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ÓÃNaOHÈÜÒºµÎ¶¨Ä³ÈõËáHA,ÈôÁ½ÕßŨ¶ÈÏàͬ,µ±µÎ¶¨ÖÁ50£¥Ê±ÈÜÒºpH = 5.00; µ±µÎ¶¨ÖÁ100£¥Ê±ÈÜÒºpH = 8.00;µ±µÎ¶¨ÖÁ200£¥Ê±ÈÜÒºpH = 12.00,Ôò¸ÃËápKaÖµÊÇ---( A ) (A) 5.00 (B) 8.00 (C) 12.00 (D) 7.00 53

½ñÓû²â¶¨pH¡Ö9µÄÈÜÒºµÄpH,ÏÂÁÐÈÜÒºÖÐÊÊÒËУÕýpH¼Æ(¶¨Î»)µÄÊÇ------( C ) (A) 0.050mol/L ÁÚ±½¶þ¼×ËáÇâ¼ØÈÜÒº

(B) 0.025mol/L KH2PO4-0.025mol/L Na2HPO4ÈÜÒº (C) 0.01mol/L ÅðɰÈÜÒº

(D) 0.10mol/L NH3-0.10mol/L NH4ClÈÜÒº 54

HCl¡¢H2SO4¡¢HNO3¼°HClO4µÄÇø·ÖÐÔÈܼÁÊÇ--------------------------------( C ) (A) Ë® (B) ÒÒ´¼ (C) ±ù´×Ëá (D) Òº°± 55

ÏÂÁÐÄÄÖÖÈܼÁ,ÄÜʹHAc¡¢H3BO3¡¢HClºÍH2SO4ËÄÖÖËáÏÔʾ³öÏàͬµÄÇ¿¶ÈÀ´?--( B ) (A) ´¿Ë® (B) Òº°± (C) ¼×»ùÒ춡ͪ (D) ÒÒ´¼ 1

±û¶þËá[CH2(COOH)2]µÄpKa1 = 3.04, pKa2 = 4.37, Æä¹²éî¼îµÄKb1 = _ 2.3¡Á10-10 (10-9.63) __, Kb2 = _1.1¡Á10-11 (10-10.96) __¡£ 2

ßÁà¤Å¼µª¼ä±½¶þ·Ó(PAR)µÄËá½âÀë³£ÊýpKa1 ,pKa2 ,pKa3·Ö±ðΪ3.1, 5.6, 11.9, ÔòÆäÖÊ×Ó»¯³£ÊýK1 = __8¡Á1011___, __4¡Á105___, ÀÛ»ýÖÊ×Ó»¯³£Êý?2 = ___3¡Á1017___, 3

pKa(HCOOH) = 3.77, pKb(HCOO-) = __10.23__; NaOHµÎ¶¨HCOOH·´Ó¦µÄKt = __1010.23___; HClµÎ¶¨HCOO-·´Ó¦µÄKt = __103.77___¡£ 4

ÏÂÁÐÎïÖÊÖÐÊôÓÚËáµÄÓÐ__ B,D,F __, ÊôÓÚ¼îµÄÓÐ__ A,C,E,H ___, ÊôÓÚÁ½ÐÔÎïÖʵÄÓÐ_ G __¡£(Ó÷ûºÅA,B,¡­,±íʾ) (A) ßÁठ(B) ßÁà¤ÑÎËáÑÎ (C) Áù´Î¼×»ùËİ·[(CH2)6N4]

(D) Áù´Î¼×»ùËİ·ÑÎËáÑÎ

(E) ôǰ·(NH2OH) (F) ÑÎËáôǰ· (G) °±»ù¼×Ëá(NH3+COO-) (H) °±»ù¼×ËáÄÆ 5

ij¶þÔªËáH2AµÄpKa1ºÍpKa2·Ö±ðΪ4.60ºÍ8.40,ÔÚ·Ö²¼ÇúÏßÉÏH2AÓëHA-ÇúÏß½»µãpHΪ__4.60__,HA-ÓëA2-ÇúÏß½»µãµÄpHΪ__8.40__,H2AÓëA2-µÄ½»µãpHΪ__6.50__,HA- ´ï×î´óµÄpH ÊÇ__6.50__¡£ 6

²ÝËá(H2C2O4)µÄpKa1ºÍpKa2·Ö±ðÊÇ1.2ºÍ4.2¡£ÇëÌîдÒÔÏÂÇé¿öµÄpH»òpH·¶Î§¡£

C2O42-ΪÖ÷ pH > 4.2 7

±È½ÏÒÔϸ÷¶ÔÈÜÒºµÄpH´óС(Ó÷ûºÅ >¡¢ = ¡¢< ±íʾ) (1) ͬŨ¶ÈµÄNaH2PO4(a)ºÍNH4H2PO4(b): (2) ͬŨ¶ÈµÄNa2HPO4(c)ºÍ(NH4)2HPO4(d): 8

25

(a)__ =__(b)

(c)__ >__(d)

[HC2O4-]Ϊ×î´óÖµ pH = 2.7 [HC2O4-] = [C2O42-] pH = 4.2 [H2C2O4] = [C2O42-] pH = 2.7 ???K2 =

?3? = __4¡Á1020___¡£

[ÒÑÖªpKb(NH3) = 4.74,H3PO4µÄpKa1~pKa3·Ö±ðÊÇ 2.16,7.20,12.36]

60 mL 0.10 mol/L Na2CO3Óë40 mL 0.15 mol/L HClÏà»ìºÏ, ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ___[H+]+[H2CO3] = [OH-]+[CO32-] ____¡£ 9

ÓÃNaOHµÎ¶¨¶þÂÈÒÒËá(HA, pKa = 1.3)ºÍ NH4Cl »ìºÏÒºÖеĶþÂÈÒÒËáÖÁ»¯Ñ§¼ÆÁ¿µãʱ, ÆäÖÊ×ÓÌõ¼þʽÊÇ___[H+]+[HA] = [NH3]+[OH-] ____¡£ 10

ÓÃÏ¡ H2SO4 ÈÜÒºµÎ¶¨ Na2CO3 ÈÜÒºÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µãʱ,ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ: ____[H+]+[HSO4-] = [OH-]+[HCO3-]+2[CO32-]____¡£ 11

ij(NH4)2HPO412

0.1 mol/L H2SO4ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ__[H+] = [SO42-]+[OH-]+0.1mol/L ____¡£ 13

Ñ¡ÔñÏÂÁÐÈÜÒº[H+]µÄ¼ÆË㹫ʽ(ÇëÌîA,B,C) (1) 0.10 mol/L ¶þÂÈÒÒËá (2) 0.10 mol/L NH4Cl (3) 0.10 mol/L NaHSO4 (4) 1.0¡Á10-4mol/L H3BO3 £¨A£© 14

ÒÑÖªEDTAµÄpKa1~pKa6·Ö±ðÊÇ0.9, 1.6, 2.0, 2.67, 6.16ºÍ10.26, 0.10 mol/L EDTA¶þÄÆÑÎ(Na2H2Y¡¤2H2O)ÈÜÒºµÄpHÊÇ___4.42___, [Y]Ϊ___10-8.58____mol/L¡£ 15

»º³åÈÜÒºÓ¦ÓÐ×ã¹»µÄ»º³åÈÝÁ¿,ͨ³£»º³å×é·ÖµÄŨ¶ÈÔÚ__0.01~1mol/L __Ö®¼ä¡£ 16

½ñÓûÅäÖÆpH = 5µÄ»º³åÈÜÒº¿ÉÑ¡Ôñ___ NaAc-HAc»ò(CH2)6N4-HCl ___ÅäÖÆ¡£ 17

ÏÂÁÐÇéÐÎÖÐ,ÈÜÒºµÄpH½«·¢Éúʲô±ä»¯(Ôö´ó¡¢¼õС»ò²»±ä): ÔÚ50mL 0.1mol/L HClO4ÈÜÒºÖмÓÈë10mL 0.1mol/L HCl 18

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20 mL 0.50 mol/L H3PO4ÈÜÒºÓë5.0 mL 1.0 mol/LµÄNa3PO4ÈÜÒºÏà»ìºÏºó, ÆäpHÊÇ__4.66 ___¡£(H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12, 7.20, 12.36)

×ÜPO4£½20¡Á0.5+5¡Á1.0=15 ×ÜH£½30£¬ËùÒÔÉú³ÉH2PO4- 20

10 g(CH2)6N4¼ÓÈëµ½4.0 mL 12 mol/L HClÈÜÒºÖÐ,Ï¡ÊÍÖÁ100 mLºó, Æä pH ֵΪ__4.85____¡£ {Mr[(CH2)6N4] = 140.0, pKb[(CH2)6N4] = 8.85}

10£¯140£½0.0715 4¡Á12£¯1000£½0.048 0.0715-0.048=0.0235 21

25 mL 0.40 mol/L H3PO4ÈÜÒººÍ30 mL 0.50 mol/L Na3PO4ÈÜÒº»ìºÏ²¢Ï¡ÊÍÖÁ 100mL,´ËÈÜÒºµÄpHÊÇ__7.8___¡£ (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12, 7.20, 12.36)

×ÜPO4£½25¡Á0.40+30¡Á0.50£½25 ×ÜH£½30 HPO42-£½20 H2PO4-£½5 22

(CH2)6N4(Áù´Î¼×»ùËİ·)µÄpKb = 8.9,ÓÉ(CH2)6N4-HCl×é³ÉµÄ»º³åÈÜÒºµÄ»º³å·¶Î§ÊÇ__4.1©¤6.1 __,ÓûÅäÖÆ¾ßÓÐ?maxµÄ¸ÃÖÖ»º³åÈÜÒº,Ó¦ÓÚ100 mL 0.1 mol/L(CH2)6N4ÈÜÒºÖмÓÈë___5___mL 1 mol/L HCl¡£

26

___²»±ä___

(pKa = 1.30) __B__ (pKa = 9.26) __A__ (pKa2 = 1.99) __B___ (pKa = 9.24)

__C___

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c[(NH4)2HPO4]

=

c

mol/L,ÆäÎïÁÏÆ½ºâʽΪ___2c = [NH3]+[NH4+] = 2([H3PO4]+[H2PO4-]+[HPO42-]+[PO43-])_____; µçºÉƽºâʽΪ___[NH4+]+[H+] = [H2PO4-]+ 2[HPO42-] +3[PO43-]+[OH-]____¡£

Ka?c £¨B£© Ka(c?[??]) £¨C£© Ka?c?KW23

±ê×¼»º³åÈÜÒºÊÇÖ¸___ pH¾­×¼È·ÊµÑé²âµÃµÄ»º³åÈÜÒº___,Æä×÷ÓÃÊÇ__×÷Ϊ²â¶¨pHʱµÄ±ê×¼²ÎÕÕÈÜÒº_____¡£ 24

-da/dpH»òdb/dpH³ÆÎªÈÜÒºµÄ__»º³åÈÝÁ¿__,µ±___¹²éîËá¼î¶ÔŨ¶È±ÈΪ1__,¼´µ±pH = __ pKa __ ʱÓÐ×î´óÖµ,ÆäֵΪ__0.575c __¡£ 25

1.0mol/L NH4HF2ÈÜÒºµÄpHΪ__ 3.18__¡£ [pKa(HF) = 3.18, pKb(NH3) = 4.74]

»º³åÈÜÒº£ºË᣺HF£¬¼î£ºF-£¬NH4+µÄµçÀë¿ÉºöÂÔ 26

½«Ï±íÖÐËÄÖÖÈÜÒºÒÔˮϡÊÍ10±¶,ÇëÌîдÆäpH±ä»¯´óСµÄ˳Ðò,±ä»¯×î´óÕßΪ¡°1¡±¡¢×îСÕßΪ¡°4¡±¡£

ÈÜÒº 0.1 mol/L HAc 0.1 mol/L HAc + 0.1 mol/L NaAc 1.0 mol/L HAc + 1.0 mol/L NaAc 0.1 mol/L HCl 27

Ñ¡Ôñ[H+]µÄ¼ÆË㹫ʽ(ÇëÌîA,B,C,D)

1. 0.1000 mol/L HClµÎ¶¨0.1000 mol/L Na2CO3ÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µã------( B ) 2. 0.05000 mol/L NaOHµÎ¶¨0.05000 mol/L H3PO4ÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µã----( D ) 3. 0.1 mol/L HAc-0.1 mol/L H3BO3»ìºÏÒº------------------------------------( A ) 4. 0.1 mol/L HCOONH4ÈÜÒº------------------------------------------( C ) (A)[H+] = (B)[H+] = (C)[H+] = (D)[H+] = 28

±È½ÏÒÔϸ÷¶ÔÈÜÒºµÄpHµÄ´óС(ÓÃ>¡¢ = ¡¢<·ûºÅ±íʾ) (1)Ũ¶ÈÏàͬµÄNaAcºÍNH4AcÈÜÒº [pKa(HAc) = 4.74, pKb(NH3) = 4.74] 29

ÒÑÖª¼×»ù³ÈpK(HIn) = 3.4,µ±ÈÜÒºpH = 3.1ʱ[In-]/[HIn]µÄ±ÈֵΪ__0.5 __; ÈÜÒºpH = 4.4ʱ[In-]/[HIn]µÄ±ÈֵΪ___10___; ÒÀͨ³£¼ÆËãָʾ¼Á±äÉ«·¶Î§Ó¦ÎªpH = pK(HIn)¡À1,µ«¼×»ù³È±äÉ«·¶Î§Óë´Ë²»·û,ÕâÊÇÓÉÓÚ___ÓÉÓÚÈËÃǵÄÑÛ¾¦¶ÔºìÉ«±È»ÆÉ«ÁéÃô____¡£ 30

ÔÚÏÂÁеζ¨ÌåϵÖÐ,ÇëÑ¡ÓÃÊÊÒ˵Äָʾ¼Á:(´Ó¼×»ù³È¡¢¼×»ùºì¡¢·Ó̪ÖÐÑ¡Ôñ) (1) 0.01 mol/L HClµÎ¶¨0.01 mol/L NaOH (2) 0.1 mol/L HClµÎ¶¨0.1 mol/L NH3

˳Ðò 2 3 4 1 Ka(?)?c

Ka1?Ka2

Ka(?)?Ka(??)

Ka1Ka2?c/(Ka1+c)

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(2)Ũ¶ÈÏàͬµÄHAc-NaAcºÍHAc-NH4Ac»º³åÈÜÒº ___£½____

__¼×»ùºì»ò·Ó̪____ __¼×»ùºì(»ò¼×»ù³È)____ ___·Ó̪___

(3) 0.1 mol/L HClµÎ¶¨Na2CO3ÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µã (4) 0.1 mol/L NaOHµÎ¶¨H3PO4ÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µã 31

__¼×»ù³È____

ijËá¼îָʾ¼ÁHIn,µ±[HIn]/[In-] = 5ʱ,¿´µ½µÄÊÇ´¿ËáÉ«,µ±[In-]/[HIn] = 3ʱ, ¿´µ½µÄÊÇ´¿¼îÉ«¡£Èô¸Ãָʾ¼ÁµÄ½âÀë³£ÊýΪpKa, 27

Ôò±äÉ«·¶Î§ÊÇ___ pKa-0.70 < pH < pKa+0.48 »ò£º

ÏÂÏÞ pKa+lg(1/5) = pKa-0.70 £¬ ÉÏÏÞ pKa+lg3 = pKa+0.48 ___¡£ 32

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(1) bµã __A___ £¨A£© [HCO3-] = [CO32-] (2) cµã __B___ £¨B£© [HCO3-]

(3) dµã __C___ £¨C£© [HCO3-] = [H2CO3] (4) eµã __D___ £¨D£© [H2CO3]

33

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A 0.00 NH3 [??]?KwKb?c B 19.98 NH4++NH3 c(???4)[?]??Ka c(??3)?C 34

20.02 H++NH4+ [??]?28

0.02?010. 40.02

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