0421 ÓûÅäÖÆpH=9µÄ»º³åÈÜÒº,ӦѡÓÃ---------------------------------------------------------------( ) (A) NH2OH(ôÇ°±) (Kb = 9.1¡Á10-9) (B) NH3¡¤H2O (Kb = 1.8¡Á10-5) (C) CH3COOH (Ka = 1.8¡Á10-5) (D) HCOOH (Ka = 1.8¡Á10-4) 0422 Ñ¡ÔñÏÂÁÐÈÜÒº[H+]µÄ¼ÆË㹫ʽ(ÇëÌîA,B,C) (1) 0.10 mol/L ¶þÂÈÒÒËá (pKa = 1.30) __________ (2) 0.10 mol/L NH4Cl (pKa = 9.26) __________ (3) 0.10 mol/L NaHSO4 (pKa2 = 1.99) __________ (4) 1.0¡Á10-4mol/L H3BO3 (pKa = 9.24) __________ £¨A£© Ka?c £¨B£© Ka(c?[??]) £¨C£© Ka?c?KW 0423 Ñ¡ÔñÏÂÁÐÈÜÒº[H+]¼ÆË㹫ʽ(ÇëÌîA,B) (1) 0.04 mol/L H2CO3ÈÜÒº __________ (2) 0.05 mol/L NaHCO3ÈÜÒº __________ (3) 0.10 mol/L °±»ùÒÒËáÈÜÒº __________ (4) 0.10 mol/L °±»ùÒÒËáÑÎËáÑÎÈÜÒº __________ (A) [H+] = Ka1?c (B) [H+] = Ka1?Ka2 0424 Ñ¡Ôñ[H+]µÄ¼ÆË㹫ʽ(ÇëÌîA,B,C,D) 1. 0.1000 mol/L HClµÎ¶¨0.1000 mol/L Na2CO3ÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µã--------------------( 2. 0.05000 mol/L NaOHµÎ¶¨0.05000 mol/L H3PO4ÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µã----------------( 3. 0.1 mol/L HAc-0.1 mol/L H3BO3»ìºÏÒº---------------------------------------------------( 4. 0.1 mol/L HCOONH4ÈÜÒº--------------------------------------------------------------------( (A)[H+] = Ka(?)?c (B)[H+] = Ka1?Ka2 (C)[H+] = Ka(?)?Ka(??) (D)[H+] = Ka1Ka2?c/(Ka1+c) ) ) ) ) 0425 д³ö¼ÆËãÒÔÏÂÈÜÒº[H+]»ò[OH-]µÄ¹«Ê½ 0.10 mol/L ÈýÒÒ´¼°·(pKb = 6.24) ______________ 0.10 mol/L ÁÚ±½¶þ¼×ËáÇâ¼Ø(pKa1 = 2.95,pKa2 = 5.41) ______________ 0.10 mol/L H2C2O4(pKa1 = 1.22, pKa2 = 4.19) ______________ 0426 ijÈýÔªËá(H3A)µÄ½âÀë³£ÊýΪpKa1 = 3.96, pKa2 = 6.00, pKa3 = 10.02, Ôò0.10 mol/L H3AÈÜÒºµÄpHΪ_______, 0.10 mol/L Na3AÈÜÒºµÄpHΪ ___________¡£ 0427 ÓÃHClµÎ¶¨ÅðÉ°(Na2B4O7¡¤10H2O)µÄ·´Ó¦Ê½Îª_____________________, Æ仯ѧ¼ÆÁ¿µã[H+]µÄ¼ÆËãʽΪ__________________¡£Èôc(Na2B4O7) = 0.050 mol/L, c(HCl) = 0.10 mol/L,Æ仯ѧ¼ÆÁ¿µãpHÊÇ_______________________________¡£ [pKa(H3BO3) = 9.24] 0428 ijÈýÔªËá(H3A)µÄ½âÀë³£ÊýΪKa1 = 1.1¡Á10-2, Ka2 = 1.0¡Á10-6, Ka3 = 1.2¡Á10-11, Ôò0.10 mol/L NaH2AÈÜÒºµÄpHΪ________, 0.10 mol/L Na2HA ÈÜÒºµÄpHΪ_______¡£(Ö»ÒªÇó½üËƼÆËã) 0429 0.10 mol/L NH4AcÈÜÒºµÄpHΪ________________________ 0.10 mol/L NH4CNÈÜÒºµÄpHΪ________________________ [ÒÑÖªpKa(HAc) = 4.74, pKb(NH3) = 4.74, pKa(HCN) = 9.21] 0430 ÒÑÖªEDTAµÄpKa1~pKa6·Ö±ðÊÇ0.9,1.6,2.0,2.67,6.16ºÍ10.26, 0.10 mol/L EDTA¶þÄÆÑÎ(Na2H2Y¡¤2H2O)ÈÜÒºµÄpHÊÇ______________, [Y]Ϊ______________mol/L¡£ 0431 0.10 mol/L Na2HPO4ÈÜÒºµÄ½üËÆpHΪ_________________ 0.10 mol/L NaH2PO4ÈÜÒºµÄ½üËÆpHΪ_________________ (ÒÑÖª H3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36) 0432 ¶ÔÓÚijһ¸ø¶¨µÄ»º³åÌåϵ,»º³åÈÝÁ¿µÄ´óСÓë__________ºÍ_________Óйء£ 0433 ijÈÜÒºÖÐÈõËáµÄŨ¶ÈΪc(HA),Æä¹²éî¼îµÄŨ¶ÈΪc(A-), ¸ÃÈÜÒºµÄ×î´ó»º³åÈÝÁ¿(?max)Ó¦µÈÓÚ______________________________¡£(д³ö¼ÆËãʽ) 0434 0.20 mol/L NH3 [pKb(NH3) = 4.74]ÈÜÒºÓë0.10 mol/L HClÈÜÒºµÈÌå»ý»ìºÏºó,ÈÜÒºµÄpHÊÇ_________________¡£ 0435 20 mL 0.50 mol/L H3PO4ÈÜÒºÓë5.0 mL 1.0 mol/LµÄNa3PO4ÈÜÒºÏà»ìºÏºó, ÆäpHÊÇ_____¡£(H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) 0436 10 g(CH2)6N4¼ÓÈëµ½4.0 mL 12 mol/L HClÈÜÒºÖÐ,Ï¡ÊÍÖÁ100 mLºó, Æä pH ֵΪ__________¡£ {Mr[(CH2)6N4] = 140.0, pKb[(CH2)6N4] = 8.85} 0437 ÔÚ1 L 0.10 mol/L HAcÈÜÒºÖмÓÈë_________ g NaAcºó, ÈÜÒºpH = 5.04¡£ [pKa(HAc) = 4.74, Mr(NaAc) = 82.03] 0438 25 mL 0.40 mol/L H3PO4ÈÜÒººÍ30 mL 0.50 mol/L Na3PO4ÈÜÒº»ìºÏ²¢Ï¡ÊÍÖÁ 100mL,´ËÈÜÒºµÄpHÊÇ__________¡£ (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) 0439 ±È½ÏÒÔϸ÷¶ÔÈÜÒºµÄpHµÄ´óС(ÓÃ>¡¢ = ¡¢<·ûºÅ±íʾ) (1)Ũ¶ÈÏàͬµÄNaAcºÍNH4AcÈÜÒº ____________ (2)Ũ¶ÈÏàͬµÄHAc-NaAcºÍHAc-NH4Ac»º³åÈÜÒº ____________ [pKa(HAc) = 4.74, pKb(NH3) = 4.74] 0440 (CH2)6N4(Áù´Î¼×»ùËÄ°·)µÄpKb = 8.9,ÓÉ(CH2)6N4-HCl×é³ÉµÄ»º³åÈÜÒºµÄ»º³å·¶Î§ÊÇ_____________,ÓûÅäÖƾßÓÐ?maxµÄ¸ÃÖÖ»º³åÈÜÒº,Ó¦ÓÚ100 mL 0.1 mol/L(CH2)6N4ÈÜÒºÖмÓÈë____________mL 1 mol/L HCl¡£ 0441 ÓûÅäÖÆpH = 5.5 ×ÜŨ¶ÈΪ 0.20 mol/L µÄÁù´Î¼×»ùËÄ°·[(CH2)6N4]»º³åÈÜÒº 500 mL,Ó¦³ÆÈ¡(CH2)6N4___________ g,Á¿È¡12 mol/L HCl __________mL¡£ {Mr[(CH2)6N4] = 140.0, pKb[(CH2)6N4] = 8.85} 0442 ÔÚÒÔÏ´¿Á½ÐÔÎïÈÜÒºÖÐ,ÄÜ×÷Ϊ±ê×¼»º³åÈÜÒºµÄÊÇ----------------------------------------( ) (A) NaH2PO4 (B) Na2HPO4 (C) °±»ùÒÒËáż¼«Àë×Ó (D) ¾ÆʯËáÇâ¼Ø ÒÑÖª: pKa1 pKa2 pKa3 H3PO4 2.12 7.20 12.36 2.35 9.60 °±»ùÒÒËá 3.04 4.37 ¾ÆʯËá 0443 ½«5 mmol ÒÒ¶þ°·ËÄÒÒËá(H4Y)¼ÓÈëµ½1 L 5.0¡Á10-3mol/LÒÒ¶þ°·ËÄÒÒËáÄÆ(Na4Y)ºÍ5.0¡Á10-3 mol/L NaOHÈÜÒºÖС£¼ÆËãÈÜÒºµÄpHºÍÒÒ¶þ°·ËÄÒÒËáÎåÖÖÐÎʽµÄŨ¶ÈÖ®±È(ºöÂÔH5Y+ºÍH6Y2+) (H4YµÄlg?1~lg?4·Ö±ðÊÇ10.26,16.42,19.09ºÍ21.09) 0444 ijһÈÜÒºÓÉHCl,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) 0445 ½«0.12 mol/L HClÓë0.10 mol/L ClCH2COONaµÈÌå»ý»ìºÏ, ÊÔ¼ÆËã¸ÃÈÜÒºµÄ pH¡£(ÒÑÖªClCH2COOHµÄKa = 1.4¡Á10-3) 0446 ÔÚ400 mLË®ÖмÓÈë6.2 g NH4Cl(ºöÂÔÆäÌå»ý±ä»¯)ºÍ45 mL 1.0 mol/L NaOH ÈÜÒº, ´Ë»ìºÏÈÜÒºµÄpHÊǶàÉÙ? »º³åÈÝÁ¿¶à´ó? (Mr(NH4Cl) = 53.5, NH3µÄ pKb = 4.74)