(A) H2S (Ka1 = 1.3¡Á10-7, Ka2 = 7.1¡Á10-15) (B) H2C2O4 (Ka1 = 5.9¡Á10-2, Ka2 = 6.4¡Á10-5)
(C) H3PO4 (Ka1 = 7.6¡Á10-3, Ka2 = 6.3¡Á10-8,Ka3 = 4.4¡Á10-13 ) (D) HCl+Ò»ÂÈÒÒËá (Ò»ÂÈÒÒËáµÄKa = 1.4¡Á10-3)
31.ÓÃ0.1 mol/LNaOHÈÜÒºµÎ¶¨0.1 mol/L pKa = 4.0µÄÈõËá, ͻԾ·¶Î§Îª7.0~9.7, ÔòÓÃ0.1 mol/L NaOHµÎ¶¨0.1 mol/L pKa = 3.0µÄÈõËáʱͻԾ·¶Î§Îª-( )
(A) 6.0~9.7 (B) 6.0~10.7 (C) 7.0~8.7 (D) 8.0~9.7
32.ÒÔͬŨ¶ÈNaOHÈÜÒºµÎ¶¨Ä³Ò»ÔªÈõËá(HA),Èô½«ËáºÍ¼îµÄŨ¶È¾ùÔö´ó10±¶, Á½Öֵζ¨pHÏàͬʱËùÏàÓ¦µÄÖкͰٷÖÊýÊÇ-( )
(A) 0 (B) 50 (C) 100 (D) 150
33.²â¶¨(NH4)2SO4ÖеĵªÊ±,²»ÄÜÓÃNaOH±ê×¼ÈÜÒºÖ±½ÓµÎ¶¨,ÕâÊÇÒòΪ ( )
(A) NH3 µÄKb̫С (B) (NH4)2SO4²»ÊÇËá (C) NH4+µÄKa̫С (D) (NH4)2SO4Öк¬ÓÎÀëH2SO4 34.ÒÔ¼×»ùºìΪָʾ¼Á,ÄÜÓÃNaOH±ê×¼ÈÜҺ׼ȷµÎ¶¨µÄËáÊÇ-( ) (A) ¼×Ëá (B) ÁòËá (C) ÒÒËá (D) ²ÝËá
35.ÏÖÓÐÒ»º¬H3PO4ºÍNaH2PO4µÄÈÜÒº,ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨ÖÁ¼×»ù³È±äÉ«, µÎ¶¨Ìå»ýΪa(mL)¡£Í¬Ò»ÊÔÒºÈô¸ÄÓ÷Ó̪×÷ָʾ¼Á, µÎ¶¨Ìå»ýΪb(mL)¡£Ôòa ºÍbµÄ¹ØÏµÊÇ-( ) (A) a>b (B) b = 2a (C) b>2a (D) a = b
36. (0614)ÒÆÈ¡º¬H2SO4ºÍH3PO4µÄ»ìºÏÈÜÒº50.00 mL, ÒÔ0.1000 mol/L NaOHÈÜÒº½øÐеçλµÎ¶¨,´ÓµÎ¶¨ÇúÏßÉϲéµÃÏÂÁÐÊý¾Ý:
V(NaOH)/mL 22.00 30.00 ÈÜÒºpH 4.66 9.78
Ôò»ìºÏÈÜÒºÖÐH3PO4µÄŨ¶ÈÊÇ-------( ) (H3PO4µÄpKa1~pKa3Ϊ 2.12¡¢7.20¡¢12.36) (A) 0.00800 mol/L (B) 0.0160 mol/L (C) 0.0240 mol/L (D) 0.0320 mol/L 37.ÏÂÁÐÄÄÖÖÈܼÁ,ÄÜʹHAc¡¢H3BO3¡¢HClºÍH2SO4ËÄÖÖËáÏÔʾ³öÏàͬµÄÇ¿¶ÈÀ´?--( ) (A) ´¿Ë® (B) Òº°± (C) ¼×»ùÒ춡ͪ (D) ÒÒ´¼
38.º¬0.10mol/L HClºÍ0.20mol/L H2SO4µÄ»ìºÏÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ ( )
(A) [H+] = [OH-]+[Cl-]+[SO42-] (B) [H+]+0.3 = [OH-]+[SO42-] (C) [H+]-0.3 = [OH-]+[SO42-]
(D) [H+]-0.3 = [OH-]+[SO42-]+[HSO4-]
39.¶àÔªËá·Ö²½µÎ¶¨Ê±,ÈôŨ¶È¾ùÔö¼Ó10±¶,µÎ¶¨pHͻԾ´óС±ä»¯--( ) (A) 1 (B) 2 (C) 10 (D) ²»±ä»¯
40.½«PO43-ÒÔMgNH4PO4¡¤6H2OÐÎʽ³Áµí,¾¹ýÂË¡¢Ï´µÓºóÏÈÓùýÁ¿HCl±ê×¼ÈÜÒºÈܽâ, ¶øºóÒÔNaOH±ê×¼ÈÜÒº·µµÎ¶¨,´ËʱӦѡµÄָʾ¼ÁÊÇ-( )
(A) ¼×»ù³È (B) ʯÈï (C) ·Ó̪ (D) °ÙÀï·Ó̪ 41.ÒÔÏÂʵÑéÖÐËùÓõÄNaOH»òNH3ÐèÒª³ýÈ¥CO32-µÄÊÇ-( )
(A) ÓÃÒÔÖкÍËá²¢µ÷ÖÁpH=4³ÁµíCaC2O4 (B) ÓÃÒÔÖкÍËá²¢µ÷ÖÁpH=5ÓÃEDTAµÎ¶¨Pb2+
(C) ÓÃÒÔÖкÍËá²¢µ÷ÖÁpH=13ÓÃEDTAµÎ¶¨Ca2+ (D) ÅäÖÆNaOH±ê×¼ÈÜÒºÓÃÒԵζ¨H3PO4(¼×»ù³Èָʾ¼Á) ¶þ£®Ìî¿ÕÌâ
1.ÑÇÁ×Ëá(H3PO3)µÄpKa1ºÍpKa2·Ö±ðÊÇ1.3ºÍ6.6, Æä´æÔÚÐÎʽÓÐH3PO3, H2PO3-ºÍHPO32-ÈýÖÖ¡£Çë˵Ã÷ÔÚÏÂÁÐpHÌõ¼þÏÂÑÇÁ×Ëá´æÔÚµÄÖ÷ÒªÐÎʽ¡£ pH = 0.3 pH = 1.3 pH = 3.95 pH > 6.6
2. pHΪ7.20µÄÁ×ËáÑÎÈÜÒº(H3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36), Á×ËáÑδæÔÚµÄÖ÷ÒªÐÎʽΪ__________ºÍ____________; ÆäŨ¶È±ÈΪ__________¡£ 3.±È½ÏÒÔϸ÷¶ÔÈÜÒºµÄpH´óС(Ó÷ûºÅ >¡¢ = ¡¢< ±íʾ)
(1) ͬŨ¶ÈµÄNaH2PO4(a)ºÍNH4H2PO4(b): (a)____(b) (2) ͬŨ¶ÈµÄNa2HPO4(c)ºÍ(NH4)2HPO4(d): (c)____(d)
20
[ÒÑÖªpKb(NH3) = 4.74,H3PO4µÄpKa1~pKa3·Ö±ðÊÇ 2.16,7.20,12.36]
4.ÒÑÖª¼×»ù³ÈpK(HIn) = 3.4,µ±ÈÜÒºpH = 3.1ʱ[In-]/[HIn]µÄ±ÈֵΪ_____; ÈÜÒºpH = 4.4ʱ[In-]/[HIn]µÄ±ÈֵΪ____; ÒÀͨ³£¼ÆËãָʾ¼Á±äÉ«·¶Î§Ó¦ÎªpH = pK(HIn)¡À1,µ«¼×»ù³È±äÉ«·¶Î§Óë´Ë²»·û,ÕâÊÇÓÉÓÚ___¡£ 5.¸ù¾ÝϱíËù¸øÊý¾Ý,ÅжÏÔÚÒÔϵζ¨Öл¯Ñ§¼ÆÁ¿µã¼°Æäǰºó0.1%µÄpH¡£ p£È µÎ¶¨Ìåϵ NaOHµÎ¶¨HCl(0.1 mol/L) NaOHµÎ¶¨HCl(1 mol/L) HClµÎ¶¨NaOH(0.01 mol/L) »¯Ñ§¼ÆÁ¿µã ǰ0.1% 4.3 »¯Ñ§ ¼ÆÁ¿µã 7.0 p£È µÎ¶¨Ìåϵ NaOHµÎ¶¨HCl NaOHµÎ¶¨HA HClµÎ¶¨B p£È Ũ¶È 0.1 mol/L 1 mol/L »¯Ñ§¼ÆÁ¿µã ǰ0.5% 4.3 »¯Ñ§ ¼ÆÁ¿µã 4.6 »¯Ñ§¼ÆÁ¿µã ºó0.5% 4.9 »¯Ñ§¼ÆÁ¿µã ǰ0.1% 4.3 »¯Ñ§ ¼ÆÁ¿µã 7.0 8.7 5.3 »¯Ñ§¼ÆÁ¿µã ºó0.1% 9.7 »¯Ñ§¼ÆÁ¿µã ºó0.1% 9.7 6.¸ù¾ÝϱíËù¸øÊý¾Ý,ÍÆ¶ÏÔÚÏÂÁеζ¨Öл¯Ñ§¼ÆÁ¿µã¼°Æäǰºó0.1%µÄpH¡£ (Ũ¶È¾ùΪ0.1 mol/L)
7.¸ù¾ÝϱíËù¸øÊý¾Ý,ÍÆ¶ÏÓÃNaOHµÎ¶¨H3AÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µã¼°Æäǰºó0.5%µÄpH¡£
8.ÓÃ0.1000 mol/L NaOHµÎ¶¨25.00 mL 0.1000 mol/L HCl,ÈôÒÔ¼×»ù³ÈΪָʾ¼ÁµÎ¶¨ÖÁpH = 4.0ΪÖÕµã,ÆäÖÕµãÎó²îEt = _______________%¡£
9.ÈçºÎ²â¶¨ÒÔÏÂÎïÖÊ(»ìºÏÒºÖÐΪ»Ïß×é·Ö)?Ö¸³ö±ØÒªÊÔ¼Á¡¢±ê×¼ÈÜÒº¼°Ö¸Ê¾¼Á¡£ ±ØÒªÊÔ¼Á ±ê×¼ÈÜÒº Ö¸ ʾ ¼Á ÎïÖÊ CH3NH2(¼×°·) NaAc H3BO3(ÅðËá) Na2B4O7¡¤10H2O(Åðɰ) C6H5NH3+Cl(ÑÎËá±½°·) H3BO3 Åðɰ ÄÜ·ñÖ±½ÓµÎ¶¨ ָʾ¼Á HCl+H3BO3 9.˵Ã÷±íÖи÷ÎïÖÊÄÜ·ñÓÃÖкͷ¨Ö±½ÓµÎ¶¨(²»¼ÓÈκÎÊÔ¼Á)¡£ÈôÄÜ, ÇëÖ¸³öָʾ¼Á¡£
[pKb(CH3NH2) = 3.38, pKb(C6H5NH2) = 9.38, pKa(HAc) = 4.74, pKa(H3BO3) = 9.24]
10. NH4+µÄËáÐÔÌ«Èõ,ÓÃNaOHÖ±½ÓµÎ¶¨Ê±Í»Ô¾Ì«Ð¡¶ø²»ÄÜ׼ȷµÎ¶¨¡£²ÉÓ÷µµÎ¶¨·¨___׼ȷ²â¶¨(ÌîÄÜ»ò²»ÄÜ),ÆäÔÒòÊÇ______ _____________¡£
11.²ÉÓÃÕôÁ󷨲ⶨNH4+ʱԤ´¦ÀíµÄ·½·¨ÊÇ_____¡£ ÈôÓÃHClÈÜÒºÎüÊÕ,²ÉÓÃNaOH±ê×¼ÈÜҺΪµÎ¶¨¼ÁʱӦѡ____Ϊָʾ¼Á, ÈôµÎ¶¨ÖÁpH = 7, ÖÕµãÎó²îΪ____Öµ(Ö¸Õý»ò¸º)¡£Èô¸ÄÓÃH3BO3ÎüÊÕ,Ó¦²ÉÓÃ_____ΪµÎ¶¨¼Á¡£ºóÒ»·½·¨ÓÅÓÚǰÕßµÄÔÒòÊÇ______¡£
21
12.ijÈËÓÃHCl±ê×¼ÈÜÒºÀ´±ê¶¨º¬CO32-µÄNaOHÈÜÒº(ÒÔ¼×»ù³È×÷ָʾ¼Á),È»ºóÓÃNaOHÈÜҺȥ²â¶¨Ä³ÊÔÑùÖÐHAc,µÃµ½µÄw(HAc)½«»á_______¡£ÓÖÈôÓÃÒԲⶨHCl-NH4ClÈÜÒºÖеÄw(HCl),Æä½á¹û»á_________¡£(ÌîÆ«¸ß¡¢Æ«µÍ»òÎÞÓ°Ïì)
13.NaOHÈÜҺŨ¶È±ê¶¨ºóÓÉÓÚ±£´æ²»Í×ÎüÊÕÁËCO2, ÒԴ˱ê×¼ÈÜÒº²â¶¨²ÝËáĦ¶ûÖÊÁ¿Ê±,½á¹û_____;ÈôÒԴ˱ê×¼ÈÜÒº²â¶¨H3PO4Ũ¶È(¼×»ù³Èָʾ¼Á)Æä½á¹û____¡£(ÌîÆ«¸ß¡¢Æ«µÍ»òÎÞÓ°Ïì)
14.ÏÂͼÊÇÓÃÕôÁ󷨲ⶨï§Ñεļòµ¥Á÷³Ì,ÊÔÌîдËùÐèÊÔ¼Á¡¢²úÎï¡¢ÏÖÏó¼°½á¹ûµÄ¼ÆËã¡£ ©³©¥©¥©¥©· ָʾ¼Á©§ ©§ ©»©¥©¥©¥©¿
©³©¥©¥©¥©· µÎ¶¨ ©¦ ©³©¥©¥©¥©¥©· ¡ý ¼ÓÈë©§ ©§ ¡ýµÎ¶¨¼Á©§ ©§ ©»©¥©¥©¥©¿ ©»©¥©¥©¥©¥©¿ ©³©¥©¥©· ©³©¥©¥©· ©³©¥©¥©¥©¥©¥©¥©¥©· ©§ÊÔÑù©§¡ª¡ª¡ú ©§ ©§¡ª¡ª¡ª¡ª¡ª¡ª¡ª¡ú©§ ©§ ©»©¥©¥©¿ÕôÁó³ö ©»©¥©¥©¿ÎüÊÕÔÚH3BO3ÖÐ ©»©¥©¥©¥©¥©¥©¥©¥©¿
©³©¥©· ©³©¥©· ÑÕÉ«±ä»¯ ©§ ©§É«ÖÁ©§ ©§É« ©»©¥©¿ ©»©¥©¿ ½á¹û¼ÆËã,w(NH4+) = __________________________. Èý£®¼ÆËãÌâ
1.¼ÙÉèijËá¼îָʾ¼ÁHInµÄ±äÉ«pH·¶Î§Îª2.60, Èô¹Û²ìµ½¸ÕÏÔËáʽ(HIn)ɫʱ±ÈÂÊ[HIn]/[In-]ºÍ¼îʽ(In-)É«
ʱ[In-]/[HIn]ÊÇÏàͬµÄ¡£µ±Ö¸Ê¾¼Á¸ÕÏÔËáÉ«»ò¼îɫʱ,HIn»òIn-ÐÎÌåËùÕ¼µÄ°Ù·Ö±ÈΪ¶àÉÙ?
2 .0.20 mol/Lij¶þÔªËá(H2A)ÈÜÒº,ÓõÈŨ¶ÈµÄNaOH±ê×¼ÈÜÒºµÎ¶¨,¼ÆËãµÎ¶¨ÖÁµÚÒ»¸ö»¯Ñ§¼ÆÁ¿µãµÄpH?(ÉèH2AµÄKa1 = 4.0¡Á10-4,Ka2 = 2.5¡Á10-9)
3.¼ÆËãÒÔ0.20 mol/L Ba(OH)2ÈÜÒºµÎ¶¨0.10 mol/L HCOOHÈÜÒºÖÁ»¯Ñ§¼ÆÁ¿µãʱ,ÈÜÒºµÄpHΪ¶àÉÙ? [Ka(HCOOH) = 2.0¡Á10-4]
4.ÓÃ0.20 mol/L NaOHµÎ¶¨0.20 mol/L HCl(ÆäÖк¬ÓÐ0.10 mol/L NH4Cl)¡£
(1) ¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH;(2) ÈôµÎ¶¨ÖÁpH = 7.0, ÎÊÖÕµãʱÓаٷÖÖ®¼¸µÄNH4+±äΪNH3; (3) ´ËʱÖÕµãÎó²îÊǶàÉÙ? [pKa(NH4+)Ϊ9.26]
5.ÅðËá(ÒÔHA±íʾ)ÊÇÒ»ÖÖºÜÈõµÄËá(pKa = 9.2),²»ÄÜÓà NaOH Ö±½ÓµÎ¶¨,µ«Èô¼ÓÈë×ãÁ¿µÄ¸Ê¶´¼(ÒÔR±íʾ)ʹ֮Éú³ÉÂçºÏÎïR2A-, Ôò¿É׼ȷµÎ¶¨¡£ ÈôÓà 0.020 mol/L NaOHµÎ¶¨0.020 mol/L ÅðËá,[R]ÖÕ= 0.20 mol/L¡£ ÊÔ¼ÆËã: (1) ´ËÌõ¼þÅðËáµÄ½âÀë³£ÊýKa?; (2) »¯Ñ§¼ÆÁ¿µãʱÈÜÒºµÄpH; (3)ÈôÖÕµã±È»¯Ñ§¼ÆÁ¿µã¸ß0.5pH, ÖÕµãÎó²î¶à´ó? (ÒÑÖªR2A-ÂçºÏÎïµÄlg?1 = 2.5, lg?2 = 4.8)
7.¼ÙÉèÓÐNaOHºÍNa2CO3µÄ»ìºÏÎï¸ÉÔïÊÔÑù(²»º¬ÆäËü×é·Ö)0.4650 g, ÈܽⲢϡÊÍÖÁ50 mL, µ±ÒÔ¼×»ù³ÈΪָʾ¼Á, ÓÃ0.200 mol/L HCl ÈÜÒºµÎ¶¨ÖÁÖÕµãʱ, ÏûºÄ50.00 mL¡£¼ÆËã»ìºÏÎïÖÐNaOHºÍNa2CO3µÄÖÊÁ¿¸÷Ϊ¶àÉÙ¿Ë? [Mr(NaOH) = 40.00, Mr(Na2CO3) = 106.0]
8. ÒÆÈ¡FeCl3-HClÊÔÒº25.00 mL, µ÷½ÚËá¶ÈÖÁpH = 2,¼ÓÈÈ, ÒԻǻùË®ÑîËáΪָʾ¼Á, ÓÃ0.02012 mol/L EDTAµÎ¶¨ÖÁÓÉ×Ϻì±äΪdz»ÆÉ«, ¼ÆºÄÈ¥20.04 mL¡£
ÁíȡͬÁ¿ÊÔÒº¼ÓÈë20.04 mL 0.02012 mol/L EDTA(Na2H2Y), ¼ÓÈÈ, ÀäÈ´ºó, ÒÔ¼×»ùºìΪָʾ¼Á, ÓÃ0.1015 mol/L NaOH µÎ¶¨, ÏûºÄÁË32.50 mL¡£ÊÔÇóÊÔÒºÖÐHClµÄŨ¶È¡£
9. 0.30mol/L H3PO4ÈÜÒºÓë0.20mol/L Na3PO4ÈÜÒºµÈÌå»ý»ìºÏ,¼ÆËã¸Ã»º³åÈÜÒºµÄpH(H3PO4µÄpKa1 ~pKa3
·Ö±ðΪ2.12¡¢7.20¡¢12.36)¡£
10. ÒÔ0.100mol/L NaOHÈÜÒºµÎ¶¨0.100mol/Lij¶þÔªÈõËáH2AÈÜÒº¡£ÒÑÖªµ±ÖкÍÖÁ pH = 1.92ʱ, x(H2A) = x (HA-);ÖкÍÖÁpH = 6.22ʱ, x (HA-) = x (A2-)¡£ ¼ÆËã: (1)ÖкÍÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µãʱ, ÈÜÒºµÄpHΪ¶àÉÙ?Ñ¡ÓúÎÖÖָʾ¼Á? (2)ÖкÍÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µãʱ, ÈÜÒºµÄpHΪ¶àÉÙ?Ñ¡ÓúÎÖÖָʾ¼Á?
22
11.½ñÒÔijÈõËáHA¼°Æä¹²éî¼îA-ÅäÖÆ»º³åÈÜÒº,ÒÑÖªÆäÖÐ[HA]= 0.25mol/L¡£ ÍùÉÏÊö 100mL»º³åÈÜÒºÖмÓÈë5.0mmol¹ÌÌåNaOHºó(ºöÂÔÆäÌå»ýµÄ±ä»¯),ÈÜÒºµÄpH = 4.60¡£ ¼ÆËãÔ»º³åÈÜÒºµÄpH(ÉèHAµÄpKa = 4.30)¡£
12.ij·ÖÎö¹¤×÷ÕßÓÃNH3ºÍNH4ClÖÆ±¸pH = 9.49µÄ»º³åÈÜÒº200mL¡£ÎÊ:
(1) ¸Ã»º³åÈÜÒºÖÐNH3ºÍNH4ClµÄƽºâŨ¶ÈÖ®±ÈΪ¶àÉÙ? (2) ÈôÏòÉÏÊö»º³åÈÜÒºÖмÓÈë1.0 mmol¹ÌÌåNaOH(ºöÂÔÌå»ý±ä»¯)¡£ÓûʹÆäpHµÄ¸Ä±ä²»´óÓÚ0.15 pH,ÔòÔ»º³åÈÜÒºÖÐ NH3ºÍNH4Cl µÄ×îµÍŨ¶È¸÷Ϊ¶àÉÙ? (NH3µÄpKb = 4.74) 12.ѪҺÊÔÑùÖÐ×ܶþÑõ»¯Ì¼??HCO3????CO??º¬Á¿¿Éͨ¹ýËữÊÔÑù²âÁ¿COµÄÌå»ý½øÐвⶨ¡£½ñ²âµÃijѪҺ
22
ÊÔÑùÖÐCO2×ÜŨ¶ÈΪ28.5 mmol/L£¬ÔÚ37¡æÊ±¸ÃѪҺµÄpHΪ7.48¡£ÊÔÎʸÃѪҺÊÔÑùÖÐHCO3-ºÍCO2µÄŨ
¶È¸÷ÊǶàÉÙ£¿(ÒÑÖªH2CO3µÄKa1=7.9¡Á10-7£¬Ka2=1.3¡Á10-10)
13.Èô½«25mL 0.20 mol/L NaOHÈÜÒº¼ÓÈë20mL 0.25 mol/L ÅðËáÈÜÒºÖУ¬ËùµÃÈÜÒºµÄpHÊǶàÉÙ£¿ [ÒÑÖªKa (H3BO3)=6.4 ¡Á10-10]
14.ΪÅäÖÆpH = 4.50£¬Àë×ÓÇ¿¶ÈΪ0.250 (Óûî¶È)µÄ»º³åÈÜÒºÐèÏò500 mL 0.100 mol/L HAcÈÜÒºÖмÓÈë¶àÉÙºÁÉý0.100 mol/L NaOHºÍ¶àÉÙ¿ËNaCl ? [ÒÑÖªpKa(HAc) = 4.74£¬ a(Ac-) = 450 pm£¬Mr(NaCl) = 58.44] 15.ÏÖÓÐijÊÔÑù0.2528 g£¬ÒÔ·Ó̪×÷ָʾ¼Á£¬ÓÃ0.0998 mol/L HCl±ê×¼ÈÜÒºµÎ¶¨£¬ÐèHClÈÜÒº14.34 mL£»ÒÔ¼×»ù³È×÷ָʾ¼Á£¬ÐèHClÈÜÒº35.68 mL¡£ÊÔÎʸÃÊÔÑùÓÉNaOH¡¢Na2CO3ºÍNaHCO3ÖеÄÄÄЩ×é·Ö×é³É£¿ÆäÖÊÁ¿·ÖÊý¸÷ÊǶàÉÙ£¿ ÏÂÁÐÊý¾Ý¿É¹©Ñ¡ÓãºMr(NaOH)=40.01, Mr(Na2CO3)=105.99, Mr(NaHCO3)= 84.01 )
16.ÏÖÓÐÒ»HAc-NaAc»º³åÈÜÒº£¬Æä[ H+ ]Ϊ9.0 ? 10-6mol/L£¬½ñ½«10.0 mmol HCl¼ÓÈë1 L¸ÃÈÜÒºÖÐʱ£¬[H+]±äΪ1.0 ? 10-5mol/L£¬ÊÔ¼ÆËãÔ»º³åÈÜÒºÖÐHAcºÍNaAcµÄŨ¶È¡£[ÒÑÖªKa(HAc)=1.8 ? 10 ¨C5 ]
17.ÏÖÓÉijÈõËáHX ( pKa = 5.30 )¼°Æä¹²éî¼îNaXÅäÖÆpHΪ5.00µÄ»º³åÈÜÒº¡£ÈôÔÚ100 mL»º³åÈÜÒºÖмÓÈë10 mmol HCl£¬ÊÔÏò×î³õ»º³åÈÜÒºµÄŨ¶ÈÐè¶à´óʱ²ÅÄÜÒòËáµÄ¼ÓÈë¶øÊ¹ÈÜÒºµÄpHÖ»¸Ä±ä0.20µ¥Î»£¿ 18.ÓûÅäÖÆpH = 5.00£¬»º³åÈÝÁ¿? = 0.28 mol / LµÄÁù´Î¼×»ùËİ·[ (CH2)6N4£ºKb = 1.4 ? 10-9£»Mr = 140.19 ]»º³åÈÜÒº200 mL£¬ÊÔÎÊÐèÒªÁù´Î¼×»ùËİ·¶àÉÙ¿Ë£¿Å¨ÑÎËá(Ũ¶ÈΪ12mol/L )¶àÉÙºÁÉý£¿ 19.ÓûÅäÖÆ500 mLÀë×ÓÇ¿¶ÈI = 0.020µÄÅðɰ»º³åÈÜÒº¡£ÊÔÎÊ£º
( 1 ) ÐèÅðɰ ( Na2B4O7¡¤10H2O£ºMr = 381.37 )¶àÉÙ¿Ë£¿ ( 2 )´Ë»º³åÈÜÒºµÄpHºÍ»º³åÈÝÁ¿ÊǶàÉÙ£¿ [ H3BO3£ºKa = 5.8 ? 10-10£»I = 0.020£¬¦Ã(H2BO3-) = 0.870 ]
20.ÓûÅäÖÆ°±»ùÒÒËá×ÜŨ¶ÈΪ0.10 mol/LµÄ»º³åÈÜÒº100 mL,ʹÆäÈÜÒºµÄpHΪ2.00,Ðè°±»ùÒÒËá¶àÉÙ¿Ë?Ðè¼ÓÈë1.0 mol/LµÄÇ¿Ëá»òÇ¿¼î¶àÉÙºÁÉý? ÒÑÖª°±»ùÒÒËáÑÎH2A+µÄKa1 = 4.5¡Á10-3, Ka2 = 2.5¡Á10-10, Mr(NH2CH3COOH) = 75.07¡£
21.ΪÅäÖÆpHΪ7.20µÄÁ×ËáÑλº³åÈÜÒº(×ÜŨ¶ÈΪ1 mol/L)500 mL, Ӧȡ1.0 mol/L H3PO4ºÍ1.0 mol/L Na2HPO4ÈÜÒº¸÷¼¸ºÁÉý? (H3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)
22.½«0.20 mol/L NH4Cl ©¤ 0.20 mol/L NH3 ÈÜÒºÓë0.020 mol/L HAc ©¤ 0.020 mol/L NaAcÈÜÒºµÈÌå»ý»ìºÏ,¼ÆËã»ìºÏºóÈÜÒºµÄpH¡£
23. È¡10 mL pH = 4.74µÄ´×ËỺ³åÈÜÒº,¼ÓÖÁij·ÖÎö²Ù×÷ÒºÖÐ, ʹÆä×ÜÌå»ýΪ100 mL,Èç¹ûÒªÇó¸Ã²Ù×÷Òº¾ßÓÐ×î´ó»º³åÈÝÁ¿0.10 mol/L,ÄÇôÓûÅäÖÆ500 mL´Ë»º³åÈÜÒºÐèÈ¡±ù´×Ëá(17 mol/L)¶àÉÙºÁÉý? Ðè´×ËáÄÆNaAc¡¤3H2O¶àÉÙ¿Ë? [Mr(NaAc¡¤3H2O) = 136, pKa(HAc) = 4.74] 24.ÓÃ0.020 mol/L EDTAµÎ¶¨25 mL pHΪ1.0µÄº¬Bi3+¡¢Zn2+µÄ»ìºÏÈÜÒº(Ũ¶È¾ùΪ0.020 mol/L),Ôڵζ¨Bi3+ºó,Ϊµ÷½ÚpHÖÁ5.5ÒԵζ¨Zn2+, Ó¦µ±¼ÓÈëÁù´Î¼×»ùËİ·¶àÉÙ¿Ë? Zn2+µÎ¶¨ÖÕÁËʱÈÜÒºpHÓÖÊǶàÉÙ? {pKb[(CH2)6N4] = 8.87, Mr[(CH2)6N4] = 140 }
25.ÓÐÒ»Á×ËáÑλìºÏÒº25.00 mL, Ñ¡·Ó̪Ϊָʾ¼ÁÐè10.00 mL HCl±ê×¼ÈÜÒº(Ũ¶ÈÔ¼0.5 mol/L), Èô¸ÄÓü׻ù³ÈΪָʾ¼ÁÔòÐè50.00 mLͬÑùŨ¶ÈµÄHClÈÜÒº¡£¼ÆËãÔÊÔÒºµÄpH¡£ (H3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)
26.³ÆÈ¡1.250 g ´¿Ò»ÔªÈõËáHA, ÈÜÓÚÊÊÁ¿Ë®ºóÏ¡ÖÁ50.00 mL, È»ºóÓÃ0.1000 mol/L NaOH ÈÜÒº½øÐеçλµÎ
23