol/L HClµÎ¶¨0.1

mol/L pKb = 2.7µÄij¼î, pHͻԾ·¶Î§Îª-( )

(A) 6.3~2.3 (B) 8.3~2.3 (C) 8.3~4.3 (D) 4.3~6.3

30.ÔÚÏÂÁжàÔªËá»ò»ìºÏËáÖÐ,ÓÃNaOHÈÜÒºµÎ¶¨Ê±³öÏÖÁ½¸öµÎ¶¨Í»Ô¾µÄÊÇ ( )

19

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