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0303 OH-µÄ¹²éîËáÊÇ------------------------------------------------------------------------------------( ) (A) H+ (B) H2O (C) H3O+ (D) O2- 0302 HPO42-µÄ¹²éî¼îÊÇ---------------------------------------------------------------------------------( ) 0301 ÔÚË®ÈÜÒºÖй²éîËá¼î¶ÔKaÓëKbµÄ¹ØÏµÊÇ---------------------------------------------------( ) (A) Ka¡¤Kb=1 (B) Ka¡¤Kb=Kw (C) Ka/Kb=Kw (D) Kb/Ka=Kw 0304 ÔÚÏÂÁи÷×éËá¼î×é·ÖÖÐ,ÊôÓÚ¹²éîËá¼î¶ÔµÄÊÇ----------------------------------------------( ) (A) H2PO4- (B) H3PO4 (C) PO43- (D) OH- 0307 Ũ¶ÈÏàͬµÄÏÂÁÐÎïÖÊË®ÈÜÒºµÄpH×î¸ßµÄÊÇ----------------------------------------------( ) (A) NaCl (B) NaHCO3 (C) NH4Cl (D) Na2CO3 0308 ÏàͬŨ¶ÈµÄCO32-¡¢S2-¡¢C2O42-ÈýÖÖ¼îÐÔÎïÖÊË®ÈÜÒº, Æä¼îÐÔÇ¿Èõ(ÓÉ´óÖÁС)µÄ˳ÐòÊÇ-------------------------------------------------------------------------------------------------------------( ) (ÒÑÖª H2CO3 pKa1 = 6.38 pKa2 = 10.25 H2S pKa1 = 6.88 pKa2 = 14.15 H2C2O4 pKa1 = 1.22 pKa2 = 4.19 ) (A) CO32->S2->C2O42- (B) S2->C2O42->CO32- (C) S2->CO32->C2O42- (D) C2O42->S2->CO32- 0309 Ë®ÈÜÒº³ÊÖÐÐÔÊÇÖ¸--------------------------------------------------------------------------------( (A) pH = 7 (B) [H+] = [OH-] (C) pH+pOH = 14 (D) pOH = 7 0310 ÔÚÒ»¶¨µÄζÈÏÂ,»î¶ÈϵÊýÓëË®ºÏÀë×Ó°ë¾¶µÄ¹ØÏµÊÇË®ºÏÀë×Ó°ë¾¶Óú´ó,Àë×ӵĻî¶ÈϵÊý--------------------------------------------------------------------------------------------------------------( ) (A) Óú´ó (B) ÓúС (C) ÎÞÓ°Ïì (D) ÏÈÔö´óºó¼õС 0311 ÔÚH2C2O4ÈÜÒºÖÐ,ÏÂÁлî¶ÈϵÊýµÄÅÅÁÐ˳ÐòÕýÈ·µÄÊÇ------------------------------------( (A) ?(HC2O4-) > ?(H+) > ?(C2O42-) (B) ?(H+) > ?(HC2O4-) > ?(C2O42-) (C) ?(C2O42-) > ?(HC2O4-) > ?(H+) (D) ?(HC2O4-) > ?(C2O42-) > ?(H+) ) ) (A) HCN-NaCN (B) H3PO4-Na2HPO4 (C) +NH3CH2COOH-NH2CH2COO- (D) H3O+-OH- 0312 0.050mol/L AlCl3ÈÜÒºµÄÀë×ÓÇ¿¶ÈΪ----------------------------------------------------------( (A) 0.60 mol/L (B) 0.30 mol/L (C) 0.15 mol/L (D) 0.10 mol/L 0313 º¬0.10 mol/L HAc-0.10 mol/L NaAc-0.20 mol/L NaClÈÜÒºÖеÄÀë×ÓÇ¿¶ÈΪ----------( ) (A) 0.60 mol/L (B) 0.40 mol/L (C) 0.30 mol/L (D) 0.20 mol/L 0314 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 0305 ÏÂÁи÷×é×é·ÖÖв»ÊôÓÚ¹²éîËá¼î¶ÔµÄÊÇ-----------------------------------------------------( ) (A) H2CO3ºÍCO32- (B) NH3ºÍNH2- (C) HClºÍCl- (D) HSO4- ºÍSO42- 0306 ÒÔϱíÊöÖдíÎóµÄÊÇ-----------------------------------------------------------------------------( (A) H2O×÷ΪËáµÄ¹²éî¼îÊÇOH- (B) H2O×÷Ϊ¼îµÄ¹²éîËáÊÇH3O+ (C) ÒòΪHAcµÄËáÐÔÇ¿,¹ÊHAcµÄ¼îÐÔ±ØÈõ (D) HAcµÄ¼îÐÔÈõ,ÔòH2Ac+µÄËáÐÔÇ¿ ) ) 0315 Ó°ÏìÆ½ºâ³£ÊýµÄÒòËØÊÇ--------------------------------------------------------------------------( ) (A) ·´Ó¦ÎïºÍ²úÎïµÄŨ¶È (B) ÈÜÒºµÄËá¶È (C) ÎÂ¶È (D) ´ß»¯¼Á 0316 ÔÚÒ»¶¨Î¶ÈÏÂ,Àë×ÓÇ¿¶ÈÔö´ó,ÈõËáµÄ±ê׼ƽºâ³£ÊýK¦Èa½«----( ) (A) Ôö´ó (B) ¼õС (C) ÎÞÓ°Ïì (D) ¼õСÖÁÒ»¶¨³Ì¶ÈºóÇ÷ÓÚÎȶ¨ 0317 ÔÚÒ»¶¨Î¶ÈÏÂ,Àë×ÓÇ¿¶ÈÔö´ó,´×ËáµÄŨ¶È³£ÊýKCa½«------------------------------------( ) (A) Ôö´ó (B) ¼õС (C) ²»±ä (D) ¼õСÖÁÒ»¶¨³Ì¶ÈºóÇ÷ÓÚÎȶ¨ 0318 Ò»ÔªÈõËáµÄ½âÀë³£ÊýµÄÁ½ÖÖÐÎʽ©¤©¤±ê׼ƽºâ³£ÊýK¦ÈMaÓë»ìºÏ³£ÊýKa,ËüÃǵĹØÏµÊÇ-------------------------------------------------------------------------------------------------------------( ) (A) K¦ÈM (B) K¦ÈMa > Ka a< Ka (C) K¦ÈMa = Ka (D) AÓëB¾ùÓпÉÄÜ 0319 ÒÑÖªH3AsO4µÄpKa1= 2.2, pKa2= 6.9, pKa3= 11.5,ÔòÔÚ pH=7.0ʱ, ÈÜÒºÖÐ[H3AsO4]/[AsO43-]µÈÓÚ--------------------------------------------------------------------------------------------------------( ) (A) 100.4 (B) 10-0.4 (C) 10-5.2 (D) 10-4.8 0320 Èô¶¯ÂöѪµÄpH=7.40, [HCO3-] = 0.024 mol/L, ÒÑÖªH2CO3µÄpKa1 = 6.38, pKa2 = 10.25, Ôò[H2CO3]Ϊ-----------------------------------------------------------------------------------------------( ) (A) 2.3¡Á10-2mol/L (B) 2.3¡Á10-3mol/L (C) 4.6¡Á10-2mol/L (D) 4.6¡Á10-3mol/L 0321 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 0322 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 0323 ÒÑÖªH3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36¡£½ñÓÐÒ»Á×ËáÑÎÈÜÒº, ²âµÃÆäpH = 7.0,ÔòÆäÖ÷Òª´æÔÚÐÎʽÊÇ----------------------------------------------------------------------------------( ) (A) H3PO4+H2PO4- (B) H2PO4-+HPO42- (C) H3PO4+HPO42- (D) HPO42-+PO43- 0324 ij²¡ÈËÍÌ·þ10g NH4Cl 1Сʱºó, ËûѪҺµÄpH = 7.38¡£ÒÑÖªH2CO3µÄ pKa1 = 6.38, pKa2 = 10.25¡£´ËʱËûѪҺÖÐ[HCO3-]/[H2CO3]Ö®±ÈΪ-------------------------------------------------( ) (A) 1/10 (B) 10 (C) 1/2 (D) 2 0325 ¶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 0326 µ±pH = 5.00ʱ,0.20 mol/L¶þÔªÈõËá(H2A)ÈÜÒºÖÐ, H2AµÄŨ¶ÈΪ---------------------( ) (ÉèH2AµÄpKa1 = 5.00, pKa2 = 8.00) (A) 0.15 mol/L (B) 0.10 mol/L (C) 0.075 mol/L (D) 0.050 mol/L 0327 Óмס¢ÒÒ¡¢±ûÈýƿͬÌå»ýͬŨ¶ÈµÄH2C2O4¡¢NaHC2O4ºÍNa2C2O4ÈÜÒº, ÈôÓÃHCl»òNaOHµ÷½ÚÖÁͬÑùpH,×îºó²¹¼ÓË®ÖÁͬÑùÌå»ý,´ËʱÈýÆ¿ÖÐ[HC2O4-]µÄ¹ØÏµÊÇ--------------------( ) (A) ¼×Æ¿×îС (B) ÒÒÆ¿×î´ó (C) ±ûÆ¿×îС (D) ÈýÆ¿ÏàµÈ 0328 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-] 0329 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- 0330 ÒÑÖª: H3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36, µ÷½ÚÁ×ËáÑÎÈÜÒºµÄpHÖÁ6.0ʱ,Æä¸÷ÓйشæÔÚÐÎʽŨ¶È¼äµÄ¹ØÏµÊÇ------------------------------------------------------------------( ) (A) [HPO42-] > [H2PO4-] > [PO43-] (B) [HPO42-] > [PO43-] > [H2PO4-] (C) [H2PO4-] > [HPO42-] > [H3PO4] (D) [H3PO4] > [H2PO4-] > [HPO42-] 0331 ½ñÓÐ(a)NaH2PO4,(b)KH2PO4ºÍ(c)NH4H2PO4ÈýÖÖÈÜÒº,ÆäŨ¶Èc(NaH2PO4) = c(KH2PO4) = c(NH4H2PO4) = 0.10mol/L, ÔòÈýÖÖÈÜÒºµÄpHµÄ¹ØÏµÊÇ--------------------------------------( ) [ÒÑÖª H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12¡¢7.20¡¢12.36; pKa(NH4+) = 9.26] (A) a = b = c (B) a c (D) a = b

0335 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-] 0336 ÓÚ60 mL 0.10mol/L Na2CO3ÈÜÒºÖмÓÈë40 mL 0.15mol/L HClÈÜÒº,ËùµÃÈÜÒºµÄ¼ò»¯ÖÊ×ÓÌõ¼þÊÇ---------------------------------------------------------------------------------------------------( ) (A) [H2CO3] = [HCO3-] (B) [HCO3-] = [CO32-] (C) [H2CO3] = [CO32-] (D) [H2CO3] = [H+] 0337 0.1 mol/L Ag(NH3)2+ÈÜÒºµÄÎïÁÏÆ½ºâʽÊÇ-------------------------------------------------( ) (A) [Ag+] = [NH3] = 0.1 mol/L (B) 2[Ag+] = [NH3] = 0.2 mol/L (C) [Ag+]+[Ag(NH3)+]+2[Ag(NH3)2+] = 0.1 mol/L (D) [NH3]+[Ag(NH3)+]+2[Ag(NH3)2+] = 0.2 mol/L 0338 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+ 0339 ±û¶þËá[CH2(COOH)2]µÄpKa1 = 3.04, pKa2 = 4.37, Æä¹²éî¼îµÄKb1 = _______________, Kb2 = ______________________¡£ 0340 ÒÑÖªH2CO3µÄpKa1 = 6.38, pKa2 = 10.25, ÔòNa2CO3µÄKb1 = ___________________, Kb2 = _________________¡£ 0341 H3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36, ÔòPO43-µÄpKb1 = ___________, pKb2 = ___________, pKb3 = ____________¡£ 0342 ßÁà¤Å¼µª¼ä±½¶þ·Ó(PAR)µÄËá½âÀë³£ÊýpKa1 ,pKa2 ,pKa3·Ö±ðΪ3.1, 5.6, 11.9, ÔòÆäÖÊ×Ó»¯0346 ij¶þÔªÈõËáH2AµÄpKa1 = 2.0, pKa2 = 5.0, ÇëÌîдÒÔÏÂÇé¿öµÄpH¡£ 2--2-[H2A] = [HA-] [HA-]Ϊ×î´óÖµ [H2A] = [A] [HA] = [A] 0347 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 0348 ÒÑÖªNH3µÄKb=1.8¡Á10-5, µ±NH3-NH4Cl»º³åÈÜÒºµÄpH=9.0ʱ, ¸ÃÈÜÒºÖÐ [NH3]/[NH4Cl]Ϊ______¡£ 0349 ??? pHΪ7.20µÄÁ×ËáÑÎÈÜÒº(H3PO4µÄpKa1 = 2.12, pKa2 = 7.20, pKa3 = 12.36), Á×ËáÑδæÔڵij£ÊýK1 = ______________, K2 = ______________, ÀÛ»ýÖÊ×Ó»¯³£Êý?2 = ____________, Ö÷ÒªÐÎʽΪ__________ºÍ____________; ÆäŨ¶È±ÈΪ___________________¡£ ?3? = ____________¡£ 0350 0 343 ±È½ÏÒÔϸ÷¶ÔÈÜÒºµÄpH´óС(Ó÷ûºÅ >¡¢ = ¡¢< ±íʾ) ÒÑÖªÄûÃÊËá(H3A)µÄpKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40¡£ÔòÆä¹²éî¼îµÄKb1 = (1) ͬŨ¶ÈµÄNaH2PO4(a)ºÍNH4H2PO4(b): (a)____(b) (2) ͬŨ¶ÈµÄNa2HPO4(c)ºÍ(NH4)2HPO4(d): (c)____(d) ?___________________________ ¡¢ Kb2 = ___________________________; ÖÊ×Ó»¯³£ÊýK2 = [ÒÑÖªpKb(NH3) = 4.74,H3PO4µÄpKa1~pKa3·Ö±ðÊÇ 2.16,7.20,12.36] ?______________¡¢K3 = ________________¡£ 0351 ÒÔÏÂÆ½ºâʽµÄÓ¢ÎÄËõд·Ö±ðÊÇ(ÇëÌîA,B,C)¡£ 0344 (1) ÎïÁÏÆ½ºâʽ ____ £¨A£© PBE HPO42-ÊÇ______µÄ¹²éîËá,ÊÇ__________µÄ¹²éî¼î,ÆäË®ÈÜÒºµÄÖÊ×ÓÌõ¼þʽ ÊÇ (2) µçºÉƽºâʽ ____ £¨B£© MBE (3) ÖÊ×ÓÆ½ºâʽ ____ £¨C£© CBE _________________________________________________¡£ 0345 0352 ²ÝËá(H2C2O4)µÄpKa1ºÍpKa2·Ö±ðÊÇ1.2ºÍ4.2¡£ÇëÌîдÒÔÏÂÇé¿öµÄpH»òpH·¶Î§¡£ ˵Ã÷ÒÔÏÂÓ¢ÎÄËõдµÄº¬Òå(ÇëÌî A,B,C)¡£ -2-2-2--[HC2O4] = [C2O4] [H2C2O4] = [C2O4] C2O4ΪÖ÷ [HC2O4]Ϊ×î´óÖµ (1) MBE ____ £¨A£© µçºÉƽºâʽ (2) CBE ____ £¨B£© ÖÊ×ÓÆ½ºâʽ (3) PBE ____ £¨C£© ÎïÁÏÆ½ºâʽ 0353 º¬0.10mol/L HClºÍ0.20mol/L H2SO4µÄ»ìºÏÈÜÒºµÄÖÊ×ÓÌõ¼þʽΪ______________ _________________________________¡£ 0354 60 mL 0.10 mol/L Na2CO3Óë40 mL 0.15 mol/L HClÏà»ìºÏ, ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ________________________________________________¡£ 0355 ÓÃNaOHµÎ¶¨¶þÂÈÒÒËá(HA, pKa = 1.3)ºÍ NH4Cl »ìºÏÒºÖеĶþÂÈÒÒËáÖÁ»¯Ñ§¼ÆÁ¿µãʱ, ÆäÖÊ×ÓÌõ¼þʽÊÇ_______________________________________¡£ 0356 ÓÃÏ¡ H2SO4 ÈÜÒºµÎ¶¨ Na2CO3 ÈÜÒºÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µãʱ,ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ: _____________________________________________________¡£ 0357 д³öÏÂÁÐÈÜÒºµÄÖÊ×ÓÌõ¼þʽ: (1) 0.1 mol/L NH4AcÈÜÒº: ____________________________________ (2) 0.1 mol/L H2SO4ÈÜÒº: ____________________________________ 0358 0.1 mol/L Na2CO3 ÈÜÒºÖÐ, ÓйØNa+µÄÎïÁÏÆ½ºâʽÊÇ______________________________ _______________________¡£ÓйØCO32-µÄÎïÁÏÆ½ºâʽÊÇ____________________________ _____________________¡£ 0359 0.1 mol/L (NH4)2HPO4ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ____________________________________ ___________________________¡£ 0.1 mol/L H2SO4ÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ_______________________________¡£ 0360 ij(NH4)2CO3ÈÜÒºc[(NH4)2CO3] = c mol/L,ÆäÎïÁÏÆ½ºâʽΪ______________________ _____________________¡£ 0361 1. ½«20mL 0.10mol/L NaOHÈÜÒººÍ10 mL 0.10 mol/L H2SO4»ìºÏ, ËùµÃÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ_________________________________________________¡£ 2. ½«20 mL 0.10 mol/L HCl ºÍ 20 mL 0.05 mol/L Na2CO3»ìºÏ, ËùµÃÈÜÒºµÄÖÊ×ÓÌõ¼þʽÊÇ_________________________________________________¡£ 0362 ijNH4HC2O4ÈÜÒºc(NH4HC2O4) = c mol/L,ÆäÎïÁÏÆ½ºâʽΪ_________________ _____________; µçºÉƽºâʽΪ_________________________________________¡£ 0363 ij(NH4)2HPO4ÈÜÒºc[(NH4)2HPO4] = c mol/L,ÆäÎïÁÏÆ½ºâʽΪ_______________ _______________________________; µçºÉƽºâʽΪ__________________________________ __________________________¡£ 0364 ¼ÆËãÒÔÏÂÈÜÒºµÄ[H+]: (1) 0.10 mol/L NH4CNÈÜÒº (2) 0.10 mol/L NH4A ÈÜÒº ÒÑÖª pKa(NH4+) = 9.26, pKa(HCN) = 9.21, pKa(HA) = 1.30¡£ 0365 ¼ÙÉèijËá¼îָʾ¼ÁHInµÄ±äÉ«pH·¶Î§Îª2.60, Èô¹Û²ìµ½¸ÕÏÔËáʽ(HIn)ɫʱ±ÈÂÊ[HIn]/[In-]ºÍ¼îʽ(In-)ɫʱ[In-]/[HIn]ÊÇÏàͬµÄ¡£µ±Ö¸Ê¾¼Á¸ÕÏÔËáÉ«»ò¼îɫʱ,HIn»òIn-ÐÎÌåËùÕ¼µÄ°Ù·Ö±ÈΪ¶àÉÙ? 0366 pKa(HCOOH) = 3.77, pKb(HCOO-) = _________; NaOHµÎ¶¨HCOOH·´Ó¦µÄKt = ______________; HClµÎ¶¨HCOO-·´Ó¦µÄKt = ______________¡£ 0401 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 0402 ÓÃNaOHµÎ¶¨Ä³Ò»ÔªËáHA,ÔÚ»¯Ñ§¼ÆÁ¿µãʱ,[H+]µÄ¼ÆËãʽÊÇ--------------------------( ) (A) Kc(??)a?c(??) (B) Ka?c(??) (C) KWKa?KWK? (D) a?c(?)c(??) 0403 EDTAµÄpKa1~pKa6·Ö±ðÊÇ0.9,1.6,2.0,2.67,6.16ºÍ10.26¡£EDTA¶þÄÆÑÎ(Na2H2Y)Ë®ÈÜÒºpHÔ¼ÊÇ---------------------------------------------------------------------------------------------------( ) (A) 1.25 (B) 1.8 (C) 2.34 (D) 4.42 0404 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 0405 pH=1.00µÄHClÈÜÒººÍpH=13.00 µÄ NaOH ÈÜÒºµÈÌå»ý»ìºÏ,ËùµÃÈÜÒºµÄpHÊÇ-------------------------------------------------------------------------------------------------------------( ) (A) 14 (B) 12 (C) 7 (D) 6 0406 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 0407 ½«µÈÌå»ýµÄpH=3µÄHClÈÜÒººÍpH=10µÄNaOHÈÜÒº»ìºÏºó,ÈÜÒºµÄpHÇø¼äÊÇ-( ) (A) 3~4 (B) 1~2 (C) 6~7 (D) 11~12 0408 0.10 mol/L NH2OH(ôǰ·)ºÍ0.20 mol/L NH4ClµÈÌå»ý»ìºÏÒºµÄpHÊÇ--------------( ) [pKb(NH2OH) = 8.04, pKb(NH3) = 4.74] (A) 6.39 (B) 7.46 (C) 7.61 (D) 7.76 0409 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 0410 ÒÔÏÂÈÜҺϡÊÍ10±¶Ê±pH¸Ä±ä×î´óµÄÊÇ--------------------------------------------------( ) (A) 0.1 mol/L NaAc-0.1 mol/L HAcÈÜÒº (B) 0.1 mol/L NH4Ac-0.1 mol/L HAcÈÜÒº (C) 0.1 mol/L NH4AcÈÜÒº (D) 0.1 mol/L NaAcÈÜÒº 0411 ½«Ï±íÖÐËÄÖÖÈÜÒºÒÔˮϡÊÍ10±¶,ÇëÌîдÆäpH±ä»¯´óСµÄ˳Ðò,±ä»¯×î´óÕßΪ¡°1¡±¡¢×îСÕßΪ¡°4¡±¡£ ÈÜÒº ˳Ðò 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 0412 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 0413 ½ñÓÐÈýÖÖÈÜÒº·Ö±ðÓÉÁ½×é·Ö×é³É: (a) 0.10 mol/L HCl-0.20 mol/L NaAcÈÜÒº (b) 0.20 mol/L HAc-0.10 mol/L NaOHÈÜÒº (c) 0.10 mol/L HAc-0.10 mol/L NH4AcÈÜÒº ÔòÈýÖÖÈÜÒºpHµÄ´óС¹ØÏµÊÇ-------------------------------------------------------------------( ) [ÒÑÖªpKa(HAc) = 4.74, pKa(NH4+) = 9.26] (A) ac (D) a = b = c 0414 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 0415 ÏÂÁÐÑεÄË®ÈÜÒº»º³å×÷ÓÃ×îÇ¿µÄÊÇ---------------------------------------------------------( ) (A) NaAc (B) Na2CO3 (C) Na2B4O7¡¤10H2O (D) Na2HPO4 0416 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 0417 ½ñÓûÓÃH3PO4ÓëNa2HPO4À´ÅäÖÆpH = 7.2µÄ»º³åÈÜÒº,ÔòH3PO4ÓëNa2HPO4ÎïÖʵÄÁ¿Ö®±Èn(H3PO4)¡Ãn(Na2HPO4)Ó¦µ±ÊÇ---------------------------------------------------------------( ) (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) (A) 1:1 (B) 1:2 (C) 1:3 (D) 3:1 0418 ½ñÓûÓÃH3PO4ÓëNaOHÀ´ÅäÖÆpH = 7.20µÄ»º³åÈÜÒº,ÔòH3PO4ÓëNaOHÎïÖʵÄÁ¿Ö®±È n(H3PO4)¡Ãn(NaOH)Ó¦µ±ÊÇ--------------------------------------------------------------------------( ) (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) (A) 1:1 (B) 1:2 (C) 2:1 (D) 2:3 0419 ½ñÓûÅäÖÆÒ»pH=7.20µÄ»º³åÈÜÒº,ËùÓà 0.10 mol/L H3PO4ºÍ0.10 mol/L NaOHÈÜÒºµÄÌå»ý±ÈÊÇ----------------------------------------------------------------------------------------------------( ) (H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) (A) 1:3 (B) 3:1 (C) 2:3 (D) 3:2 0420 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) 0421 ÓûÅäÖÆpH=9µÄ»º³åÈÜÒº,ӦѡÓÃ---------------------------------------------------------------( ) (A) NH2OH(ôǰ±) (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¶¨Åðɰ(Na2B4O7¡¤10H2O)µÄ·´Ó¦Ê½Îª_____________________, Æä»¯Ñ§¼ÆÁ¿µã[H+]µÄ¼ÆËãʽΪ__________________¡£Èôc(Na2B4O7) = 0.050 mol/L, c(HCl) = 0.10 mol/L,Æä»¯Ñ§¼ÆÁ¿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»ÒªÇó½üËÆ¼ÆËã) 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×»ùËİ·)µÄpKb = 8.9,ÓÉ(CH2)6N4-HCl×é³ÉµÄ»º³åÈÜÒºµÄ»º³å·¶Î§ÊÇ_____________,ÓûÅäÖÆ¾ßÓÐ?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×»ùËİ·[(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) ¾ÆÊ¯ËáÇâ¼Ø ÒÑÖª: pKa1 pKa2 pKa3 H3PO4 2.12 7.20 12.36 °±»ùÒÒËá 2.35 9.60 ¾ÆÊ¯Ëá 3.04 4.37 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) 0447 ij·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] 0448 ½ñÓûÅäÖÆpHΪ7.50µÄÁ×ËỺ³åÒº1 L,ÒªÇóÔÚ50 mL´Ë»º³åÒºÖмÓÈë5.0 mL 0.10 mol/LµÄHClºópHΪ7.10, ÎÊӦȡŨ¶È¾ùΪ0.50 mol/LµÄH3PO4ºÍNaOHÈÜÒº¸÷¶àÉÙºÁÉý?(H3PO4µÄpKa1~pKa3·Ö±ðÊÇ2.12,7.20,12.36) 0449 Óû½«100ml 0.10 mol/L HClÈÜÒºµÄpH´Ó1.00Ôö¼ÓÖÁ4.44ʱ,Ðè¼ÓÈë¹ÌÌå´×ËáÄÆ(NaAc)¶àÉÙ¿Ë(²»¿¼ÂǼÓÈëNaAcºóÈÜÒºÌå»ýµÄ±ä»¯)? [Mr(NaAc) = 82.0, pKa(HAc) = 4.74] 0450 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¡£ 0451 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) 0452 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¡£ 0453 ½ñÓÉijÈõËáHB¼°Æä¹²éî¼îÅäÖÆ»º³åÈÜÒº,ÒÑÖªÆäÖй²éîËá [HB] = 0.25mol/L¡£ÓÚ´Ë100 mL»º³åÈÜÒºÖмÓÈë200 mgµÄ¹ÌÌåNaOH(ºöÂÔÌå»ýµÄ±ä»¯)ºó,ËùµÃÈÜÒºµÄpHΪ5.60 ¡£ÎÊÔ­À´ËùÅäÖÆµÄ»º³åÈÜÒºµÄpHΪ¶àÉÙ? [Mr(NaOH) = 40.0, Ka(HB) = 5.0¡Á10-6] 0454 Èô½«0.10 mol/L HAcÈÜÒººÍ0.20 mol/L NaOHÈÜÒºÖ±½Ó»ìºÏ,ÅäÖÆ³ÉpHΪ5.20 µÄ»º³åÈÜÒº1 L,ÎÊÐè¼ÓÈëÉÏÊöÈÜÒº¸÷¶àÉÙºÁÉý? [ÒÑÖªpKa(HAc) = 4.74] 0455 ÒÑÖªÖÊ×Ó»¯ÒÒ¶þ°·ÑÎËáÑÎ(ÒÔH2en2+±íʾ)µÄpKa1= 6.85, pKa2= 9.93, ÓûÅäÖÆ1L pH=6.55¡¢×ÜŨ¶ÈΪ0.15 mol/LµÄÒÒ¶þ°·ÑÎËáÑλº³åÈÜÒº,Ðè¼ÓÈë¹ÌÌåNaOH¶àÉÙ¿Ë? [ÒÑÖªMr(NaOH) = 40.00] 0456 0.60 mol/L HClÈÜÒºÓë1.8 mol/L°±»ùÒÒËáÈÜÒºµÈÌå»ý»ìºÏºó,¸ÃÈÜÒºµÄpHÊǶà´ó? (pKa1 = 2.35, pKa2 = 9.78) 0457 ΪÅäÖÆpHΪ4.00ºÍ5.00µÄHAc-Ac-»º³åÈÜÒº,ÎÊ·Ö±ðÓ¦Íù100 mL 0.30 mol/L HAcÈÜÒºÖмÓÈë¶àÉÙºÁÉý2.0 mol/L NaOHÈÜÒº? (HAcµÄpKa = 4.74) 0458 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] 0459 ¼ÆËãÒÔϸ÷ÈÜÒºµÄpH: (1) 20 mL 0.10 mol/L HClÈÜÒºÓë20 mL 0.10 mol/L HAc»ìºÏ (2) 0.20 mL 0.10 mol/L HClÈÜÒºÓë20 mL 0.10 mol/L HAcÏà»ìºÏ [pKa(HAc) = 4.74]

0459 HClÓëHAc»ìºÏÒºµÄÖÊ×ÓÌõ¼þʽÊÇ [H+] = c(HCl)+[Ac-] 1.ÓÃ×î¼òʽ [H+] = c(HCl) = 20¡Á0.10/40 = 0.050 (mol/L) pH=1.30 Kac(HAc) 10-4.74-1.30 ´Ëʱ[Ac-] = ©¤©¤©¤©¤©¤ = ©¤©¤©¤©¤©¤©¤©¤©¤ [H+] 10-1.30 = 10-4.74(mol/L)<<0.05mol/L ºöÂÔAc-ÏîºÏÀí 2. 0.20¡Á0.10 c(HCl) = ©¤©¤©¤©¤©¤©¤ = 0.0010 (mol/L) c(HAc) = 0.10 mol/L 20.2 ÓýüËÆÊ½ 10-4.74-1.00 [H+] = 0.0010+©¤©¤©¤©¤©¤©¤©¤ [H+] ½âµÃ[H+] = 10-2.72mol/L pH = 2.72

0460 0.10 mol/L NaOHÓë0.050 mol/L H2SO4µÈÌå»ý»ìºÏ,¼ÆËãÈÜÒºpH¡£ (H2SO4µÄpKa2 = 1.99) 0461 ÓÃ0.1000 mol/L NaOHÈÜÒºµÎ¶¨0.05000 mol/L H2SO4,¼ÆËãµÎ¶¨·ÖÊýT ·Ö±ðΪ 0 ºÍ1.00 ʱÈÜÒºµÄpH (H2SO4µÄpKa2 = 1.99)¡£ 0462 ÓÃ0.020 mol/L EDTAµÎ¶¨Í¬Å¨¶ÈµÄ25 mL Zn2+ÈÜÒº, µÎ¶¨¿ªÊ¼Ê±pH = 5.50, Ï£ÍûµÎ¶¨ÖÕÁËʱÈÜÒºpHϽµ²»µ½0.30¡£Èô²ÉÓÃHAc-Ac-»º³åÈÜÒº,¼ÓÈëÁ¿Îª5 mL, ÎÊÈôÅä´ËÈÜÒº1 L,Ó¦¼Ó¶àÉÙ¿ËNaAc¡¤3H2OºÍ¶àÉÙºÁÉý±ù´×Ëá{c[HAc(l)] = 17 mol/L}? [pKa(HAc) = 4.74, Mr(NaAc¡¤3H2O) = 136] 0463 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 } 0464 ÐèÒªÖÆ±¸200mL pH = 9.49µÄNH3-NH4Cl»º³åÈÜÒº,ÇÒʹ¸ÃÈÜÒºÔÚ¼ÓÈë1.0 mmol µÄHCl»òNaOHʱpHµÄ¸Ä±ä²»´óÓÚ0.12, ÖÆ±¸¸Ã»º³åҺʱÐèÓöàÉٿ˵ÄNH4ClºÍ¶àÉÙºÁÉý1.0 mol/L°±Ë®? [pKb(NH3) = 4.74, Mr(NH4Cl) = 53.49] 0465 ½ñÓÐÒ»»ìºÏ¼îÒºCH3NH2-(CH2)6N4, Ũ¶È¾ùÔ¼0.1 mol/L, ÄÜ·ñÓÃËá¼îµÎ¶¨·¨Ö±½Ó²â¶¨CH3NH2Ũ¶È(ÔÊÐíÎó²î0.5£¥),˵Ã÷Åжϸù¾Ý¡£ÈôÄÜ,²ÉÓÃͬŨ¶ÈHCl±ê×¼ÈÜÒºµÎ¶¨,¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH¡£ {pKb(CH3NH2) = 3.38, pKb[(CH2)6N4] = 8.85} 0466 ÓÉijÈõ¼îBOH¼°ÆäÑÎBClÅäÖÆ³É»º³åÈÜÒº,ʹÆäpH = 10.00¡£µ±Ïò140 mL´Ë»º³åÈÜÒºÖмÓÈë60 mL 1.0 mol/L HClÈÜÒººó,»º³åÈÜÒºµÄpH±ä»¯ÖÁ9.00¡£¼ÆËãÔ­À´ËùÅäÖÆµÄ»º³åÈÜÒºÖÐ, BOH¼°BClµÄƽºâŨ¶È¸÷Ϊ¶àÉÙ?[ÉèKb(BOH) = 5.0¡Á10-5] 0501 ½«Å¨¶ÈÏàͬµÄÏÂÁÐÈÜÒºµÈÌå»ý»ìºÏºó,ÄÜʹ·Óָ̪ʾ¼ÁÏÔºìÉ«µÄÈÜÒºÊÇ-------------( ) (A) °±Ë®+´×Ëá (B) ÇâÑõ»¯ÄÆ+´×Ëá (C) ÇâÑõ»¯ÄÆ+ÑÎËá (D) Áù´Î¼×»ùËİ·+ÑÎËá 0502 ½«¼×»ù³Èָʾ¼Á¼Óµ½Ò»ÎÞɫˮÈÜÒºÖÐ,ÈÜÒº³Ê»ÆÉ«,¸ÃÈÜÒºµÄËá¼îÐÔΪ---------------( ) (A) ÖÐÐÔ (B) ¼îÐÔ (C) ËáÐÔ (D) ²»ÄÜÈ·¶¨ÆäËá¼îÐÔ 0503 Ç¿ËáµÎ¶¨Èõ¼î,ÒÔÏÂָʾ¼ÁÖв»ÊÊÓõÄÊÇ----------------------------------------------------( ) (A) ¼×»ù³È (B) ¼×»ùºì (C) ·Ó̪ (D) äå·ÓÀ¶ (pT = 4.0) 0504 ½«·Óָ̪ʾ¼Á¼Óµ½Ä³ÎÞÉ«ÈÜÒºÖÐ,ÈÜÒºÈÔÎÞÉ«,±íÃ÷ÈÜÒºËá¼îÐÔΪ---------------------( ) (A) ËáÐÔ (B) ÖÐÐÔ (C) ¼îÐÔ (D) ²»ÄÜÈ·¶¨ÆäËá¼îÐÔ 0505 ijËá¼îָʾ¼ÁµÄK(HIn)Ϊ1.0¡Á10-5,Æä±äÉ«µãpHΪ_________,ÀíÂÛ±äÉ«·¶Î§Îª____________¡£ 0506 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 0507 ÔÚÏÂÁжàÔªËá»ò»ìºÏËáÖÐ,ÓÃNaOHÈÜÒºµÎ¶¨Ê±³öÏÖÁ½¸öµÎ¶¨Í»Ô¾µÄÊÇ-------------( ) (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) 0508 ÓÃ0.1 mol/L HClµÎ¶¨0.1 mol/L NaOHµÄͻԾ·¶Î§Îª9.7~4.3, Ôò0.01 mol/L HClµÎ¶¨0.01 mol/L NaOHµÄͻԾ·¶Î§Ó¦Îª-------------------------------------------------------------------( ) (A) 9.7~4.3 (B) 8.7~4.3 (C) 8.7~6.3 (D) 10.7~3.3 0509 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 0510 ÒÔͬŨ¶ÈNaOHÈÜÒºµÎ¶¨Ä³Ò»ÔªÈõËá(HA),Èô½«ËáºÍ¼îµÄŨ¶È¾ùÔö´ó10±¶, Á½Öֵζ¨pHÏàͬʱËùÏàÓ¦µÄÖкͰٷÖÊýÊÇ----------------------------------------------------------------( ) (A) 0 (B) 50 (C) 100 (D) 150 0511 ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨Ò»ÔªÈõËáʱ,ÈôÈõËáºÍNaOHµÄŨ¶È¶¼±ÈÔ­À´Ôö´óÊ®±¶,ÔòµÎ¶¨ÇúÏßÖÐ-----------------------------------------------------------------------------------------------------( ) (A) »¯Ñ§¼ÆÁ¿µãǰºó0.1%µÄpH¾ùÔö´ó (B) »¯Ñ§¼ÆÁ¿µãǰºó0.1%µÄpH¾ù¼õС (C) »¯Ñ§¼ÆÁ¿µãǰ0.1%µÄpH²»±ä,ºó0.1%µÄpHÔö´ó (D) »¯Ñ§¼ÆÁ¿µãǰ0.1%µÄpH¼õС,ºó0.1%µÄpHÔö´ó 0512 ÓÃ0.10 mol/LNaOHÈÜÒºµÎ¶¨0.10 mol/L HA(Ka=5.0¡Á10-5), ÈôÖÕµãµÄpHΪ9.0, ÔòÖÕµãÎó²îΪ----------------------------------------------------------------------------------------------( ) (A) +0.02% (B) +0.01% (C) -0.02% (D) -0.01% 0513 ÒÔÏÂÐðÊöÕýÈ·µÄÊÇ-------------------------------------------------------------------------------( ) (A) ÓÃNaOHµÎ¶¨HCl, Ñ¡¼×»ù³ÈΪָʾ¼ÁµÄÖÕµãÎó²îÊÇÕýÖµ (B) ÓÃHClµÎ¶¨NaOH, Ñ¡·Ó̪Ϊָʾ¼ÁµÄÖÕµãÎó²îΪÕýÖµ (C) ÓÃÕôÁ󷨲âNH4+, Èô²ÉÓÃHClÎüÊÕNH3,ÒÔNaOH·µµÎÖÁpHΪ7, ÖÕµãÎó²îΪ¸ºÖµ (D) ÓÃÕôÁ󷨲âNH4+, Èô²ÉÓÃH3BO3ÎüÊÕNH3,ÒÔHClµÎ¶¨ÖÁ¼×»ù³È±äÉ«, ÖÕµãÎó²îΪ¸ºÖµ 0514 ÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨0.1mol/LHCl-0.1mol/L H3PO4»ìºÏÒº,Ôڵζ¨ÇúÏßÉϳöÏÖ¼¸¸öͻԾ-----------------------------------------------------------------------------------------------------------( ) (A) 1 (B) 2 (C) 3 (D) 4 0515 ÏÂͼΪÓÃHClµÎ¶¨Na2CO3µÄµÎ¶¨ÇúÏß¡£ÈôaµãʱÈÜÒºÖÐÕ¼ÓÅÊÆµÄÐÎÌåΪCO32-,ÊÔÖ¸³öÒÔϸ÷µãÕ¼ÓÅÊÆµÄÐÎÌå(ÌîA,B,C»òD)¡£ (1) bµã _____ £¨A£© [HCO3-] = [CO32-] (2) cµã _____ £¨B£© [HCO3-] (3) dµã _____ £¨C£© [HCO3-] = [H2CO3] (4) eµã _____ £¨D£© [H2CO3] 0516 ÓÃ0.2 mol/L NaOHÈÜÒºµÎ¶¨0.2 mol/L HClºÍ0.2 mol/LÄûÃÊËá(H3A)µÄ»ìºÏÒº(H3AµÄKa1 = 7.4¡Á10-4, Ka2 = 1.7¡Á10-5, Ka3 = 4.0¡Á10-7), Èç¹ûÔÊÐíµÎ¶¨Îó²îΪ0.2%,ÔòÖÕµãʱÈÜÒº×é³ÉӦΪ---------------------------------------------------------------------------------------------------( ) (A) NaCl+H3A (B) NaCl+NaH2A (C) NaCl+Na2HA (D) NaCl+Na3A

0517 ²â¶¨(NH4)2SO4ÖеĵªÊ±,²»ÄÜÓÃNaOH±ê×¼ÈÜÒºÖ±½ÓµÎ¶¨,ÕâÊÇÒòΪ----------------( ) (A) NH3 µÄKb̫С (B) (NH4)2SO4²»ÊÇËá (C) NH4+µÄKa̫С (D) (NH4)2SO4Öк¬ÓÎÀëH2SO4

0518 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_________________________¡£ 0519 0.1 mol/L HClµÎ¶¨20.00 mLÏàͬŨ¶ÈNH3

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µã V(NaOH)/mL ÌåϵµÄ×é³É [H+]µÄ¼ÆËãʽ A 19.98 B 20.00 C 20.02 0521 0.1 mol/L NaOHÈÜÒºµÎ¶¨20.00 mLÏàͬŨ¶ÈHAcµÄµÎ¶¨ÇúÏßÈçÏÂͼ, д³öÇúÏßÉÏA,B,CÈýµãµÎ¶¨ÌåϵµÄ×é³ÉºÍ[H+]µÄ¼ÆËãʽ¡£ µã V(NaOH)/mL ÌåϵµÄ×é³É [H+]µÄ¼ÆËãʽ A 0.00 B 19.98 C 20.00 0522 0.1 mol/L HClµÎ¶¨20.00 mL Na2CO3µÄµÎ¶¨ÇúÏßÈçÏÂͼ, д³öÇúÏßÉÏA¡¢BÁ½µãµÎ¶¨ÌåϵµÄ×é³ÉºÍ[H+]µÄ¼ÆËãʽ¡£ µã V(HCl)/mL ÌåϵµÄ×é³É [H+]µÄ¼ÆËãʽ A 20.00 B 40.00 0523 ÏÂͼµÎ¶¨ÇúÏßµÄÀàÐÍÊÇ___________________________¡£ ÒËÑ¡ÓõÄָʾ¼ÁÊÇ________________________¡£

0524 ÓÃHCl±ê×¼ÈÜÒºµÎ¶¨NH3,·Ö±ðÒÔ¼×»ù³ÈºÍ·Ó̪×÷ָʾ¼Á, ºÄÓõÄHClÌå»ý·Ö±ðÒÔV(¼×»ù³È)ÓëV(·Ó̪)±íʾ,ÔòV(¼×»ù³È)ÓëV(·Ó̪)µÄ¹ØÏµÊÇ:________; ÈôÊÇÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨HClʱÔòÊÇ__________¡£ (ÒÔ¡°?¡±£¬¡°?¡±»ò¡°¡Ö¡±À´±íʾ) 0525 ÓÃ0.1 mol/L HClµÎ¶¨0.1 mol/Lij¶þÔª¼î Na2B(pKb1 = 2, pKb2 = 3), µÎ¶¨¹ý³ÌÖÐÓÐ_____¸öͻԾ, ÖÕµã²úÎïÊÇ__________,¿ÉÓÃ___________×÷ָʾ¼Á¡£ 0526 ÓÃ0.1 mol/L NaOHÈÜÒºµÎ¶¨0.1 mol/Lij¶þÔªËáH2A ( pKa1 = 2.70, pKa2 = 6.00 ), ÓÐ______¸öͻԾ,ÕâÊÇÒòΪ________________¡£»¯Ñ§¼ÆÁ¿µãʱ¼ÆËã[H+]µÄ¹«Ê½Îª ____________________,¿ÉÑ¡ÓÃ_____________ָʾ¼Á¡£ 0527 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 ______________ (3) 0.1 mol/L HClµÎ¶¨Na2CO3ÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µã ______________ (4) 0.1 mol/L NaOHµÎ¶¨H3PO4ÖÁµÚ¶þ»¯Ñ§¼ÆÁ¿µã ______________ 0528 ÏÂÁеζ¨Öи÷ѡʲôָʾ¼ÁΪÒË? (1) ÓÃ0.1 mol/L HCl ÈÜÒºµÎ¶¨0.05 mol/L Na2CO3 ___________ (2) ÓÃ0.1 mol/L NaOHÈÜÒºµÎ¶¨0.05 mol/L ¾ÆÊ¯Ëá ___________ (3) ÓÃ0.1 mol/L NaOHÈÜÒºµÎ¶¨ 0.1 mol/L HCl ___________ (4) ÓÃ0.1 mol/L NaOHÈÜÒºµÎ¶¨0.1 mol/L H3PO4(µÎ¶¨ÖÁH2PO4-) ___________ 0529 ¸ù¾ÝϱíËù¸øÊý¾Ý,ÅжÏÔÚÒÔϵζ¨Öл¯Ñ§¼ÆÁ¿µã¼°Æäǰºó0.1%µÄpH¡£ p£È µÎ¶¨Ìåϵ »¯Ñ§¼ÆÁ¿µã »¯Ñ§ »¯Ñ§¼ÆÁ¿µã ǰ0.1% ¼ÆÁ¿µã ºó0.1% NaOHµÎ¶¨HCl(0.1 mol/L) 4.3 7.0 9.7 NaOHµÎ¶¨HCl(1 mol/L) HClµÎ¶¨NaOH(0.01 mol/L) 0530 ¸ù¾ÝϱíËù¸øÊý¾Ý,ÍÆ¶ÏÔÚÏÂÁеζ¨Öл¯Ñ§¼ÆÁ¿µã¼°Æäǰºó0.1%µÄpH¡£ (Ũ¶È¾ùΪ0.1 mol/L) p£È µÎ¶¨Ìåϵ »¯Ñ§¼ÆÁ¿µã »¯Ñ§ »¯Ñ§¼ÆÁ¿µã ǰ0.1% ¼ÆÁ¿µã ºó0.1% NaOHµÎ¶¨HCl 4.3 7.0 9.7 NaOHµÎ¶¨HA 8.7 HClµÎ¶¨B 5.3 0531 ¸ù¾ÝϱíËù¸øÊý¾ÝÍÆ¶ÏÓÃNaOHµÎ¶¨HAcÖÁÏÂÁи÷µãµÄpH¡£ p£È Ũ¶È »¯Ñ§¼ÆÁ¿µã »¯Ñ§ »¯Ñ§¼ÆÁ¿µã ǰ0.1% ¼ÆÁ¿µã ºó0.1% 0.1 mol/L 7.7 8.7 9.7 0.01 mol/L 0532 ¸ù¾ÝϱíËù¸øÊý¾Ý,ÍÆ¶ÏÓÃNaOHµÎ¶¨H3AÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µã¼°Æäǰºó0.5%µÄpH¡£ p£È Ũ¶È »¯Ñ§¼ÆÁ¿µã »¯Ñ§ »¯Ñ§¼ÆÁ¿µã ǰ0.5% ¼ÆÁ¿µã ºó0.5% 0.1 mol/L 4.3 4.6 4.9 1 mol/L 0533 ÓÃNaOHµÎ¶¨0.1 mol/LÒ»ÔªËá,ÇëÍê³ÉÏÂ±í¡£ HB p£È 99.9% 100.0% 100.1% HCl 4.3 7.0 9.7 HA(Ka¡Ö10-5) 8.8 HB(Ka¡Ö10-7) 9.8 10.0 0534 ÓÃ0.1000 mol/L NaOHµÎ¶¨25.00 mL 0.1000 mol/L HCl,ÈôÒÔ¼×»ù³ÈΪָʾ¼ÁµÎ¶¨ÖÁpH = 4.0ΪÖÕµã,ÆäÖÕµãÎó²îEt = _______________%¡£ 0535 ÈôÒÔ0.100 mol/L NaOHÈÜÒºµÎ¶¨20.0 mLŨ¶È¾ùΪ0.100 mol/LÑÎËáôǰ·(NH3+OH¡¤Cl-)ºÍNH4ClµÄ»ìºÏÈÜÒºÖеÄÑÎËáôǰ·¡£ (1) ¼ÆË㻯ѧ¼ÆÁ¿µãʱÈÜÒºµÄpH£» (2) »¯Ñ§¼ÆÁ¿µãʱÓаٷÖÖ®¼¸µÄNH4Cl²Î¼ÓÁË·´Ó¦? µÎ¶¨ÄÜ·ñ׼ȷ½øÐÐ? [ÒÑÖª ôǰ·Kb(NH2OH) = 9.1¡Á10-9, NH3µÄKb = 1.8¡Á10-5] 0536 ÓÃ0.10 mol/L NaOHÈÜÒºµÎ¶¨0.10 mol/L ¶þÂÈÒÒËá(¼òд³ÉHA), ÈôÈÜÒºÖл¹º¬ÓÐ 0.010 0544 ÓÃ0.20 mol/L NaOHµÎ¶¨0.20 mol/L HCl(ÆäÖк¬ÓÐ0.10 mol/L NH4Cl)¡£ mol/L NH4Cl, ¼ÆË㻯ѧ¼ÆÁ¿µãµÄpHºÍ¹ýÁ¿0.1%µÄpH¡£ [pKa(HA) = 1.30, pKa(NH4+) = 9.26] 0537 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) 0538 ¼ÆËãÒÔ0.20 mol/L Ba(OH)2ÈÜÒºµÎ¶¨0.10 mol/L HCOOHÈÜÒºÖÁ»¯Ñ§¼ÆÁ¿µãʱ,ÈÜÒºµÄpHΪ¶àÉÙ? [Ka(HCOOH) = 2.0¡Á10-4] 0539 ÓÃ0.10 mol/L NaOHÈÜÒºµÎ¶¨Í¬Å¨¶ÈÁÚ±½¶þ¼×ËáÇâ¼Ø(¼òд³ÉKHP)¡£¼ÆË㻯ѧ¼ÆÁ¿µã¼°Æäǰºó0.1%µÄpH¡£ (H2PµÄpKa1 = 2.95, pKa2 = 5.41) 0540 0.1000 mol/L NaOHÈÜÒºµÎ¶¨20.00 mL 0.05000 mol/L H2C2O4,¼ÆËã: (1) »¯Ñ§¼ÆÁ¿µãǰ0.1%µÄpH (2) »¯Ñ§¼ÆÁ¿µãµÄpH (3) »¯Ñ§¼ÆÁ¿µãºó0.1%µÄpH (H2C2O4µÄpKa1 = 1.25, pKa2 = 4.29) 0541 ÒÔ0.20 mol/L NaOHÈÜÒºµÎ¶¨0.20 mol/L HAcºÍ0.40 mol/L H3BO3µÄ»ìºÏÈÜÒºÖÁpHΪ6.20,¼ÆËãÖÕµãÎó²î¡£ [Ka(HAc) = 1.8¡Á10-5, Ka(H3BO3) = 5.8¡Á10-10 ]

0542 ÒÔ0.10 mol/L NaOHÈÜÒºµÎ¶¨0.10 mol/L HClºÍ0.20 mol/L H3BO3µÄ»ìºÏÈÜÒº¡£ (1) ¼ÆË㻯ѧ¼ÆÁ¿µãʱÈÜÒºµÄpH; (2) ÈôµÎ¶¨ÖÕµãpH±È»¯Ñ§¼ÆÁ¿µã¸ß0.5£¬¼ÆËãÖÕµãÎó²î¡£ [Ka(H3BO3) = 5.8¡Á10-10] 0543 ÓÃ0.10 mol/L HClµÎ¶¨0.10 mol/L NaOH, ¶øNaOHÊÔÒºÖл¹º¬ÓÐ0.10 mol/L NaAc, ¼ÆËã: (1) »¯Ñ§¼ÆÁ¿µã¼°»¯Ñ§¼ÆÁ¿µãǰºó0.1%µÄpH (2) µÎ¶¨ÖÁpH = 7.0ʱµÄÖÕµãÎó²î [pKa(HAc) = 4.74]

(1) ¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH; (2) ÈôµÎ¶¨ÖÁpH = 7.0, ÎÊÖÕµãʱÓаٷÖÖ®¼¸µÄNH4+±äΪNH3; (3) ´ËʱÖÕµãÎó²îÊǶàÉÙ? [pKa(NH4+)Ϊ9.26] 0549 ÓÃ0.10 mol/L NaOHµÎ¶¨Í¬Å¨¶ÈµÄÒ»ÔªÈõËáHA(pKa = 7.0),¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH,ÈôÒÔ·Ó̪Ϊָʾ¼Á(ÖÕµãʱpHΪ9.0), ¼ÆËãÖÕµãÎó²î¡£ 0545 Ó¦ÓÃËá¼îµÎ¶¨·¨²â¶¨º¬Á×ÎïÖÊ¡£ÊÔÑù¾­´¦Àíʹ³É¿ÉÈÜÐÔÁ×ËáÑκó, ÔÙת»¯Îª(NH4)2HPO4¡¤12MoO3¡¤H2O³Áµí, ´Ë³Áµí¾­¹ýÂËÏ´µÓºóÓùýÁ¿µÄNaOH±ê×¼ÈÜÒº(ÒÑ֪Ũ¶ÈºÍÌå»ý)Èܽâ,È»ºóÔÙÓÃHNO3±ê×¼ÈÜÒº·µµÎ¶¨,Ó÷Ó̪Ϊָʾ¼Á,µÎ¶¨ÖÁºìÉ«¸Õ¸ÕÍÊÈ¥(pH?8.0)¡£ÊÔ¼ÆËãµÎ¶¨ÖÕµãʱ, NH3¡¢HPO42-¡¢H2PO4-ºÍMoO42-¸÷×é·ÖµÄ·Ö²¼ÏµÊý¡£´ÓÉÏÊö¼ÆËã½á¹ûÖ¸³ö¸Ã·½·¨²â¶¨Á×ʱÖÕµãÎó²îÊǶàÉÙ? ÕâÒ»µÎ¶¨·´Ó¦µÄ·´Ó¦Ê½Îª: (NH4)2HPO4¡¤12MoO3+24OH- = 12MoO42-+HPO42-+2NH4++12H2O [pKb(NH3) = 4.75, pKa(HMoO4-) = 3.75, H3PO4µÄpKa1~pKa3·Ö±ðÊÇ 2.16, 7.21, 12.32] 0546 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) 0547 ÈôµÎ¶¨¼ÁÓë±»²âÎïÈÜҺŨ¶È¾ùÔö´óÊ®±¶, NaOHµÎ¶¨HClµÄµÎ¶¨Í»Ô¾__________, NaOHµÎ¶¨HAcµÄµÎ¶¨Í»Ô¾________________ , NaOHµÎ¶¨H3PO4(ÖÁH2PO4-)µÄµÎ¶¨Í»Ô¾_______________¡£(ÌîpHÔö¼Ó¶àÉÙ¡¢¼õÉÙ¶àÉÙ»ò²»±ä) 0548 ÓÃ0.10 mol/L NaOHµÎ¶¨Í¬Å¨¶ÈµÄ¼×Ëá(HA),¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH¼°Ñ¡¼×»ùºìΪָʾ¼Á(ÖÕµãʱpHΪ6.2)ʱµÄÖÕµãÎó²î¡£[pKa(HA) = 3.74] 0550 ÓÃ0.10 mol/L NaOHµÎ¶¨Í¬Å¨¶ÈµÄH2SO4,¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH¼°Ñ¡¼×»ù³ÈΪָʾ¼Á(ÖÕµãʱpHΪ4.4)µÄÖÕµãÎó²î¡£(H2SO4µÄpKa2 = 1.99) 0551 ÓÃ0.10 mol/L HClµÎ¶¨Í¬Å¨¶ÈµÄCH3NH2ÈÜÒº,¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH;ÈôÑ¡¼×»ù³ÈΪָʾ¼Á(ÖÕµãʱpHΪ4.0),¼ÆËãÖÕµãÎó²î¡£[pKb(CH3NH2) = 3.38] 0552 ÓÃ0.020 mol/L HClµÎ¶¨20.00 mL 0.020 mol/L Ba(OH)2,¼ÆË㻯ѧ¼ÆÁ¿µãǰºó0.1%µÄpH,ÈôÑ¡¼×»ù³ÈΪָʾ¼Á(ÖÕµãʱpHΪ4.0),¼ÆËãÖÕµãÎó²î¡£ 0553 ij¼×²â¶¨HAcµÄŨ¶È¡£ÒÆÈ¡25.00 mLÊÔÒº,ÓÃ0.1010 mol/L µÄ NaOH µÎ¶¨ÖÁ¼×»ù³È±ä»Æ(pH = 4.4)ÏûºÄÁË7.02 mL, Óɴ˼ÆËãHAcŨ¶ÈΪ0.02836 mol/L¡£ ijÒÒÖ¸³ö¼×µÄ´íÎóÊÇÑ¡´íָʾ¼Á,²¢×÷ÈçÏÂУÕý,ÒÔÇóµÃHAcµÄ׼ȷŨ¶È: ?4.4 pH = 4.4ʱ, x(??c)?1010?4.74?10?4.4?69% c(HAc) = 0.02836¡Á100/31 = 0.09148(mol/L) ÄãÈÏΪÒҵķ½·¨ÊÇ·ñÕýÈ·?Ϊʲô?

0601 ÔÚÏÂÁÐÈÜÒºÖÐÄÜÓÃNaOH±ê×¼ÈÜÒºÖ±½ÓµÎ¶¨µÄÊÇ (Et¡Ü0.1£¥) ------------------------( ) (A) 0.1mol/L NH2OH¡¤HCl(ÑÎËáôǰ·) [pKb(NH2OH) = 8.04] (B) 0.1mol/L NH4Cl(ÂÈ»¯ï§) [pKa(NH4+) = 9.26] (C) 0.1mol/L (CH2)6N4(Áù´Î¼×»ùËİ·) {pKb[(CH2)6N4] = 8.85} (D) 0.1mol/L C5H5N(ßÁà¤) [pKb(C5H5N) = 8.87] 0602 ÒÔ¼×»ùºìΪָʾ¼Á,ÄÜÓÃNaOH±ê×¼ÈÜҺ׼ȷµÎ¶¨µÄËáÊÇ------------------------------( ) (A) ¼×Ëá (B) ÁòËá (C) ÒÒËá (D) ²ÝËá 0603 ÏÂÁÐÈÜÒºÓÃËá¼îµÎ¶¨·¨ÄÜ׼ȷµÎ¶¨µÄÊÇ-----------------------------------------------------( ) (A) 0.1 mol/L HF (pKa = 3.18) (B) 0.1 mol/L HCN (pKa = 9.21) (C) 0.1 mol/L NaAc [pKa(HAc) = 4.74] (D) 0.1 mol/L NH4Cl [pKb(NH3) = 4.75] 0604 ÏÂÁеζ¨(Ũ¶È¾ùΪ0.1 mol/L)ÖпÉÐеÄÊÇ---------------------------------------------------( ) ÒÑÖªpKa(HA) = 4.85, pKa(HB) = 9.3, pKb(MOH) = 8.70, pKb(ROH) = 3.80 (A) HClµÎ¶¨A- (B) NaOHµÎ¶¨R+ (C) HClµÎ¶¨MOH (D) HClµÎ¶¨B- 0605 ÒÔϸ÷»ìºÏËá(»ò¼î)ÖÐ,ÄÜ׼ȷµÎ¶¨ÆäÖÐÇ¿Ëá(»òÇ¿¼î)µÄÊÇ------------------------------( ) (Ũ¶È¾ùΪ0.1 mol/L) (A) HCl+Ò»ÂÈÒÒËá (pKa = 2.86) (B) HCl+HAc (pKa = 4.74) (C) HCl+NH4Cl (pKa = 9.26) (D) NaOH+NH3 0606 ÒÑÖª¼×°·µÄpKbΪ3.38, ±½°·pKbΪ9.38, ¼×ËáµÄpKaΪ3.74,ÒÔÏ»ìºÏÒº(Ũ¶È¾ùΪ0.1 mol/L)ÄÜÓÃÖкͷ¨×¼È·µÎ¶¨»­Ïß×é·ÖµÄÊÇ-------------------------------------------------------( ) (A) NaOH - ¼×°· (B) NaOH - ±½°· (C) HCl - ¼×Ëá (D) HCl - FeCl3 0607 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 0608 ÓÃ˫ָʾ¼Á·¨²â¶¨¿ÉÄܺ¬ÓÐNaOH¼°¸÷ÖÖÁ×ËáÑεĻìºÏÒº¡£ÏÖȡһ¶¨Ìå»ýµÄ¸ÃÊÔÒº,ÓÃHCl±ê×¼ÈÜÒºµÎ¶¨,ÒÔ·Ó̪Ϊָʾ¼Á,ÓÃÈ¥HCl 18.02 mL¡£È»ºó¼ÓÈë¼×»ù³Èָʾ¼Á¼ÌÐøµÎ¶¨ÖÁ³Èɫʱ,ÓÖÓÃÈ¥20.50mL,Ôò´ËÈÜÒºµÄ×é³ÉÊÇ-------------------------------------------------------( ) (A) Na3PO4 (B) Na2HPO4 (C) NaOH+Na3PO4 (D) Na3PO4+Na2HPO4 0609 ijÈÜÒº¿ÉÄܺ¬ÓÐNaOHºÍ¸÷ÖÖÁ×ËáÑÎ,½ñÓÃÒ»HCl±ê×¼ÈÜÒºµÎ¶¨,ÒÔ·Ó̪Ϊָʾ¼Áʱ, ÓÃÈ¥ 12.84 mL, Èô¸ÄÓü׻ù³ÈΪָʾ¼ÁÔòÐè 20.24 mL, ´Ë»ìºÏÒºµÄ×é³ÉÊÇ----------------( ) (A) Na3PO4 (B) Na3PO4+NaOH (C) Na3PO4+Na2HPO4 (D) Na2HPO4+NaH2PO4 0610 ijһNaOHºÍNa2CO3»ìºÏÒº,ÓÃHClÈÜÒºµÎ¶¨,ÒÔ·Ó̪Ϊָʾ¼Á,ºÄÈ¥HCl V1(mL), ¼ÌÒÔ¼×»ù³ÈΪָʾ¼Á¼ÌÐøµÎ¶¨,ÓÖºÄÈ¥HCl V2(mL),ÔòV1ÓëV2µÄ¹ØÏµÊÇ-------------------------( ) (A) V1 = V2 (B) V1 = 2V2 (C) 2V2 = V2 (D) V1> V2 0611 ij¼îÒº25.00 mL, ÒÔ0.1000 mol/L HCl±ê×¼ÈÜÒºµÎ¶¨ÖÁ·Ó̪ÍÊÉ«,ÓÃÈ¥15.28 mL,ÔÙ¼Ó¼×»ù³È¼ÌÐøµÎ¶¨, ÓÖÏûºÄHCl 6.50 mL,´Ë¼îÒºµÄ×é³ÉÊÇ-------------------------------------------( ) (A) NaOH+NaHCO3 (B) NaOH+Na2CO3 (C) NaHCO3 (D) Na2CO3 0612 ¼×È©·¨²â¶¨NH4+,»ùÓÚÒÔÏ·´Ó¦Öû»³ö¡°Ëᡱ,ÔÙÓÃNaOHµÎ¶¨ 4NH4++ 6HCHO = (CH2)6N4H++ 3H++6H2O NH4+ÓëNaOHµÄ¼ÆÁ¿¹ØÏµn(NH4+):n(NaOH)ÊÇ-------------------------------------------( ) {pKb[(CH2)6N4] = 8.87} (A) 4:3 (B) 4:4(1:1) (C) 4:6 (D) 2:1 0613 Óü×È©·¨²â¶¨w[(NH4)2SO4]¡Ý98£¥µÄ·ÊÌï·Û{Mr[(NH4)2SO4] = 132}ÖÐNH4+º¬Á¿Ê±, Èô½«ÊÔÑùÈܽâºóÓÃ250 mL ÈÝÁ¿Æ¿¶¨ÈÝ, ÓÃ25 mL ÒÆÒº¹ÜÎüÈ¡Èý·ÝÈÜÒº×÷ƽÐвⶨ,·Ö±ðÓÃ0.2mol/L NaOHÈÜÒºµÎ¶¨,ÔòÓ¦³ÆÈ¡¶àÉÙ¿ËÊÔÑù?-----------------------------------------------( ) [Ìáʾ: 4NH4+ + 6HCHO = (CH2)6N4H++ 3H++ 3H2O ] (A) 2.6 g~4.0 g (B) 1.3 g~2.0 g (C) 5.2 g~8.0 g (D) 1.0 g~1.5 g 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

0615 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) 5.00 (B) 8.00 (C) 12.00 (D) 7.00 0616 ÓûÓÃËá¼îµÎ¶¨·¨ÔÚË®ÈÜÒºÖвⶨNaAcÊÔ¼ÁµÄ´¿¶È,²ÉÓÃָʾ¼ÁÈ·¶¨ÖÕµã,´ïµ½0.2£¥×¼È·

¶È,ÒÔϺÎÖÖ·½·¨¿ÉÐÐ------------------------------------------------------------------------------( ) [pKa(HAc) = 4.74]

(A) Ìá¸ß·´Ó¦ÎïŨ¶ÈÖ±½ÓµÎ¶¨ (B) ²ÉÓ÷µµÎ¶¨·¨²â¶¨

(C) Ñ¡ºÃָʾ¼Á,ʹ±äÉ«µãÇ¡Ó뻯ѧ¼ÆÁ¿µãÒ»Ö (D) ÒÔÉÏ·½·¨¾ù´ï²»µ½

0617 HCl¡¢H2SO4¡¢HNO3¼°HClO4µÄÇø·ÖÐÔÈܼÁÊÇ------------------------------------------( ) (A) Ë® (B) ÒÒ´¼

(C) ±ù´×Ëá (D) Òº°±

0618 ÏÂÁÐÄÄÖÖÈܼÁ,ÄÜʹHAc¡¢H3BO3¡¢HClºÍH2SO4ËÄÖÖËáÏÔʾ³öÏàͬµÄÇ¿¶ÈÀ´?--( ) (A) ´¿Ë® (B) Òº°± (C) ¼×»ùÒ춡ͪ (D) ÒÒ´¼

0619ÏÂÁжàÔªËá(¼î)Ũ¶È¾ùΪ0.1 mol/L, ÓÃÏàͬŨ¶ÈµÄNaOH(»òHCl)ÄÜ·ñ½øÐзֲ½µÎ¶¨(ÌîÄÜ»ò²»ÄÜ)

(1) Na2HPO4

(pKa1~pKa3:2.12¡¢7.20¡¢12.36) _______ (2) H2SO3

(pKa1~pKa2:1.81¡¢6.91)

_______ (3) CH2(COOH)2(±û¶þËá) (pKa1~pKa2:2.8¡¢6.16) _______ (4) NH2COONa(°±»ùÒÒËáÄÆ) (NH2COOHµÄpKa1~pKa2:2.35¡¢9.60) _______

0620 ÔÚÏÂÁÐÎïÖÊÖÐ,

NH4Cl [pKb(NH3) = 4.74] C6H5OH(±½·Ó) [pKa(C6H5OH) = 9.96]

Na2CO3 (H2CO3µÄpKa1 = 6.38,pKa2 = 10.25) NaAc [pKa(HAc) = 4.74] HCOOH

[pKa(HCOOH) = 3.74]

ÄÜÓÃÇ¿¼î±ê×¼ÈÜÒºÖ±½ÓµÎ¶¨µÄÎïÖÊÊÇ________________________ ; ÄÜÓÃÇ¿Ëá±ê×¼ÈÜÒºÖ±½ÓµÎ¶¨µÄÎïÖÊÊÇ________________________¡£

0621 ÈçºÎ²â¶¨ÒÔÏÂÎïÖÊ(»ìºÏÒºÖÐΪ»­Ïß×é·Ö)?Ö¸³ö±ØÒªÊÔ¼Á¡¢±ê×¼ÈÜÒº¼°Ö¸Ê¾¼Á¡£

H3BO3 Åðɰ HCl+H3BO3 ±ØÒªÊÔ¼Á ±ê×¼ÈÜÒº Ö¸ ʾ ¼Á 0622 ˵Ã÷±íÖи÷ÎïÖÊÄÜ·ñÓÃÖкͷ¨Ö±½ÓµÎ¶¨(²»¼ÓÈκÎÊÔ¼Á)¡£ÈôÄÜ, ÇëÖ¸³öָʾ¼Á¡£

ÎïÖÊ ÄÜ·ñÖ±½ÓµÎ¶¨ ָʾ¼Á CH3NH2(¼×°·) NaAc H3BO3(ÅðËá) Na2B4O7¡¤10H2O(Åðɰ) C6H5NH3+Cl(ÑÎËá±½°·) [pKb(CH3NH2) = 3.38, pKb(C6H5NH2) = 9.38, pKa(HAc) = 4.74, pKa(H3BO3) = 9.24] 0623 ÅжÏÒÔÏÂÁ½ÐÔÎïÖÊÓÃÖкͷ¨Ö±½ÓµÎ¶¨µÄ¿ÉÄÜÐÔ¡£

ÎïÖÊ KHS ¾ÆÊ¯ËáÇâ¼Ø(¼òд³ÉKHA) °±»ùÒÒËáż¼«Àë×Ó ÄÜ·ñµÎ¶¨¼°²úÎï µÎ¶¨¼Á ָʾ¼Á Һȥ²â¶¨Ä³ÊÔÑùÖÐHAc,µÃµ½µÄw(HAc)½«»á_______¡£ÓÖÈôÓÃÒԲⶨHCl-NH4ClÈÜÒºÖеÄw(HCl),Æä½á¹û»á_________¡£(ÌîÆ«¸ß¡¢Æ«µÍ»òÎÞÓ°Ïì)

0633 ij»ìºÏ¼î(¿ÉÄܺ¬ÓÐNaOH¡¢Na2CO3¡¢NaHCO3)ÈÜÒº,ÓÃHClÈÜÒºµÎ¶¨ÖÁ·Ó̪ÖÕµã, ÏûºÄHCl V1(mL),Ôٵζ¨ÖÁ¼×»ù³ÈÖÕµã,ÓÖÏûºÄHCl V2(mL),ÈôV1>V2, Ôò»ìºÏ¼îµÄ×é³ÉΪ__________________;ÈôV1£¼V2,Ôò»ìºÏ¼îµÄ×é³ÉΪ__________________¡£

0634 ÓÐÒ»¼îÒº,¿ÉÄÜΪNaOH¡¢Na2CO3¡¢NaHCO3»òÆäÖÐijÁ½ÕߵĻìºÏÎï, Óñê×¼ËáµÎ¶¨ÖÁ·Ó̪ÖÕµãËùÏûºÄµÄËáµÄÌå»ýΪV1 (mL),¼ÌÒÔ¼×»ù³ÈΪָʾ¼Á, ÓÖÏûºÄ±ê×¼ËáV2 (mL),ÊÔÓÉV1ºÍV2Çé¿öÅжϴ˼îÒºµÄ×é³É:

1 2 V1< V2 3 V1 = V2 4 V1 = 0 5 V2 = 0 ÒÑÖª: pKa1 pKa2 H2S 6.88 14.15 ¾ÆÊ¯Ëá 3.04 4.37 °±»ùÒÒËá 2.35 9.60 0624 Ìîдϱí 0.1 mol/L C2H5NH2(pKb = 3.25) 0.1 mol/L HA(pKa = 2.86) + 0.1 mol/L H3BO3(pKa = 9.24) 0.1 mol/L NaAc[pKa(HAc) = 4.74] ÄÜ·ñÖ±½ÓµÎ¶¨ µÎ¶¨¼Á ָʾ¼Á V1> V2 0635 ÒÆÈ¡25.00 mL¿ÉÄܺ¬HClºÍ¸÷ÖÖÁ×ËáÑεĻìºÏÈÜÒº,ÓÃ0.1000 mol/L NaOH ±ê×¼ÈÜÒºµÎ¶¨, ϱíÖÐV1(NaOH)ΪÓü׻ù³È×÷ָʾ¼ÁËùºÄÌå»ý, V2(NaOH)ΪѡÓ÷Ó̪×÷ָʾ¼ÁËùºÄÌå»ý¡£ÇëÌîдÒÔÏÂÈÜÒº×é³É¼°Å¨¶È¡£

ÊÔÒº 1 2 V1(NaOH)/mL 18.72 0.00 V2(NaOH)/mL 23.60 16.77 ×é³É Ũ¶ÈcB/(mol/L) 0625 NH4+µÄËáÐÔÌ«Èõ,ÓÃNaOHÖ±½ÓµÎ¶¨Ê±Í»Ô¾Ì«Ð¡¶ø²»ÄÜ׼ȷµÎ¶¨¡£²ÉÓ÷µµÎ¶¨·¨___________׼ȷ²â¶¨(ÌîÄÜ»ò²»ÄÜ),ÆäÔ­ÒòÊÇ________________________ _____________¡£ 0626 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_________¡£ 0627 Óü×È©·¨²â¶¨ÁòËáï§ÖеªÊ±,ΪÖкÍÊÔÑùÖеÄÓÎÀëËá,Ó¦µ±Ñ¡__________Ϊָʾ¼Á¡£ÔÙ¼ÓÈë¼×È©²â¶¨°±Ê±Ó¦µ±Ñ¡________________Ϊָʾ¼Á¡£Ôڴ˵ζ¨ÖÐÑÕÉ«µÄ±ä»¯ÊÇ_____________________¡£µªÓëNaOHÎïÖʵÄÁ¿Ö®±Èn(N):n(NaOH)ÊÇ_________¡£ 0628 ²ÉÓÃÕôÁ󷨲ⶨï§ÑÎʱ,ÕôÁó³öÀ´µÄNH3¿ÉÓýӽü±¥ºÍµÄH3BO3ÈÜÒºÎüÊÕ, È»ºóÓÃHCl±ê×¼ÈÜÒºµÎ¶¨,µ«_______(Ìî¿ÉÒÔ»ò²»¿ÉÒÔ)ÓÃHAcÈÜÒº´úÌæH3BO3×÷ÎüÊÕÒº,ÒòΪ__________________________________________¡£ 0629 Óü×È©·¨²â¶¨¹¤Òµ(NH4)2SO4{Mr [(NH4)2SO4]=132}Öа±µÄÖÊÁ¿·ÖÊýw(NH3), °ÑÊÔÑùÈܽ⠺óÓÃ250 mLÈÝÁ¿Æ¿¶¨ÈÝ,ÒÆÈ¡25 mL,ÓÃ0.2 mol/L NaOH±ê×¼ÈÜÒºµÎ¶¨, ÔòÓ¦³ÆÈ¡ÊÔÑùÔ¼_________ g¡£ 0630 ¿ËÊÏ·¨²â¶¨µªÊ±, ³ÆÈ¡0.2800 gÓлúÎï, ¾­Ïû»¯´¦ÀíºóÕô³öµÄNH3ÕýºÃÖкÍ20.00 mL 0.2500 mol/L µÄH2SO4, Ôò¸ÃÓлúÎïÖеªµÄÖÊÁ¿·ÖÊýw(N)[Ar(N)=14.00]Ϊ ____________¡£ 0631 ÓÃNaOHÈÜÒºµÎ¶¨HClÈÜÒºÒԲⶨNaOHÓëHClµÄÌå»ý±È¡£½ñÑ¡¼×»ù³ÈΪָʾ¼Á²âµÃ V(NaOH)/V(HCl) = 1.005, ¶øÑ¡·Ó̪Ϊָʾ¼Á²âµÃV(NaOH)/V(HCl) = 1.012 , ÆäÖ÷ÒªÔ­ÒòÊÇ____________________¡£ 0632 ijÈËÓÃHCl±ê×¼ÈÜÒºÀ´±ê¶¨º¬CO32-µÄNaOHÈÜÒº(ÒÔ¼×»ù³È×÷ָʾ¼Á),È»ºóÓÃNaOHÈÜ0636 ÒÆÈ¡25.00 mL¿ÉÄܺ¬HClºÍ¸÷ÖÖÁ×ËáÑεĻìºÏÈÜÒº, ÓÃ0.1000 mol/L NaOH±ê×¼ÈÜÒºµÎ¶¨, Óü׻ù³ÈΪָʾ¼ÁºÄÈ¥ V1(NaOH) (mL), Èô¸ÄÓ÷Ó̪Ϊָʾ¼ÁÔòºÄÈ¥V2(NaOH) (mL),ÇëÌîдÒÔÏÂÈÜÒºµÄ×é³É¼°Å¨¶È

ÊÔÒº 1 2 V1(NaOH)/mL 13.12 13.33 V2(NaOH)/mL 35.19 26.65 ×é³É Ũ¶ÈcB/(mol/L) 0637 ijһÁ×ËáÑÎÊÔÒº,¿ÉÄÜΪNa3PO4¡¢Na2HPO4¡¢NaH2PO4»òijÁ½Õß¿ÉÄܹ²´æµÄ»ìºÏÎï,Óñê×¼ËáµÎ¶¨ÖÁ·Ó̪ÖÕµãËùÏûºÄµÄËáΪV1(HCl) (mL), ¼ÌÒÔ¼×»ù³ÈΪָʾ¼ÁÓÖÏûºÄ±ê×¼ËáΪV2(HCl) (mL), ÊÔ¸ù¾ÝV1ºÍV2ÅÐ¶ÏÆä×é³É¡£

1 2 3 4 V1 = V2 V10 V1 = 0,V2 = 0 0638 Ò»ÊÔÑùÈÜÒº¿ÉÄÜÊÇNaOH¡¢NaHCO3¡¢Na2CO3»òÊÇËüÃǵĻìºÏÈÜÒº,ÓÃ20.00 mL 0.1000 mol/L HClÈÜÒº¿ÉµÎ¶¨ÖÁ·Ó̪Öյ㡣

(1)Èô¸ÃÊÔÒºÖк¬ÓÐÏàͬÎïÖʵÄÁ¿µÄNaOHºÍNa2CO3,ÔòÐèÔÙ¼ÓÈë________mL HCl¿ÉµÎ¶¨ÖÁ¼×»ù³ÈÖÕµã;

(2)Èô¸ÃÊÔÒºÖк¬ÓÐÏàͬÎïÖʵÄÁ¿µÄNaHCO3ºÍNa2CO3,ÔòÐèÔÙ¼ÓÈë______mL HCl ¿ÉµÎ¶¨ÖÁ

¼×»ù³ÈÖյ㡣

0639 ÓÃ˫ָʾ¼Á·¨ÒÔHCl±ê×¼ÈÜÒºµÎ¶¨Ì¼ËáÑλìºÏÎï,Éè·Ó̪±äɫʱHClËùÏûºÄÌå»ýΪV1,ÓÉ·Ó̪ÖÕµãÖÁ¼×»ù³ÈÖÕµãËùÏûºÄHClÌå»ýΪV2¡£

(1)Èç¹ûÊÔÑùÈÜÒºÖÐNa2CO3µÄŨ¶ÈÁ½±¶ÓÚNaHCO3,ÔòV1: V2 = _________

(2)Èç¹û¹ÌÌåÊÔÑùº¬3 mmol µÄNaOHºÍ4 mmol NaHCO3,ÔòV1: V2 = _________ 0640 ij¼îÒº¿ÉÄÜÊÇNaOH¡¢Na2CO3¡¢NaHCO3»òËüÃǵĻìºÏÎï,½ñÈ¡Á½µÈ·Ý¼îÒº, ·Ö±ðÓÃHClÈÜÒºµÎ¶¨,µÚÒ»·ÝÓ÷Ó̪×÷ָʾ¼Á,ÏûºÄV1(HCl) (mL),µÚ¶þ·ÝÓü׻ù³ÈΪָʾ¼Á, ÏûºÄV2(HCl) (mL),¸ù¾Ýϱí,È·¶¨¼îÒºµÄ×é³É:

V1(HCl)/mL V2(HCl)/mL ×é³É 1 22.42 22.44 2 15.60 41.33 3 28.84 35.16 4 18.02 36.03 5 0.00 24.87 0641 ÔÚ´¿CH3OHÈܼÁÖÐÓÃ0.1 mol/L Ç¿¼îµÎ¶¨0.1 mol/L Ç¿Ëá,»¯Ñ§¼ÆÁ¿µãµÄpHÊÇ______, »¯Ñ§¼ÆÁ¿µãǰ0.1£¥µÄpHÊÇ______, »¯Ñ§¼ÆÁ¿µãºó0.1£¥µÄpHÊÇ______¡£ [pKs(CH3OH) = 16.7]

0642 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]

0643 4.000 g NH4NO3ÅäÖÆ³ÉÈÜÒº, ÓÃÈÝÁ¿Æ¿¶¨ÈÝÖÁ500 mL, ÒÆÈ¡25.00 mL´ËÈÜÒº, ¼ÓÈë 10mL¼×È©ÈÜÒº, ·´Ó¦Îª: 4NH4++ 6HCHO = (CH2)6N4H++ 3H++ 6H2O¡£ ÓÃ0.1000 mol/L NaOHµÎ¶¨ËùÉú³ÉµÄËá, ¼ÆºÄÈ¥24.25 mL, ¼ÆËãNH4NO3µÄÖÊÁ¿·ÖÊý, Èô´Ë NH4NO3 ÊÔÑùÖк¬ÓÐ 2.20 % µÄÎüʪˮ, Ôò¸ÉÊÔÑùÖÐNH4NO3µÄÖÊÁ¿·ÖÊýÓÖÊǶàÉÙ? {Mr(NH4NO3) = 80.04, pKb[(CH2)6N4] = 8.85}

0644 ³ÆÈ¡Á×ôû(P4O10)ÊÔÑù0.0240 g, ¼ÓË®±ä³ÉH3PO4, È»ºó³ÁµíΪÁ×îâËáï§, ³Áµí¾­¹ýÂËÏ´µÓºó, ÈÜÓÚ40.00 mL 0.2160 mol/L NaOHÈÜÒºÖÐ, ÓÃ0.1284 mol/L HNO3ÈÜÒº·µµÎ¶¨ºÄÈ¥20.00 mL ,×ܵķ´Ó¦Îª: (NH4)2HPO4¡¤12MoO3¡¤H2O+24OH- = 12MoO42-+HPO42-+2NH4++13H2O ¼ÆËãÁ×ôûÖÊÁ¿·ÖÊýw(P4O10)¡£ [Mr(P4O10) = 283.9]

0645²â¶¨Ä³¹¤ÒµÉÕ¼îÖÐNaOHºÍNa2CO3µÄÖÊÁ¿·ÖÊý, ³ÆÈ¡ÊÔÑù2.546 g,ÈÜÓÚË®²¢¶¨ÈÝÓÚ250 mLÈÝÁ¿Æ¿ÖÐ, È¡³ö25.00 mL, ÒÔ¼×»ù³ÈΪָʾ¼Á, µÎ¶¨µ½³Èɫʱ, ÓÃÈ¥HCl±ê×¼ÈÜÒº24.86 mL¡£ÁíÈ¡25.00 mLÈÜÒº, ¼ÓÈë¹ýÁ¿BaCl2, ÒÔ·Ó̪Ϊָʾ¼Á, µÎ¶¨µ½ºìÉ«¸ÕÍÊ, ÓÃÈ¥HCl±ê×¼ÈÜÒº23.74 mL, ÓÖÖªÖкÍ0.4852 g Åðɰ(Na2B4O7¡¤10H2O)ÐèÒª´ËHCl±ê×¼ÈÜÒº24.37 mL¡£¼ÆËã¸ÃÊÔÑùNaOH

ºÍNa2CO3µÄÖÊÁ¿·ÖÊý¡£[Mr(Na2B4O7¡¤10H2O) = 381.4, Mr(NaOH) = 40.00, Mr(Na2CO3) = 106.0] 0646 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¡£

0647 ÔÚ¼×Ëá(HCOOH)½éÖÊÖÐ, ÓÃ0.1 mol/L Ç¿¼îµÎ¶¨0.1 mol/L Ç¿Ëá¡£¼ÆË㻯ѧ¼ÆÁ¿µãµÄpH(¼´pHCOOH2+)ÖµÒÔ¼°»¯Ñ§¼ÆÁ¿µãǰºó0.1£¥µÄpH¡£ [pKs(HCOOH)Ϊ6.2]

0648 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)

0649 ´¿¾»¸ÉÔïµÄNaOHºÍNaHCO3ÒÔ2:1µÄÖÊÁ¿±È»ìºÏ,³ÆÈ¡Ò»¶¨Á¿µÄ¸Ã»ìºÏÎïÈÜÓÚË®ÖÐ,ÈôʹÓÃͬһHCl±ê×¼ÈÜÒºµÎ¶¨,¼ÆËãÒÔ·Ó̪×÷ָʾ¼ÁʱËùÐèËáµÄÌå»ýÓëÒÔ¼×»ù³È×÷ָʾ¼ÁʱËùÐèËáµÄÌå»ýÖ®±È¡£[Mr (NaOH) = 40.00, Mr (NaHCO3) = 84.01]

0650 ³ÆÈ¡1.250 g ´¿Ò»ÔªÈõËáHA, ÈÜÓÚÊÊÁ¿Ë®ºóÏ¡ÖÁ50.00 mL, È»ºóÓÃ0.1000 mol/L NaOH ÈÜÒº½øÐеçλµÎ¶¨, ´ÓµÎ¶¨ÇúÏß²é³öµÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µãʱ, NaOHÈÜÒºÓÃÁ¿Îª37.10 mL¡£µ±µÎÈë7.42 mL NaOHÈÜҺʱ,²âµÃpH = 4.30¡£¼ÆËã:(1)Ò»ÔªÈõËáHAµÄĦ¶ûÖÊÁ¿; (2)HAµÄ½âÀë³£ÊýKa; (3)µÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µãʱÈÜÒºµÄpH¡£

0651 ³ÆÈ¡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¡£

0652 ³ÆÈ¡Ä³´¿Ò»ÔªÈõËáHA 0.8150 g, ÈÜÓÚÊÊÁ¿Ë®ºó, ÒÔ·Ó̪Ϊָʾ¼Á, ÒÔ0.1100 mol/L NaOHÈÜÒºµÎ¶¨ÖÁÖÕµãʱ, ÏûºÄ24.60 mL,µ±¼ÓÈëNaOHÈÜÒº11.00 mL ʱ, ¸ÃÈÜÒºµÄpH = 4.80¡£¼ÆËãÈõËáHAµÄpKaÖµ¡£

0653 Ìî±í˵Ã÷ÒÔÏÂÎïÖÊÓÃËá¼îÖкͷ¨Ö±½Ó²â¶¨µÄ¿ÉÄÜÐÔ¼°Ìõ¼þ¡£ ÎïÖÊ ÄÜ·ñ²â¶¨ µÎ¶¨¼Á ָʾ¼Á 0.1 mol/L C6H5NH3+Cl- 0.1 mol/L Na2SO3 0.1 mol/L NH4Ac ÒÑÖª: pKb(C6H5NH2) = 9.38, pKa(NH4+) = 9.26, pKa(HAc) = 4.74, H2SO3µÄpKa1 = 1.90, pKa2 = 7.20

0654 ÅäÖÆNaOH±ê×¼ÈÜҺʱδ³ý¾»CO32-,½ñÒÔ²ÝËá(H2C2O4¡¤2H2O)±ê¶¨ÆäŨ¶Èºó, ÓÃÒԲⶨHAcŨ¶È, ²âµÃ½á¹û___________;ÈôÓÃÒԲⶨHCl-NH4Cl»ìºÏÒºÖÐHCl Ũ¶È,Æä½á¹û_______¡£(ÌîÆ«¸ß¡¢Æ«µÍ»òÎÞÓ°Ïì)

0655 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)

0656 ÔÚËá¼îµÎ¶¨ÖÐ,µÎ¶¨¼ÁÓë±»²âÎïŨ¶Èͨ³£ÔÚ0.1 mol/L×óÓÒ¡£ÈôŨ¶È̫ϡ,Ôò______________ ;¶øÅ¨¶ÈÌ«´ó, Ôò__________________¡£

0657 ±û¶þËá HOOC-CH2-COOH(¼òдΪH2M)ÓÐÏÂÁÐÁ½¸öƽºâ H2M HM-+ H+

pKa1 = 2.86

HM-

M2- + H+

pKa2 = 5.70

ÏÖÈ¡20.00 mLijº¬±û¶þËá¶þÄÆ(Na2M)ºÍ±û¶þËáÇâÄÆ(NaHM)ʱ»ìºÏÈÜÒº,Óà 0.01000 mol/L HClµÎ¶¨¡£Ôڵζ¨¹ý³ÌÖеõζ¨ÇúÏßÉÏÁ½¸öµãµÄpH:

V(HCl)/mL pH 1.00 5.70 10.00 4.28 ÊÔ¼ÆËã´ïµ½±û¶þËữѧ¼ÆÁ¿µãʱ,×ܹ²ÐèÒª¶àÉÙºÁÉýÑÎËá(2λÓÐЧÊý×Ö)?

0657 ÉèNa2MΪ x (mol), NaHMΪ y (mol) ©° x - 0.01000¡Á1.00 y + 0.01000¡Á1.00 ©¦ ©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤ = ©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤©¤ ©È 21.00 21.00 ©¦ ©¸ x = 0.01¡Á10.00 x = 0.1 mol y = 0.08 mol 0.1¡Á2+0.08 ×ܹ²ÐèV(HCl) = ©¤©¤©¤©¤©¤©¤ = 28 (mL) 0.01000

0658 Éè¼Æ²â¶¨Na2B4O7ºÍH3BO3»ìºÏÎïµÄ·½°¸¡£(ÒªÇó:дÃ÷Ö÷Òª²½Öè¡¢µÎ¶¨¼Á¡¢Ö¸Ê¾¼Á¡¢ÆäËüÖ÷ÒªÊÔ¼ÁºÍ·ÖÎö½á¹ûµÄ¼ÆË㹫ʽ)

0659 ²â¶¨H2SO4+(NH4)2SO4»ìºÏÒºÖжþÕßµÄŨ¶È¡£(ÒªÇó:дÃ÷Ö÷Òª²½Öè¡¢µÎ¶¨¼Á¡¢Ö¸Ê¾¼ÁµÈºÍ·ÖÎö½á¹ûµÄ¼ÆË㹫ʽ)

0660 ÓÃËá¼îµÎ¶¨·¨²â¶¨H2SO4ºÍH3PO4»ìºÏÒºÖÐÁ½ÖÖËáµÄŨ¶È¡£Óüòµ¥Á÷³Ì±íÃ÷ʵÑé²½Öè(±êÒº¡¢Ö¸Ê¾¼Á),²¢Ð´³ö½á¹ûµÄ¼ÆËãʽ¡£

0661 ÊÔÉè¼ÆÒ»ÖֲⶨÑÎËáºÍÅðËá»ìºÏÈÜÒºÁ½×é·ÖŨ¶ÈµÄ¼òµ¥·½°¸(°üÀ¨µÎ¶¨¼Á¡¢±ØÒªÊÔ¼Á¡¢Ö¸Ê¾¼Á¼°ÖÕµãʱÑÕÉ«µÄ¸Ä±ä)

0661 HCl 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 0662 Éè¼Æ²â¶¨º¬ÓÐÖÐÐÔÔÓÖʵÄNa2CO3ÓëNa3PO4»ìºÏÎïÖжþ×é·ÖÖÊÁ¿·ÖÊýµÄ·ÖÎö·½°¸¡£Óüòµ¥Á÷³Ì±íÃ÷Ö÷Òª²½Öè¡¢µÎ¶¨¼Á¡¢Ö¸Ê¾¼Á¡¢½á¹û¼ÆË㹫ʽ¡£

0663 ÈôÔÚHCOOH(pKs = 6.2)½éÖÊÖÐ, ÓÃ0.10 mol/LÇ¿ËáSH2+µÎ¶¨0.10 mol/LÇ¿¼îS-, ijͬѧ¼ÆË㻯ѧ¼ÆÁ¿µã¼°»¯Ñ§¼ÆÁ¿µãºó0.1%µÄ[SH2+]µÃ: »¯Ñ§¼ÆÁ¿µã: [SH2+] = [S-] =

K-3.1s = 10

»¯Ñ§¼ÆÁ¿µãºó0.1%: [SH.?10?32+] =

012 = 10-4.3

Îʴ˽á¹ûÊÇ·ñºÏÀí,´íÔÚÄÄÀï?ÈçºÎ¸ÄÕý?ÓɼÆËã½á¹ûÍÆ³ö´ËµÎ¶¨Í»Ô¾ÊǶàÉÙ? ÄÜ·ñ׼ȷµÎ¶¨? 0664 ʵÑéÊÒ±¸Óбê×¼Ëá¼îÈÜÒººÍ³£ÓÃËá¼îָʾ¼ÁÈç¼×»ù³È¡¢¼×»ùºì¡¢·Ó̪,ÁíÓй㷺pHÊÔÖ½ºÍ¾«ÃÜpHÊÔÖ½¡£ÏÖÓÐһδ֪һԪÈõËá¿ÉÓüî±ê×¼ÈÜÒºµÎ¶¨,ÇëÓÃÉÏÊöÎïÆ·ÒÔ×î¼òµ¥µÄʵÑé·½·¨²â¶¨¸ÃÒ»ÔªÈõËáµÄ½âÀë³£ÊýKaµÄ½üËÆÖµ¡£

2301 ijµç½âÖÊMA(M2+,A2-)ÈÜÒº,ÆäŨ¶Èc(MA) = 0.10mol/L,Ôò¸ÃÈÜÒºµÄÀë×ÓÇ¿¶ÈΪ----( ) (A) 0.10mol/L (B) 0.30mol/L (C) 0.40mol/L (D) 0.60mol/L

2302 0.10mol/L M2A(M+,A2-)µç½âÖÊÈÜÒºµÄÀë×ÓÇ¿¶ÈΪ----------------------------------------( ) (A) 0.10mol/L (B) 0.30mol/L (C) 0.40mol/L (D) 0.60mol/L

2303 0.10mol/L MA(M+,A-)µç½âÖÊÈÜÒºµÄÀë×ÓÇ¿¶ÈΪ-------------------------------------------( ) (A) 0.10mol/L (B) 0.30mol/L (C) 0.40mol/L (D) 0.60mol/L

2304 0.10mol/L Na3PO4ÈÜÒº,ÆäÀë×ÓÇ¿¶ÈΪ-------------------------------------------------------( ) (A) 0.10mol/L (B) 0.30mol/L (C) 0.40mol/L (D) 0.60mol/L

2305 ¸ù¾ÝÖÊ×ÓÀíÂÛ,¼È¿ÉÆðËáµÄ×÷ÓÃÓÖ¿ÉÆð¼îµÄ×÷ÓõÄÈܼÁ³ÆÎª----------------------------( ) (A) À­Æ½ÐÔÈܼÁ (B) ·Ö±æÐÔÈܼÁ (C) Á½ÐÔÈܼÁ (D) ¶èÐÔÈܼÁ

2306 ¸ù¾ÝËá¼îÖÊ×ÓÀíÂÛ,ÏÂÁбíÊöÖÐÕýÈ·µÄÊÇ----------------------------------------------------( ) (A) ¶ÔÓÚËùÓеÄËáÀ´Ëµ,H2OÊÇÒ»ÖÖÀ­Æ½ÐÔÈܼÁ (B) H2OÊÇÒ»ÖÖ¶èÐÔÈܼÁ (C) NH4+ÆðÈõËáµÄ×÷ÓÃ

(D) ´×Ëá¸ùÀë×ÓÊÇ´×ËáµÄ¹²éîËá

2307 ½«´¿Ëá¼ÓÈë´¿Ë®ÖÐÖÆ³ÉÈÜÒº,ÔòÏÂÁбíÊöÖÐÕýÈ·µÄÊÇ-------------------------------------( )

(A) ËáµÄŨ¶ÈÔ½µÍ, ½âÀëµÄÈõËáµÄ°Ù·ÖÊýÔ½´ó (B) ËáµÄ\Ç¿\ºÍ\Èõ\ÓëËáµÄÎïÖʵÄÁ¿Å¨¶ÈÓÐ¹Ø (C) Ç¿ËáµÄ½âÀë°Ù·ÖÊýËæÅ¨¶È¶ø±ä»¯

(D) ÿÉýº¬1.0¡Á10-7molÇ¿Ëá(ÀýÈçHCl),Ôò¸ÃÈÜÒºµÄpHΪ7.0

2308 ÏÂÁÐÎïÖÊÖÐÊôÓÚÁ½ÐÔÎïÖʵÄÊÇ-----------------------------------------------------------------( ) 2322 ½ñÓÐ1Lº¬0.1mol H3PO4ºÍ0.3mol Na2HPO4µÄÈÜÒº,ÆäpHÓ¦µ±ÊÇ----------------( ) (H3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)

(A) 2.12 (B) (2.12+7.20)/2 (C) 7.20 (D) (7.20+12.36)/2

2323 ij1Lº¬0.2mol Na3PO4ºÍ0.3mol HClµÄÈÜÒº,ÆäpHÓ¦µ±ÊÇ------------------------( ) (H3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)

(A) H2CO3 (B) °±»ùÒÒËáÑÎËáÑÎ (C) °±»ùÒÒËá (D) °±»ùÒÒËáÄÆ

2309 ÏÂÁÐÎïÖÊÖÐÊôÓÚËáµÄÓÐ__________, ÊôÓÚ¼îµÄÓÐ__________, ÊôÓÚÁ½ÐÔÎïÖʵÄÓÐ__________¡£(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

2310 ij¶þÔªËáH2AµÄpKa1ºÍpKa2·Ö±ðΪ4.60ºÍ8.40,ÔÚ·Ö²¼ÇúÏßÉÏH2AÓëHA-ÇúÏß½»µãpHΪ____,HA-ÓëA2-ÇúÏß½»µãµÄpHΪ____,H2AÓëA2-µÄ½»µãpHΪ____,HA- ´ï×î´óµÄpH ÊÇ____¡£ 2311 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-]

2312 ¼Ó 40 mL 0.15mol/L HClÈÜÒºÖÁ60mL 0.10mol/L Na2CO3ÈÜÒºÖÐ,¸ÃÈÜÒºµÄºÏÀíµÄ¼ò»¯ÖÊ×ÓÌõ¼þÊÇ__________________________________________¡£

2313 ÓÃNaOHÈÜÒºµÎ¶¨H3PO4ÖÁpH = 4.7ʱ, ÈÜÒºµÄºÏÀíµÄ¼ò»¯ÖÊ×ÓÌõ¼þÊÇ ________ __¡£ (H3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)¡£

2314 c(Na2CO3) = cµÄNa2CO3ÈÜÒºµÄÎïÁÏÆ½ºâ·½³ÌʽÊÇ______________________________¡£ 2315 0.1mol/L NaClÈÜÒºµÄÖÊ×ÓÆ½ºâʽÊÇ_____________________________________¡£ 2316 0.1mol/L NaClÈÜÒºµÄµçºÉƽºâʽÊÇ_____________________________________¡£ 2317 NH4H2PO4 ÈÜÒºÖÐ×îÖ÷ÒªµÄÖÊ×Ó´«µÝ·´Ó¦ÊÇ_____________________________ (ÒÑÖªH3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36,NH3µÄpKbÊÇ4.74)¡£

2318 ijÈýÔªËáH3AµÄpKa1 = 3.96, pKa2 = 6.00, pKa3 = 10.02,Ôò0.10mol/L H3AµÄpHÊÇ---( ) (A) 1.00 (B) 2.48 (C) 3.96 (D) 4.98

2319ijÈýÔªËáH3AµÄpKa1 = 3.96¡¢pKa2 = 6.00¡¢pKa3 = 10.02, Ôò 0.10 mol/L Na3AÈÜÒºµÄpHÊÇ () (A) 8.01 (B) 10.02 (C) 11.51 (D) 12.51

2320 ÒÔÏÂÈÜҺϡÊÍ10±¶Ê±,pH¸Ä±ä×î´óµÄÊÇ-------------------------------------------------( ) (A) 0.1mol/L NH4Ac (B) 0.1mol/L NaAc (C) 0.1mol/L HAc (D) 0.1mol/L HCl

2321 ÒÔÏÂÈÜҺϡÊÍ10±¶Ê±,pH¸Ä±ä×îСµÄÊÇ-------------------------------------------------( ) (A) 0.1mol/L HAc (B) 0.1mol/L NH4Ac (C) 0.1mol/L NaAc

(D) 0.1mol/L NH4Cl

(A) 2.12 (B) (2.12+7.20)/2 (C) 7.20 (D) (7.20+12.36)/2 2324 ½ñÓÐÈýÖÖÈÜÒº·Ö±ðÓÉÁ½×é·Ö×é³É: (a) 0.10mol/L HCl ~ 0.20mol/L NaAc (b) 0.20mol/L HAc ~ 0.10mol/L NaOH (c) 0.10mol/L HAc ~ 0.10mol/L NH4Ac

ÒÑÖªpKa( HAc) = 4. 74, pKa(NH4+) = 9. 26, ÔòÕâÈýÖÖÈÜÒºpH´óСµÄ¹ØÏµÊÇ_________ __________________________(Ó÷ûºÅa¡¢b¡¢c±íʾ)¡£

2325 ΪÏÂÁÐÈÜҺѡÔñ¼ÆËã[H+]»ò[OH-]µÄºÏÀí¹«Ê½(ÇëÌîдA,BµÈ): (1) 0.10mol/LÈýÒÒ´¼°·(pKb = 6.24)

__________ (2) 0.10mol/LÁÚ±½¶þ¼×ËáÇâ¼Ø(pKa1 = 2.95¡¢pKa2 = 5.41) __________ (3) 0.10mol/L H2C2O4(pKa1 = 1.22¡¢pKa2 = 4.19) __________ (4) 0.10mol/L±½¼×Ëá(pKa = 4.21)

__________

A. [H+] = Ka1(c-[H?]) B. [H+] = Ka1Ka2 C. [H+] =

Kac D. [OH-] =

Kb c

2326 ¸ø³ö¼ÆËãÏÂÁи÷ÈÜÒº[H+]µÄºÏÀí¹«Ê½(½öÓ÷ûºÅ±íʾ): (A) 0.10mol/L ¶þÂÈÒÒËá(pKa = 1.30)

________________ (B) 0.10mol/L NH4Cl[pKb(NH3) = 4.74] ________________ (C) 0.10mol/L NaHSO4[pKa(HSO4-) = 1.99]

________________ (D) 1.0¡Á10-4mol/L H3BO3(pKa = 9.24)

________________

2327 ¸ø³ö¼ÆËãÏÂÁи÷ÈÜÒº[H+]µÄºÏÀí¹«Ê½(½öÓ÷ûºÅ±íʾ): (A) 0.04mol/L H2CO3(pKa1 = 6.38¡¢pKa2 = 10.25)

________________ (B) 0.05mol/L NaHCO3

________________ (C) 0.10mol/L °±»ùÒÒËá(NH3+CH2COO-)(pKa1 = 2.35¡¢pKa2 = 9.70) ________________ (D) 0.10mol/L °±»ùÒÒËáÑÎËáÑÎ(NH3+CH2COOHCl-)

________________

2328 Çë¸ø³öÏÂÁÐÇéÐμÆËã[H+]µÄºÏÀí¹«Ê½:

(A) 0.1000mol/L HClµÎ¶¨0.1000mol/L Na2CO3ÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µãʱ, [H+] = ________ (B) 0.05000mol/L NaOHµÎ¶¨0.05000mol/L H3PO4ÖÁµÚÒ»»¯Ñ§¼ÆÁ¿µãʱ, [H+] = ___________ (C) 0.1mol/L HAc ¡ª 0.1mol/L H3BO3»ìºÏÒº, [H+] = _____________ (D) 0.1mol/L HCOONH4, [H+] = _________________________

2329 ÏÂÁбíÊöÖÐ,ÕýÈ·µÄÊÇ[ÒÑÖªpKa(HAc) = 4.74,pKb(NH3) = 4.74]-------------------------( ) (A) Ũ¶ÈÏàͬµÄNaAcÈÜÒººÍNH4AcÈÜÒºµÄpHÏàͬ

(B) Ũ¶ÈÏàͬµÄNaAcÈÜÒººÍNH4AcÈÜÒº,ǰÕßµÄpH´óÓÚºóÕß

(C) ¸÷×é·ÖŨ¶ÈÏàͬµÄHAc-NaAcÈÜÒººÍHAc-NH4AcÈÜÒº,ǰÕßµÄpH´óÓÚºóÕß (D) ¸÷×é·ÖŨ¶ÈÏàͬµÄHAc-NaAcÈÜÒººÍHAc-NH4AcÈÜÒº,ǰÕßµÄpHСÓÚºóÕß 2330 ÔÚÏÂÁÐÈÜÒºÖлº³åÈÝÁ¿×î´óµÄÊÇ____,»º³åÈÝÁ¿×îСµÄÊÇ____(Ìî·ûºÅA,B,?)¡£ (A) 0.1mol/L HAc (B) 0.1mol/L HAc-0.1mol/L NaAc (C) 1.0mol/L HAc-1.0mol/L NaAc (D) 0.1mol/L HCl

2331 ½«13.5gÁù´Î¼×»ùËİ·¼Óµ½4.0mL 12mol/L HClÖÐ,Ï¡ÊÍÖÁ100mL,ÆäpHΪ------( ) {pKb[(CH2)6N4] = 8.85, Mr[(CH2)6N4] = 140.0} (A) 0.32 (B) 2.57 (C) 4.43 (D) 5.15

2332 ÏÂÁÐÎïÖÊ¿ÉÒÔÓÃÀ´Ö±½ÓÅäÖÆ±ê×¼»º³åÈÜÒºµÄÊÇ-------------------------------------------( ) (A) NaAc (B) Na2CO3 (C)Na2B4O7¡¤10H2O (D) Na2HPO4¡¤12H2O

2333 ÏÂÁи÷×éÈÜÒºÖпÉÓÃ×÷±ê×¼»º³åÈÜÒºµÄÊÇ--------------------------------------------------( ) (A) 0.05mol/L ÁÚ±½¶þ¼×ËáÇâ¼Ø (B) ¼×Ëá--NaOH (C) ÁÚ±½¶þ¼×ËáÇâ¼Ø-HCl (D) Na2B4O7-HCl

2334 ÏÂÁи÷×éÈÜÒºÖпÉÓÃ×÷±ê×¼»º³åÈÜÒºµÄÊÇ--------------------------------------------------( ) (A) ±¥ºÍ¾ÆÊ¯ËáÇâ¼Ø (B) °±»ùÒÒËá-ÑÎËá (C) ÈýÒÒ´¼°·-ÑÎËá

(D) °±»ùÒÒËá-ÇâÑõ»¯ÄÆ

2335 ÏÂÁи÷×éÎïÖÊÖпÉÓÃ×÷±ê×¼»º³åÈÜÒºµÄÊÇ--------------------------------------------------( ) (A) HAc-NaAc (B) Áù´Î¼×»ùËİ· (C) Åðɰ (D) NH4Cl-NH3

2336 ±ê×¼»º³åÈÜÒºÊÇÖ¸_________________,Æä×÷ÓÃÊÇ________ _____________________¡£ 2337 »º³åÈÜÒºÓ¦ÓÐ×ã¹»µÄ»º³åÈÝÁ¿,ͨ³£»º³å×é·ÖµÄŨ¶ÈÔÚ____________Ö®¼ä¡£ 2338 ͨ³£º¬Óй²éîËá¼î¶ÔµÄÈÜÒº³ÆÎª__________,Ëü¿ÉÒÖÖÆÈÜÒº______µÄ±ä»¯¡£ 2339 ÓÃÁù´Î¼×»ùËİ·ÅäÖÆµÄ»º³åÈÜҺʱ,ÆäpHԼΪ____¡£

2340 -da/dpH»òdb/dpH³ÆÎªÈÜÒºµÄ________,µ±__________________,¼´µ±pH = ____ ʱÓÐ×î´óÖµ,ÆäֵΪ____¡£

2341 ½ñÓûÅäÖÆpH = 5µÄ»º³åÈÜÒº¿ÉÑ¡Ôñ__________ÅäÖÆ¡£ 2342 ½ñÓûÅäÖÆpHÔ¼10µÄ»º³åÈÜÒº¿ÉÑ¡ÓÃ______________¡£

2343 Áù´Î¼×»ùËİ·(CH2)6N4µÄpKb = 8.85,ÔòÓÃÁù´Î¼×»ùËİ·»º³åÈÜÒºµÄ»º³å·¶Î§Ó¦ÊÇpH____________¡£

2344 ÏÂÁÐÇéÐÎÖÐ,ÈÜÒºµÄpH½«·¢Éúʲô±ä»¯(Ôö´ó¡¢¼õС»ò²»±ä): (1) ÔÚ50mL 0.1mol/L HFÈÜÒºÖмÓÈë1g NaF __________ (2) ÔÚ50mL 0.1mol/L HAcÈÜÒºÖмÓÈë1g NaAc __________ (3) ÔÚ50mL 0.1mol/L NH3ÈÜÒºÖмÓÈë1g NH4Cl

__________

(4) ÔÚ50mL 0.1mol/L HClO4ÈÜÒºÖмÓÈë10mL 0.1mol/L HCl __________ 2345 ½«0.1mol/L HAc + 0.1mol/L NaAcÈÜҺϡÊÍ10±¶ºó,ÔòpHΪ_____ ¡£ [pKa(HAc) = 4.74]

2346 ÓûÅäÖÆpH = 9µÄ»º³åÈÜÒº,Ò»°ã¿ÉÑ¡ÓÃ________________Ìåϵ¡£

2347 1.0mol/L NH4HF2ÈÜÒºµÄpHΪ________¡£ [pKa(HF) = 3.18, pKb(NH3) = 4.74] 2348 ÏÖÓÐÒ»º¬0.2mol/L H3PO4ºÍ0.3mol/L NaOHµÄÈÜÒº,ÆäpHÓ¦µ±ÊÇ_______ ¡£ (H3PO4µÄpKa1~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)

2349 ÓûÅäÖÆpH = 9µÄ»º³åÈÜÒº,ÏÂÁÐÁ½ÖÖÎïÖÊÖпÉͨ¹ý¼ÓÈëÑÎËáÀ´ÅäÖÆµÄÊÇ-----------( ) (A) ôǰ± NH2OH (Kb = 9.1¡Á10-9) (B) NH3¡¤H2O (Kb = 1.8¡Á10-5)

2350 0.30mol/L H3PO4ÈÜÒºÓë0.20mol/L Na3PO4ÈÜÒºµÈÌå»ý»ìºÏ,¼ÆËã¸Ã»º³åÈÜÒºµÄpH(H3PO4µÄpKa1 ~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)¡£

2351 ijÈõËáHYµÄKaΪ8.5¡Á10-6¡£ÔÚ50.00mL 0.1000mol/L¸ÃËáÈÜÒºÖÐÐè¼ÓÈë¶àÉÙºÁÉý0.1000mol/L NaOHÈÜÒº·½ÄÜʹÈÜÒºµÄpHΪ5.00?

2352 0.30mol/L NaH2PO4ÈÜÒºÓë0.20mol/L Na3PO4ÈÜÒºµÈÌå»ý»ìºÏ,¼ÆËã¸Ã»º³åÈÜÒºµÄpH(H3PO4µÄpKa1 ~pKa3·Ö±ðΪ2.12¡¢7.20¡¢12.36)¡£

2353 ½ñÓÐpH = 5.00µÄHAc-Ac-»º³åÈÜÒº300mL,ÒÑÖªÆä×ÜŨ¶Èc(HAc + Ac-) = 0.50mol/L,½ñ¼ÓÈë300mL 0.040mol/L NaOHÈÜÒº,¼ÆËãÆäpH¡£ (ÒÑÖªHAcµÄpKa = 4.74) 2354 0.20mol/L HClÓë0.30mol/L Na2CO3ÈÜÒºµÈÌå»ý»ìºÏ,¼ÆËã¸ÃÈÜÒºµÄpH¡£ (H2CO3µÄpKa1 = 6.38, pKa2 = 10.25)¡£

2355 ijÁ×ËáÑλº³åÈÜÒº,ÿÉýº¬0.080mol Na2HPO4ºÍ0.020mol Na3PO4¡£½ñÓÐ 1.0mmolÓлú»¯ºÏÎïRNHOHÔÚ100mLÉÏÊöÁ×ËáÑλº³åÈÜÒºÖнøÐеç½âÑõ»¯, Æä·´Ó¦ÈçÏÂ: RNHOH+H2O¡úRNO2+4H++4e- µ±Ñõ»¯·´Ó¦½øÐÐÍêÈ«ºó,¼ÆË㻺³åÈÜÒºµÄpH¡£

(H3PO4µÄpKa1~pKa3·Ö±ðΪ 2.12¡¢7.20¡¢12.36)¡£

2356 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?

2357 ½ñÒÔijÈõËáHA¼°Æä¹²éî¼îA-ÅäÖÆ»º³åÈÜÒº,ÒÑÖªÆäÖÐ[HA]= 0.25mol/L¡£ ÍùÉÏÊö 100mL»º³åÈÜÒºÖмÓÈë5.0mmol¹ÌÌåNaOHºó(ºöÂÔÆäÌå»ýµÄ±ä»¯),ÈÜÒºµÄpH = 4.60¡£ ¼ÆËãÔ­»º³åÈÜÒºµÄpH(ÉèHAµÄpKa = 4.30)¡£

2358 ½ñÓÃijÈõ¼îB (pKb = 4.30)¼°Æä¹²éîËáBH+ÅäÖÆ³ÉpH = 9.40µÄ»º³åÈÜÒº200mL, Ïò´Ë»º³åÈÜÒºÖмÓÈë30.0mmol¹ÌÌåNaOHºó,ÈÜÒºµÄpH = 10.0(ºöÂÔÌå»ýµÄ±ä»¯)¡£¼ÆËãÔ­»º³åÈÜÒºÖÐBºÍBH+µÄŨ¶È¡£

2359 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)

2360 ÈôÓÃ0.1000mol/L KOHÈÜÒº·Ö±ðµÎ¶¨25.00mL ijH2SO4ºÍHAcÈÜÒº,ÈôÏûºÄµÄÌå»ýÏà µÈ, Ôò±íʾÕâÁ½ÖÖÈÜÒºÖÐ------------------------------------------------------------------------------( ) (A) [H+]ÏàµÈ (B) c(H2SO4) = c(HAc) (C) c(H2SO4) = 2c(HAc) (D) 2c(H2SO4) = c(HAc)

2361 ÏÖÓÐͬÌå»ýµÄËÄÖÖÈÜÒº (1) 0.01mol/L NH4Cl-0.01mol/L NH3 (2) 0.1mol/L NH4Cl-0.1mol/L NH3 (3) 0.01mol/L HAc-0.01mol/L NaAc (4) 0.1mol/L HAc-0.1mol/L NH3

·Ö±ð¼ÓÈë0.05mL 1mol/L HClÒýÆðpH±ä»¯×î´óµÄÊÇ---------------------------------( ) (A) (2) (B) (4) (C) (1)ºÍ(3) (D) (2)ºÍ(4)

2362 ÒÔÏÂÊÇÒ»¸öÊÔÌâ¼°Æä½â´ð,ÔĺóÇëÅж¨Æä½á¹ûÕýÈ·Óë·ñ,ÈçÓдíÎó£¬Ö¸³ö´íÔÚÄÄÀï? ³ÆÈ¡Î´ÖªËáHA(Mr = 82.00)ÊÔÑù1.600 g,ÈܽⲢϡÊÍÖÁ60.00mL,ÓÃ0.2500mol/L NaOH ×÷µçλµÎ¶¨, ²âµÃÖкÍÒ»°ëʱÈÜÒºpH = 5.00,Öк͵½»¯Ñ§¼ÆÁ¿µãʱpH = 9.00, Çów(HA)¡£ ½â: Öк͵½Ò»°ëʱ[HA] = [A-], ¹ÊpKa = pH = 5.00¡£ ÉèÖкÍÖÁ»¯Ñ§¼ÆÁ¿µãʱÓÃÈ¥V mL NaOH,Ôò c(NaA) = 0.2500¡ÁV/(60.00+V) .00 Òò[OH-] =

Kb?c, ¹Ê 10?5.00?10?1410?5.00?0.2500V60.00?V

½âµÃV = 40.00mL w(HA) =

0.2500?40.00?82.001600.?1000?100% = 51.25%

2363 Ëá¼îָʾ¼ÁµÄÀíÂÛ±äÉ«µãÊÇ______,±äÉ«·¶Î§¿ÉÓù«Ê½±íʾΪ__________¡£

2364 ijËá¼îָʾ¼ÁHIn,µ±[HIn]/[In-] = 5ʱ,¿´µ½µÄÊÇ´¿ËáÉ«,µ±[In-]/[HIn] = 3ʱ, ¿´µ½µÄÊÇ´¿¼îÉ«¡£Èô¸Ãָʾ¼ÁµÄ½âÀë³£ÊýΪpKa,Ôò±äÉ«·¶Î§ÊÇ____________¡£

2365 ÒÑÖª¼×»ù»ÆµÄpK(HIn) = 3.25,±äÉ«·¶Î§Îª2.9~4.0 (ºì-»Æ),Ôòµ±Ö¸Ê¾¼Á·Ö±ðÏÔºìÉ«ºÍ»ÆÉ«Ê±,[HIn]/[In-]·Ö±ðÊÇ____ºÍ____¡£

2366 ½«¼×»ùºìָʾ¼Á[K(HIn)=7.9¡Á10-6]¼ÓÈëijδ֪µÄpH»º³åÈÜÒºÖÐ, Ó÷ֹâ¹â¶È·¨²âµÃ¸ÃÈÜÒºÖÐָʾ¼ÁµÄ¼îÉ«ÓëËáɫ֮±ÈΪ2.15:1,ÔòÈÜÒºµÄpHΪ________¡£ 2367 Ëá¼îָʾ¼ÁµÄ½âÀëÆ½ºâ¿É±íʾΪ: HIn = H+ + In-

Ôò±ÈÖµ[In-]/[HIn]ÊÇ______________µÄº¯Êý¡£Ò»°ã˵À´,¿´µ½µÄÊǼîɫʱ,¸Ã±ÈֵΪ____;¿´µ½µÄÊÇËáɫʱ,¸Ã±ÈֵΪ_____;¿´µ½»ìºÏɫʱ,¸Ã±ÈֵΪ________¡£

2368 ʲôÊÇËá¼îָʾ¼Á?¼òÊöËá¼îָʾ¼ÁµÄ×÷ÓÃÔ­Àí¡£

ͨ³£ÓÐÁ½Àà»ìºÏ(Ëá¼î)ָʾ¼Á,Ò»ÀàÊÇÓÉÁ½ÖÖ»òÁ½ÖÖÒÔÉϵÄָʾ¼Á»ìºÏ¶ø³É;ÁíÒ»ÀàÓÉijÖÖָʾ¼ÁºÍÒ»ÖÖ¶èÐÔȾÁÏ×é³É¡£ÊÔ¼òÒª±È½ÏËüÃǵÄÒìͬµã¡£

2369 ͨ³£ÓÐÁ½Àà»ìºÏ(Ëá¼î)ָʾ¼Á,Ò»ÀàÊÇÓÉÁ½ÖÖ»òÁ½ÖÖÒÔÉϵÄָʾ¼Á»ìºÏ¶ø³É;ÁíÒ»ÀàÓÉijÖÖָʾ¼ÁºÍÒ»ÖÖ¶èÐÔȾÁÏ×é³É¡£ÊÔ¼òÒª±È½ÏËüÃǵÄÒìͬµã¡£

2370 Ç¿ËáµÎ¶¨Ç¿¼îʱ,ÈôŨ¶È¾ùÔö´ó10±¶,ÔòÆäpHͻԾÔö´ó----------------------------------( ) (A) 1 (B) 2 (C) 10 (D) ²»±ä»¯

2371 ¶àÔªËá·Ö²½µÎ¶¨Ê±,ÈôŨ¶È¾ùÔö¼Ó10±¶,µÎ¶¨pHͻԾ´óС±ä»¯-------------------------( ) (A) 1 (B) 2 (C) 10 (D) ²»±ä»¯

2372 ÏÖÓÐ50mLij¶þÔªËáH2X, c(H2X) = 0.1000mol/L,ÓÃ0.1000mol/L NaOH µÎ¶¨¡£ÔÚ¼ÓÈë25mL NaOHʱ,pH = 4.80;¼ÓÈë50mL NaOH,¼´ÔÚµÚÒ»»¯Ñ§¼ÆÁ¿µãʱ, pH = 7.15,ÔòKa2ֵΪ( ) (A) 5.62¡Á10-10 (B) 3.2¡Á10-10 (C) 8.4¡Á10-9 (D) 7.1¡Á10-8

2373 ÓÃÇ¿¼îµÎ¶¨ÈõËá,µ±ËáµÄŨ¶ÈÒ»¶¨Ê±,ËáÓúÇ¿(KaÖµÓú´ó),ËüµÄ¹²éî¼îÓú______, µÎ¶¨·´Ó¦µÄÍêÈ«³Ì¶ÈÓú____,ͻԾ·¶Î§Ò²Óú____¡£

2374 ÓÃÇ¿¼îµÎ¶¨ÈõËá,µ±Á½ÖÖËáµÄŨ¶ÈÒ»¶¨Ê±,ÈôÈõËáÓúÈõ(KaÖµÓúС),ËüµÄ¹²éî¼îÓú______, µÎ¶¨·´Ó¦µÄÍêÈ«³Ì¶ÈÓú____,ͻԾ·¶Î§Ò²Óú____¡£

2375 ÔÚÇ¿¼îµÎ¶¨Ç¿Ëáʱ,ÈôËáºÍ¼îµÄŨ¶ÈÔö´ó10±¶,µÎ¶¨pHͻԾ½«______________; ·´Ö®ÈôËüÃǵÄŨ¶È¼õС10±¶,ÔòµÎ¶¨pHͻԾ½«________________¡£

2376 ÔÚÇ¿¼îµÎ¶¨ÈõËáʱ,ÈôËáºÍ¼îµÄŨ¶ÈÔö´ó10±¶,pHͻԾ·¶Î§½«______________; ·´Ö®µ±ËüÃǵÄŨ¶È¼õС10±¶,ÔòpHͻԾ·¶Î§½«________________¡£ 2377 ÔÚËá¼îµÎ¶¨ÖÐ,Ñо¿µÎ¶¨ÇúÏßµÄÒâÒåÖ÷ÒªÔÚÓÚ:

(1)______________________________________________________________ (2)______________________________________________________________ (3)______________________________________________________________

2378 ÓÃNaOHµÎ¶¨Ò»ÔªÈõËáHA,ÈôNaOHºÍHAµÄŨ¶È¾ùÔö´ó10±¶,Ôò»¯Ñ§¼ÆÁ¿µãǰ0.1% pH_________; »¯Ñ§¼ÆÁ¿µãpH________;»¯Ñ§¼ÆÁ¿µãºó0.1% pH_________(Ö¸Ôö¼Ó»ò¼õÉÙ¶àÉÙ)¡£ 2379 Ò»ÔªÈõËá(HA)µÎ¶¨µÄÖÕµãÎó²î¿ÉÓù«Ê½Et = _______________________¼ÆËã¡£ Éè?pH = pH

ÖÕ

-pH

¼Æ

, Kt ΪµÎ¶¨·´Ó¦³£Êý, ÔòÉÏʽÓֿɱíʾΪ Et = _________________________

____________¡£

2380 ÓÃͬŨ¶ÈÇ¿¼îµÎ¶¨ÈõËáͻԾ·¶Î§µÄ´óСÓë__________________ÓйØ, ÈôÒªÄÜ׼ȷµÎ¶¨(Et<0.2%),ÔòÒªÇóÂú×ã_________________Ìõ¼þ

2381 ÓÃͬŨ¶ÈÇ¿ËáµÎ¶¨Èõ¼îͻԾ·¶Î§µÄ´óСÓë__________________ÓйØ, ÈôÒªÄÜ׼ȷµÎ¶¨(Et<0.2%),ÔòÒªÇóÂú×ã_________________Ìõ¼þ¡£

2382 ÏÂÁÐËáÈÜÒºÄÜ·ñÓÃNaOH׼ȷµÎ¶¨»ò׼ȷ·Ö²½µÎ¶¨(Ũ¶È¾ùΪ0.1mol/L), ÇëÔÚ±¸Ôñ´ð°¸(A,B,?)ÖÐÑ¡ÔñÕýÈ·´ð°¸:

(1) ¾ÆÊ¯Ëá(pKa1 = 3.04¡¢pKa2 = 4.37)

________

(2) ÄûÃÊËá(pKa1 = 3.13¡¢pKa2 = 4.76¡¢pKa3 = 6.60) ________ (3) ÑÇÁ×Ëá(pKa1 = 1.30¡¢pKa2 = 6.60)

________

(4) ÅðËá(pKa = 9.24)+±½¼×Ëá(pKa = 4.21)

________

A. ÄÜ׼ȷ·Ö²½µÎ¶¨

B. ÄÜ×÷Ϊ¶àÔªËáͬʱµÎ¶¨,µ«²»ÄÜ·Ö²½ C. µÚÒ»²½ÄÜ׼ȷµÎ¶¨,µ«µÚ¶þ²½²»Äܵζ¨

D. µÚÒ»¡¢¶þ²½ÄÜ·Ö²½×¼È·µÎ¶¨,µ«µÚÈý²½²»Äܵζ¨

2383 Á×ÒÔMgNH4PO4¡¤6H2OÐÎʽ³Áµí,¾­¹ýÂË¡¢Ï´µÓºóÓÃÊÊÁ¿HCl±ê×¼ÈÜÒºÈܽâ,¶øºóÒÔ NaOH±ê×¼ÈÜÒº·µµÎ¶¨,Ñ¡¼×»ù³ÈΪָʾ¼Á,ÕâʱÁ×ÓëHClµÄÎïÖʵÄÁ¿±Èn(P):n(HCl)Ϊ-( ) (A) 1:3 (B) 3:1 (C) 1:1 (D) 1:2

2384 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

2385 ½ñÓû²â¶¨pH¡Ö9µÄÈÜÒºµÄpH,ÏÂÁÐÈÜÒºÖÐÊÊÒËУÕýpH¼Æ(¶¨Î»)µÄÊÇ--------------( ) (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

2386 ÊÔ¾ÙÁ½Öֱ궨NaOHÈÜÒºµÄ»ù×¼ÎïÖÊ:______________ºÍ_____________¡£ 2387 ¾Ù³öÁ½Öֱ궨HClÈÜÒºµÄ»ù×¼ÎïÖÊ:______________ºÍ_____________¡£

2388 Á×ÒÔMgNH4PO4¡¤6H2OÐÎʽ³Áµí,¾­¹ýÂË¡¢Ï´µÓºóÓùýÁ¿µÄHCl±ê×¼ÈÜÒºÈܽâ,ÒÔNaOH±ê×¼ÈÜÒº·µµÎ¶¨ÖÁ¼×»ù³È±ä»Æ,´ËʱÁ×ÓëHClµÄÎïÖʵÄÁ¿±È n(P):n(HCl)ÊÇ________¡£ 2389 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+) = __________________________.

2390

½«0.300mol/L H3PO4ÈÜÒº50.0mLÓë0.200mol/L Na3PO4ÈÜÒº50.0mL»ìºÏ¡£ (1) ÒÔ¼×»ùºìΪָʾ¼Á,ÓÃ0.100mol/LHClÈÜÒºµÎ¶¨ÖÁÖÕµã,ÐèÏûºÄHClÈÜÒº¶àÉÙºÁÉý? (2) Èô¸ÄÓÃ0.400mol/LNaOHÈÜÒºµÎ¶¨ÖÁ°ÙÀï·Ó̪ΪÖÕµã,ÐèÏûºÄNaOHÈÜÒº¶àÉÙºÁÉý? 2391 ³ÆÈ¡´¿NaHCO3 1.008g,ÈÜÓÚÊÊÁ¿Ë®ÖÐ,È»ºóÍù´ËÈÜÒºÖмÓÈë´¿¹ÌÌåNaOH 0.3200g,×îºó½«ÈÜÒºÒÆÈë250mLÈÝÁ¿Æ¿ÖÐ,Ï¡ÊÍÖÁ±êÏß¡£ÒÆÈ¡´ËÈÜÒº50.0mL,ÒÔ0.100mol/L HClÈÜÒºµÎ¶¨¡£¼ÆËã:(1)ÒÔ·Ó̪Ϊָʾ¼ÁµÎ¶¨ÖÁÖÕµãʱ,ÏûºÄHClÈÜÒº¶àÉÙºÁÉý? (2)¼ÌÐø¼ÓÈë¼×»ù³Èָʾ¼ÁµÎ¶¨ÖÁÖÕµãʱ,ÓÖÏûºÄHClÈÜÒº¶àÉÙºÁÉý? [Mr(NaHCO3) = 84.00, Mr(NaOH) = 40.00]

2392 ¼òÊö¼×È©·¨²â¶¨ï§ÑεĻù±¾Ô­Àí,ÒªÇóд³ö·´Ó¦Ê½¡¢µÎ¶¨¼Á¡¢Ö¸Ê¾¼Á¼°»¯Ñ§¼ÆÁ¿±ÈµÈ¡£ 2393 Á׿ɳÁµíΪMgNH4PO4¡¤6H2O,³Áµí¾­¹ýÂË¡¢Ï´µÓºó¿ÉÓùýÁ¿HClÈܽâÒÔNaOH ·µµÎ¶¨,ÎÊ: (1) ´ËµÎ¶¨Ó¦Ñ¡¼×»ù³È»¹ÊÇ·Ó̪Ϊָʾ¼Á,Ϊʲô? (2) n(P)¡Ãn(HCl)ÊǶàÉÙ?

2394 ÒÔ˫ָʾ¼Á·¨½øÐлìºÏ¼îµÄ·ÖÎöʱ,È¡Á½·ÝÌå»ýÏàͬµÄÊÔÒº,µÚÒ»·ÝÓ÷Ó̪×÷ָʾ¼Á,ÏûºÄ±ê×¼HClÈÜÒºµÄÌå»ýΪV1,µÚ¶þ·ÝÓü׻ù³È×÷ָʾ¼Á,ÏûºÄHClΪV2, ÊÔ¸ù¾ÝHClÌå»ýµÄ¹ØÏµ,ÅжϻìºÏ¼îµÄ×é³É¡£

V1 V2 ×é³É V1 = V2 V1 = 0 2V1 = V2 2V1>V2 2V1

V1 V2 ×é³É V1 = 0 0 = V2 V1 = V2 V1 > V2 V1 < V2 2396 ÊÔ¾ÙÒ»ÖÖÅäÖÆ²»º¬CO32-µÄNaOHÈÜÒºµÄ·½·¨¡£

2397 ÒÒ´¼×÷ΪÈܼÁʱµÄÖÊ×Ó×Եݷ´Ó¦Îª______________________________, ÔÚÒÒ´¼ÈÜÒºÖÐ,×îÇ¿µÄ¼îÊÇ_____________¡£ 4301 ¼òÊöÏÂÁл¯Ñ§¼ÒµÄÖ÷Òª¹±Ï×

²¼ÀÊË¹ÌØµÂ£­ÀÍÀ³£¨J.N.Br?nsted-T.M.Lowry£©____________________________

µÂ°Ý£­Óȸñ¶û£¨P.Debye-W.H¨¹ckel£© ___________________________________ ÄÜË¹ÌØ£¨H.W.Nernst£©________________________________________________

·Ò×Ê¿£¨G.N.Lewis£©_________________________________________________

4302 Ö¸³öÏÂÁй«Ê½Öи÷·ûºÅµÄÒâÒå

?lg¦Ã?0.512z2I1?BaI

4303 0.002 mol/L CuCl2 + 0.002 mol/L (NH4)2SO4 + 0.002 mol/L Ga(NO3)3ÈÜÒºµÄÀë×ÓÇ¿¶ÈΪ________________¡£

4304 ÔÚÏÂÁйØÓÚ»î¶ÈϵÊýµÄ±íÊöÖÐÕýÈ·µÄÊÇ??????????????????( ) ÔÚÀë×ÓÇ¿¶È0~0.1mol/L·¶Î§ÄÚ£º( 1 )ËæÀë×ÓÇ¿¶ÈÔö¼Ó¶ø¼õÉÙ£»( 2 )ËæÀë×ÓµçºÉÔö¼Ó¶ø¼õÉÙ£»( 3 )ËæË®ºÏÀë×Ó°ë¾¶½µµÍ¶ø¼õС¡£ ( A ) (1)

( B ) (1) ºÍ (2) ( C ) (2) ºÍ (3)

( D ) (1)¡¢(2) ºÍ (3)

4305 ¼ÆËã³öÏÂÁÐÈÜÒºµÄÀë×ÓÇ¿¶È£º

( 1 ) 0.002 mol/L La(IO3)3 ___________________________¡£

( 2 ) 0.002 mol/L CuSO4 ______________________________¡£

4306 0.25 mol/LijÈõËáHAÈÜÒºµÄpHΪ3.20,¸ÃËáµÄ½âÀë³£ÊýÊÇ________ ¡£

4307 ijһÈõËáHAÊÔÑùÓÃͬŨ¶ÈNaOHÈÜÒºµÎ¶¨£¬¼ÓÈë10.00 mL NaOHºó£¬pHΪ5.70¡£ÈôµÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µãʱ£¬¹²Ðè30.00 mL NaOH¡£¸ÃËáµÄ½âÀë³£ÊýÊÇ_____________________ ¡£ 4308 ³ÆÈ¡0.3600g±½¼×Ëá(C7H6O2)ÊÔÑùÈÜÓÚ180 mLË®£¬¼ÓÈë10.00mL 0.08690 mol/L NaOH£¬È»ºó£¬½«ÆäÓÃˮϡÊÍÖÁ200.0 ml¡£²âµÃ¸ÃÈÜÒº25¡æÊ±pHΪ3.82¡£ÊÔ¼ÆËã±½¼×ËáµÄ½âÀë³£Êý¡£(ÒÑÖªMr (C7H6O2)=122.12)

4309 ÓÃNaOHÈÜÒºµÎ¶¨Ä³Ò»ÔªÈõËáHA£¬µÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µãʱ£¬Ðè37.20 mLNaOH¡£Ôڸõãʱ£¬¼ÓÈë18.60 mLÓëNaOHͬŨ¶ÈµÄHClÈÜÒº£¬²âµÃÆäpHΪ4.30£¬ÊÔ¼ÆËãHAµÄKa _________¡£ 4310 È¡50.0mLÒ»·Ý0.120mol/LijһԪÈõËáHA£¬ÓÃ0.200mol/L NaOHÈÜÒºµÎ¶¨¡£¼ÓÈë17.5 mL¼îÒÔºó£¬ÈÜÒºµÄpHΪ4.80¡£ÊÔ¼ÆËãHAµÄKa¡£

4311ij0.20 mol/LÒ»Ôª°·RNH2ÈÜÒºµÄpHΪ8.42£¬¸ÃÒ»Ôª¼îµÄKbÊǶàÉÙ£¿

4312 ÏÖÓÐ25.0mLijÈõËáHAÈÜÒºÓÃ0.100mol/LNaOHÈÜÒºµÎ¶¨£¬¼ÓÈë17.5mL NaOHºó£¬ÈÜÒºµÄpH£½5.80¡£ÊÔ¼ÆËã¸ÃÈõËáHAµÄ½âÀë³£ÊýKa¡£

4313 ʵÑé²âµÃ0.100mol/L HCOONaÈÜÒºµÄpHΪ8.34£¬ÊÔ¼ÆËãHCOOHµÄ½âÀë³£Êý¡£ 4314ÊÔ¼ÆËã0.070mol/L AlCl3ÈÜÒºµÄpHÒÔ¼°¸÷ÖÖ·Ö×ÓºÍÀë×ÓµÄŨ¶È£¬ÒÑÖª ?Al(H3?2O)6??H???Al?H2O?5?OH??2?,Ka?[Al(H2O)6]}?1.12?10?5

4315 ÒÑÖªpKb(C6H5CH2NH2)=4.65£¬¸ø³ö±½°·(C6H5CH2NH2)ÓëHClÖ®¼ä·´Ó¦µÄƽºâ³£Êý£¬ ______________ ¡£

4316 ÔÚ200 mL 0.100 mol/L HAc-1.0¡Á10-4 mol/Läå¼×·ÓÀ¶Ö¸Ê¾¼Á[pK(HIn)=4.9]ÈÜÒºÖУ¬¼ÓÈë¶àÉÙºÁÉý1.00mol/L NaOHÈÜÒº²ÅÄÜʹ£ÛHIn£Ý/£ÛIn-£Ý= 1/10 ?ÒÑÖªpKa(HAc) = 4.74¡£ 4317 ijËá¼îָʾ¼ÁHInÔÚÆäÓÐ1/5ת±ä³ÉÀë×ÓÐÎʽʱ·¢ÉúÑÕÉ«±ä»¯£¬ÈôÑÕÉ«±ä»¯Ê±µÄ pH = 6.40£¬ÊÔ¼ÆËãָʾ¼ÁµÄ½âÀë³£ÊýKa(HIn)£¬_______________________¡£

4318 ÔÚ200mL 0.100mol/L HAc-1.0¡Á10-4mol/Läå¼×·ÓÀ¶Ö¸Ê¾¼Á[pKa(HIn)=4.9]ÈÜÒºÖУ¬¼ÓÈë¶àÉÙ

ºÁÉý1.00mol/L NaOHÈÜÒº²ÅÄÜʹ£ÛHIn£Ý/£ÛIn-£Ý=10/1 ? [ÒÑÖªpKa(HAc)=4.74] 4319 д³öÏÂÁÐÈÜÒºµÄÖÊ×ÓÆ½ºâ·½³Ìʽ£º (1) H3PO4 _________________________________________¡£ (2) NaHCO3 ________________________________________¡£ 4320 д³öÏÂÁÐÈÜÒºµÄÖÊ×ÓÆ½ºâ·½³Ìʽ:

(1) NH4H2PO4 ____________________________________¡£

(2) NaAc ________________________________________¡£ 4321 д³öÏÂÁÐÈÜÒºµÄÖÊ×ÓÆ½ºâ·½³Ìʽ:

(1) Na2HAsO4 ______________________________________¡£

(2) (NH4)2HPO4 ____________________________________¡£

4322 ¶ÔÓÚijº¬H2O¡¢H+¡¢OH-¡¢ClO4-¡¢Fe(CN)63-¡¢CN-¡¢Fe3+¡¢Mg2+¡¢HCN¡¢NH3ºÍNH4+ÈÜÒº£¬µçºÉƽºâ·½³ÌʽÊÇ____________________________________________¡£

4323 CaF2ÈÜÒºµÄµçºÉƽºâ·½³Ìʽ(ÈÜÒºÖÐÓÐCa2+¡¢F-¡¢ºÍCaF+µÈÀë×Ó´æÔÚ)ÊÇ____________¡£ 4324 (1)д³öCaCl2ÈÜÒºµÄÎïÁÏÆ½ºâ·½³Ìʽ£¬ÈôÈÜÒºÖÐÒÔÎïÖÖCa2+ºÍCl-´æÔÚ£º_________________________________________________¡£

(2)д³öCaCl2ÈÜÒºµÄÎïÁÏÆ½ºâ·½³Ìʽ£¬ÈôÈÜÒºÖÐÒÔÎïÖÖCa2+¡¢CaCl+ºÍCl-´æÔÚ£º _________________________________________________¡£

4325 д³öijº¬H+¡¢OH-¡¢Ca2+¡¢HCO3-¡¢CO32-¡¢Ca(HCO3)+¡¢Ca(OH)+¡¢K+ºÍClO4-ÈÜÒºµÄµçºÉƽºâ·½³Ìʽ£º______________________________________________¡£ 4326 (1)д³ö±¥ºÍCaF2ÈÜÒºµÄÎïÁÏÆ½ºâ·½³Ìʽ£¬Èô·´Ó¦Îª CaF2(s) = Ca2+ + 2F-

F- + H+ = HF(aq)

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(2)д³öCaF2ÈÜÒºµÄÎïÁÏÆ½ºâ·½³Ìʽ£¬ÈôÔÚÉÏÊö·´Ó¦Ö®Í⻹·¢ÉúÏÂÁз´Ó¦

HF(aq) + F- = HF2-

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4327 д³öCa3(PO4)2ÈÜÒºµÄÎïÁÏÆ½ºâ·½³Ìʽ£¬ÈôÈÜÒºÖдæÔÚÎïÖÖCa2+¡¢PO43-¡¢HPO42-¡¢H2PO4-ºÍH3PO4 ¡£__________________________________________________________¡£ 4328 д³öijº¬

CaCO3

ºÍ

KClO4

µÄË®ÈÜÒºµÄµçºÉƽºâ·½³Ìʽ¡£ ____________________________________________¡£

4329 µÈÌå»ý»ìºÏpHΪ2.00ºÍ3.00µÄHClÈÜÒº£¬ËùµÃÈÜÒºµÄpHΪ ___________________¡£

4330 ÏÖÓÐÒ»´×ËáÈÜÒº£¬ÈôÓÐ3.0%µÄ´×Ëá½âÀ룬ÊÔÎʸô×ËáÈÜÒºµÄŨ¶ÈÊǶàÉÙ£¿ÒÑÖª Ka(HAc)=1.8¡Á10-5

4331 ¼ÆËã0.0100 mol/L H2SO4ÈÜÒºÖи÷ÎïÖÖµÄŨ¶È£¬ÒÑÖªH2SO4 µÄpKa2

=1.2¡Á10-2

4332 ѪҺÊÔÑùÖÐ×ܶþÑõ»¯Ì¼??HCO?CO3???CO2??º¬Á¿¿Éͨ¹ýËữÊÔÑù²âÁ¿

2µÄÌå»ý½øÐвâ

¶¨¡£½ñ²âµÃijѪҺÊÔÑùÖÐCO2×ÜŨ¶ÈΪ28.5 mmol/L£¬ÔÚ37¡æÊ±¸ÃѪҺµÄpHΪ7.48¡£ÊÔÎʸÃѪҺÊÔÑùÖÐHCO3-ºÍCO2µÄŨ¶È¸÷ÊǶàÉÙ£¿(ÒÑÖªH2CO3µÄKa1=7.9¡Á10-7£¬Ka2=1.3¡Á10-10) 4333 ½ñ½«5.00 mL 0.100 mol/L NaOH ÈÜÒº¼ÓÈëijһº¬204 mg KH2PO4µÄÈÜÒºÖУ¬ÓÃˮϡÊÍÖÁ200mL£¬ÊÔ¼ÆËãËùµÃÈÜÒºµÄpH ¡£

[ÒÑÖªH3PO4µÄpKa1

=2.12¡¢pKa2

=7.21¡¢pKa3

=12.0£¬Mr(KH2PO4)=136.09]

4334 Èô½«25mL 0.20 mol/L NaOHÈÜÒº¼ÓÈë20mL 0.25 mol/L ÅðËáÈÜÒºÖУ¬ËùµÃÈÜÒºµÄpHÊǶàÉÙ£¿ [ÒÑÖªKa (H3BO3)=6.4 ¡Á10-10]

4335 ½«112 mL 0.1325 mol/L H3PO4ºÍ136 mL 0.1450 mol/L Na2HPO4»ìºÏ£¬ÊÔ¼ÆËãËùµÃ»ìºÏÈÜÒºµÄpHºÍ»º³åÈÝÁ¿¡£ (ÒÑÖªÁ×ËáµÄpKa1

= 2.12£¬pKa2

= 7.21£¬pKa3

= 12.0)

4336 ½«pHΪ2.00µÄÇ¿ËáÈÜÒººÍpHΪ13.00µÄÇ¿¼îÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpHÊÇ______________________________________________________________¡£

4337 ÏÖÓÐpH·Ö±ðΪ2.00µÄÇ¿ËáÈÜÒººÍ13.00µÄÇ¿¼îÈÜÒºµÈÌå»ý»ìºÏ£¬ÆäpHΪ???( ) ( A ) 5.5 ( B ) 6.5 ( C ) 11.0

( D ) 12.65

4338

0.010 mol/L ÁÚ±½¶þ¼×Ëá[C6H4(COOH)2]

ÈÜÒºµÄpH [ÒÑÖªC6H4(COOH)2µÄKa1

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Ka-6

2

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4339½«ôǰ·ºÍÑÎËá¸÷0.10 mol¼ÓÈë500 mLË®ÖУ¬ËùµÃÈÜÒºµÄpH [ÒÑÖªKb(HONH2) = 1.3 ¡Á10-6] ÊÇ_____________________________________________¡£(ÒªÇóд³ö¼ÆËãʽ)

4340 ¼ÆËã0.200 mol/L NH4NO3ÈÜÒºµÄpH¼°¸÷ÖÖÀë×ӺͷÖ×ÓµÄŨ¶È£¬ÒÑÖªpKa(NH4+) = 9.25¡£ 4341 0.010 mol/L ÁÚ±½¶þ¼×Ëá¶þ¼ØÈÜÒºµÄpH [ÒÑÖªC6H4(COOH)2µÄKa=1.2 ? 10-3, Ka2

=3.9 ? 10-6 1

]

ÊÇ______________________________________________¡£(ÒªÇóд³ö¼ÆËãʽ) 4342 µ±ÈÜÒºµÄpHÔö¼Ó1.60µ¥Î»Ê±£¬ÈÜÒºµÄ£ÛH+

£Ý¸Ä±ä¶àÉÙ£¿

_______________________________________________________¡£ 4343 Èç¹ûÒ»ÈÜÒºµÄ[H+]= 10 [OH-]£¬ÄÇô¸ÃÈÜÒºµÄpHÊÇ

( A ) 6.0 ( B ) 6.5 ( C ) 7.5

( D ) 8.0

4344 0.10 mol/L HIO3ÈÜÒºµÄpH [ÒÑÖªKa(HIO3) = 2 ? 10-1] ÊÇ _______________________ ___________________________________________¡£(ÒªÇóд³ö¼ÆËãʽ)

4345 ijÓлú°·RNH2µÄpKbΪ4.20£¬ÏÖÓÐ0.20 mol/L¸ÃÓлú°·ÈÜÒº£¬ÆäpHÊÇ_________ ___________________________________£¨ÒªÇóд³ö¼ÆËãʽ)

4346 0.25 mol/LÑÎËáßÁà¤(C5H5NH+Cl-)ÈÜÒºµÄpH [ÒÑÖªKb(C5H5N) = 1.7 ? 10-9]ÊÇ£º____________________________________________¡£(ÒªÇóд³ö¼ÆË㹫ʽ)

4347 ÊÔ¼ÆË㽫25.0 mL 0.0840 mol/L NH3ÈÜÒººÍ20.0 mL 0.1050 mol/L HAc»ìºÏËùµÃÈÜÒºµÄpH¡£[ÒÑÖªpKa(NH4+) = 9.26£¬pKa(HAc) = 4.74]

4348 ½«ÏÂÁÐÈÜÒº¸÷75.0 mL»ìºÏ£º0.500 mol/L NH3 £¬0.0500 mol/L HCl£¬0.100 mol / L NH4Cl£¬ÊÔ¼ÆËãËùµÃÈÜÒºµÄpH¡£[ÒÑÖªpKa(NH4+) = 9.26]

4349 ¼ÆË㽫4.92 g NaAcÈܽâÓÚ200 mL 0.200 mol/L HClÈÜÒºÖÐÖÆ³ÉµÄ»º³åÈÜÒºµÄpH¡£Éè×îÖÕµÄÌå»ýV×îÖÕ = 200 mL¡£[ÒÑÖªKa(HAc) = 1.8 ?10-5£¬Mr(NaAc) = 82.03]

4350 ½ñ½«10.70 g NH4Cl¼ÓÈë200 mL 0.250 mol/L NaOHÈÜÒºÖУ¬ÊÔ¼ÆËãËùµÃ»º³åÈÜÒºµÄpH¡£

[ÒÑÖªpKa (NH4+) = 9.26£¬Mr (NH4Cl) = 53.49 ]

4351 ÓÉ0.20 mol / LÁÚ±½¶þ¼×Ëá(H2P)ºÍ0.10 mol / LÁÚ±½¶þ¼×ËáÇâ¼Ø(KHP)¹¹³ÉµÄÈÜÒºµÄpH [ÒÑÖªC6H4(COOH)2µÄKa1

= 1.2 ? 10-3£¬Ka2

= 3.9 ? 10-6]ÊÇ____________¡£(д³ö¼ÆËãʽ)

4352 ÉèijEDTAÈýÄÆÑÎÈÜÒºµÄŨ¶ÈΪ0.050 mol/L£¬Ôò¸ÃÈÜÒºµÄpHΪ__________________¡£

[ÒÑÖªEDTA(H4Y)µÄKa = 1.0 ? 10-2£¬Ka2

= 2.2 ? 10-3£¬Ka3

= 6.9 ? 10-7£¬Ka4

= 5.5 ? 10-111

]

4353 ÉèijEDTA¶þÄÆÑÎ(Na2H2Y)ÈÜÒºµÄŨ¶ÈΪc mol/L£¬Ôò¸ÃÈÜÒºµÄpH¿É°´ºÎÖÖ¹«Ê½¼ÆËã¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­( ) ?

?A? ?H??K?a1c ?B? ?H??Ka2c?C? ?H???K?

a1Ka2 ?D? ?H??Ka2Ka34354 ÉèijEDTAÈýÄÆÑÎ(Na3HY)ÈÜÒºµÄŨ¶ÈΪc mol / L£¬Ôò¸ÃÈÜÒºµÄpH¿É°´ºÎÖÖ¹«Ê½¼ÆËã¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­¡­( ) ?A?

?H???Ka2Ka3 ?B? ?H???Ka3Ka4?C? ?H???Kac ?D? ?H

?2??Ka3c4355 ÅäÖÆ1L pH = 6.90µÄ»º³åÈÜÒºÐè0.100 mol / L H3PO4ºÍ0.100 mol/L NaOH¸÷¶àÉÙºÁÉý£¿

(ÒÑÖªÁ×ËáµÄpKa1

= 2.12£¬pKa2

= 7.21£¬pKa3

= 12.0)

4356ÔÚ250 mL 0.100 mol/L NH3ÈÜÒºÖмÓÈë¶àÉÙ¿ËNH4Cl²ÅÄÜʹ£ÛH+£ÝÔö´ó100±¶£¿

[ÒÑÖªKb(NH3)=1.8 ? 10-5£¬Mr(NH4Cl) = 53.49]

4357 ¶ÔÓÚ50 mL 0.100 mol/L HAcÈÜÒº£¬ÔÚÏÂÁд¦ÀíÖÐÄÄÒ»ÖÖ´¦ÀíÄÜʹÈÜÒºµÄpHÔö¼Ó×î´ó£¿ÊÔͨ¹ý¼ÆËã¼ÓÒÔ˵Ã÷¡£(1) ¼ÓˮϡÊÍÖÁ1L£»(2) ¼ÓÈë20.00 mL 0.0500 mol/L NaOH£¬(3) ¼ÓÈë0.082 g NaAc¡£[ÒÑÖªpKa(HAc) = 4.74£¬Mr(NaAc) = 82.03]

4358ÓÃÇ¿ËáµÎ¶¨Èõ¼îʱ£¬ÔÚÄÄÒ»µã»º³åÈÝÁ¿´ïµ½¼«´óÖµ£¿____________________¡£ 4359 ÔÚ200 mL 0.100 mol/L HAcÈÜÒºÖмÓÈë¶àÉÙ¿ËNaAc²ÅÄÜʹÆäpHÔö´ó2¸öµ¥Î»£¿

[ÒÑÖªKa(HAc)=1.8 ? 10-5£¬Mr(NaAc?3H2O) = 136.00]

4360 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]

4361 ÔÚ1L pH = 10.00£¬Ã¿ºÁÉýº¬21.4 mg NH3µÄ»º³åÈÜÒºÖУ¬ÓжàÉÙ¿ËNH4Cl£¿

[ÒÑÖªpKa = 9.26£¬Mr(NH3) = 17.03£¬Mr(NH4Cl) = 53.49]

4362½«0.5mol/LµÄ¸Ê°±Ëá(H2NCH2COOH)ÈÜÒºÓë0.25mol/LµÄHClÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHÊǶàÉÙ£¿(ÒÑÖª¸Ê°±ËáµÄpKa1

= 2.35£¬pKa2

= 9.78) __________________________¡£

4363ÏÖÓÐÒ»¸Ê°±Ëá (H2NCH2COOH)-NaOH»º³åÈÜÒº£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ2:1ʱ£¬ÈÜÒºµÄpHÊǶàÉÙ£¿(ÒÑÖª¸Ê°±ËáµÄpKa1

= 2.35£¬pKa2

= 9.78)__________________________¡£

4364

ÏÖÓÐÁ½ÖÖHAc-NaAcÈÜÒº£¬Ã¿ÖÖÈÜÒºµÄŨ¶È¾ùΪc(HAc + Ac- ) = 0.100 mol/L£¬ÆäpH

·Ö±ðΪ4.00ºÍ5.00¡£Èô½«Á½ÖÖÈÜÒºµÈÌå»ý»ìºÏ£¬ÊÔÎÊËùµÃÈÜÒºµÄpHÊǶàÉÙ£¿

[ÒÑÖªpKa( HAc ) = 4.74]

4365 ÖÆ±¸500 mL pH = 1.62 µÄ»º³åÈÜÒº£¬ÐèÈ¡NaHSO4ºÍNa2SO4¸÷¶àÉÙ¿Ë£¿

Éèc(NaHSO4 + Na2SO4)=1.00 mol/L¡£

[ÒÑÖªH2SO4µÄKa2 = 1.2 ? 10-2£¬Mr(Na2SO4) = 142.04£¬Mr(NaHSO4)=120.06] 4366

ÊԱȽÏÏÂÁÐÈýÖÖ»º³åÈÜÒº»º³åÈÝÁ¿µÄ´óС£º( 1 ) 0.010 mol/L HAc - 0.10 mol/L NaAc; (2)

0.010 mol/L HAc - 0.0040 mol/L NaAc; (3) 0.010 mol/L HAc - 0.0010 mol/L NaAc¡£Í¨¹ý±È½Ï¿ÉµÃµ½Ê²Ã´½áÂÛ£¿

4367 ½«1.10g NaHCO3Óë3.18 g Na2CO3 ÈÜÓÚË®²¢Ï¡ÊÍÖÁ500 mLÖÆ³É»º³åÈÜÒº£¬¼ÆËãÆäpH£»ÏÖ¼ÓÈë1.00 mL 1.00mol/L HClÈÜÒº£¬Ôò¸Ã»º³åÈÜÒºµÄpH½«ÊǶàÉÙ£¿

[ÒÑÖªH2CO3µÄKa1

=4.2?10-7, Ka2

=4.8?10-11¡£Mr(Na2CO3)=105.9, Mr(NaHCO3)=84.01]

4368 Èý(ôǼ׻ù)°±»ù¼×Íé£Û(HOCH2)3CNH2 - Tris»òTHAM£ÝÊÇÉúÎﻯѧÖо­³£ÓÃÀ´ÅäÖÆ»º³åÈÜÒºµÄÒ»ÖÖÈõ¼î£¬ÆäKbΪ1.2 ? 10-6£¬ÊÔÎÊÈôÅäÖÆ1 L pHΪ7.40µÄ»º³åÈÜÒºÐèÈ¡¶àÉÙ¿ËTHAMÓë100 mL 0.50 mol / L HClÈÜÒºÏà»ìºÏ£¿[Mr (THAM)=121.1]

4369 ÊÔ¼ÆËã180 mL 0.500 mol/L NaHSO4-0.500 mol/L Na2SO4»º³åÈÜÒºµÄ»º³åÈÝÁ¿ ________________________________________________________(д³ö¼ÆËãʽ)¡£

4370 ΪÅäÖÆpH = 4.00µÄ»º³åÈÜÒº£¬ÔÚ250 mL 0.100mol/L¾ÆÊ¯ËáÈÜÒºÖÐÓ¦¼ÓÈë¶àÉÙ¿ËNaOH£¿[ÒÑÖª¾ÆÊ¯ËáµÄKa1

=1.1 ? 10-3£¬Ka2

=6.9 ? 10-5£¬Mr(NaOH)=40.00 ]

4371 ijѪҺÊÔÑùÖÐÁ×ËáÑεÄ×ÜŨ¶È¾­·Ö¹â¹â¶È·¨²â¶¨Îª3.0 ? 10-3 mol/L£¬Èô¸ÃѪҺÊÔÑùµÄpHΪ7.45¡£ÊÔÎÊѪҺÖÐH2PO4-ºÍHPO42-µÄŨ¶È¸÷ÊǶàÉÙ£¿

£¨ÒÑÖªH3PO4µÄKa1

=1.1 ? 10-2£¬Ka2

=7.5 ? 10-8£¬Ka3

=4.8 ? 10-13)

4372 ½«¶àÉÙºÁÉý1.00 mol/L NaOHÈÜÒº¼ÓÈëµ½º¬0.9004 g ²ÝËáµÄÈÜÒºÖУ¬²¢Ï¡ÊÍÖÁ100 mL£¬¿ÉµÃµ½pHΪ4.43µÄ»º³åÈÜÒº£¿[ÒÑÖªH2C2O4µÄpKa1

=1.42£¬pKa2

=4.30£¬Mr(H2C2O4)=90.04]

4373 ÓÉHA-A-×é³ÉµÄ»º³åÈÜÒºÆä»º³åÈÝÁ¿

??2.303??Kw?cKa?H?????H????H????H???K2? a??É軺³å¶ÔHA-A-µÄ·ÖÎöŨ¶ÈΪc £¬ÇÒc > 10-3 mol/L£¬ÊÔÖ¤Ã÷

??2.303c?HA??c?A??c?HA??c?A?? 4374 ÓÉHA-A-×é³ÉµÄÈÜÒº£¬µ±c ( HA + A- )>10-3 mol/Lʱ£¬Æä»º³åÈÝÁ¿¿É±íʾΪ???( ) 2.303c?HA??c?A??c?HA?c?HA??c?A?? ( B ) ??2.303?c?A?( A ) ???c?HA??c?A??

c?HA??c?A?( C ) ??2.303?c?HA??c?A?? ( D ) ??2.303c?HA??c?A??c?HA??c?A?? 4375 ½«0.200 mol/L NaOH¡¢0.160 mol/L HCl¡¢0.120 mol/L Na2HPO4ºÍ0.120 mol/L NaH2PO4ÈÜÒºµÈÌå»ý»ìºÏ£¬ÊÔ¼ÆËãËùµÃÈÜÒºµÄpH¡£(ÒÑÖªÁ×ËáµÄpKa1

= 2.12£¬pKa2

=7.21£¬pKa3

= 12.0)

4376 ¼ÆËã0.225 mol/L Na2SO3ÈÜÒºÖеÄpH¼°¸÷ÖÖÀë×ӺͷÖ×ÓµÄŨ¶È¡£

£¨ÒÑÖªH2SO3µÄpKa1

=1.89£¬pKa2

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4377 ½«Ï൱ÓÚ0.600 mol HCl¡¢0.0500 mol H3PO4¡¢0.0800 mol NaH2PO4¡¢0.350 mol Na3PO4ºÍ0.110 mol NaOHµÄ¼¸ÖÖÈÜÒº»ìºÏÔÚÒ»¸öÊÔ¼ÁÆ¿ÖУ¬ÊÔ¼ÆËãËùµÃÈÜÒºµÄpH¡£

(ÒÑÖªÁ×ËáµÄpKa1

= 2.12£¬pKa2

= 7.21£¬pKa3

= 12.36)

4378 ÊÔ¼ÆËã0.200 mol/L AlCl3ÈÜÒºµÄpH¡£ (д³ö¼ÆËãʽ) ÒÑÖª·´Ó¦ [Al(H2O)6]3+=H++[Al(H2O)5(OH)]2+ Ka{[Al(H2O)6]3+}=1.12¡Á10-5

4379 ¼×»ù³ÈµÄÀíÂÛ±äÉ«µãΪ????????????????????????£¨ £© ( A ) 3.1 ( B ) 3.4 ( C ) 4.4 ( D ) 5.2 4380 Ãû´Ê½âÊÍ Ëá¼îָʾ¼ÁµÄÀíÂÛ±äÉ«µã ___________________________________________¡£ µÎ¶¨·ÖÎö·¨ _______________________________________________________¡£ ÖÕµãÎó²î _________________________________________________________¡£

»ù×¼ÎïÖÊ _________________________________________________________¡£

4381 ½ñÓÐÒ»º¬2.00 mmolëºÍ16.00 mmol NaHCO3µÄË®ÈÜÒº£¬ÏÖÓõâ±ê×¼ÈÜÒº½øÐеζ¨ ( N2H4 + 2I3- + 4HCO3- = N2 + 6I- + 4H2CO3 )£¬É軯ѧ¼ÆÁ¿µãʱÈÜÒºµÄÌå»ýΪ50.00 mL£¬ÊÔÎÊÔÚ»¯Ñ§¼ÆÁ¿µãʱÈÜÒºµÄpHÊǶàÉÙ£¿(ÒÑÖªH2CO3µÄpKa1

=6.38£¬pKa2

=10.32)

4382 ¼ÆË㽫50.0 mL 0.0500 mol/L HClÈÜÒººÍ50.0 mL 0.02500 mol/L Ba(OH)2ÈÜÒº»ìºÏËùµÃÈÜÒºµÄpH¡£

4383 ¼ÆËã100mL 0.200 mol/L NaOHºÍ150 mL 0.400 mol/L HAc »ìºÏËùµÃÈÜÒºµÄpH¡£

[ÒÑÖªpKa(HAc) = 4.74]

4384 ¼Ù¶¨ÓÃָʾ¼ÁÅж¨ÖÕµãµÄ¦¤pH = 0.2£¬Èô¶àÔªËá( HnA )µÄKa¡Ý 10-8£¬µ±Ka/ Ka ¡Ý 1041

1

2

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/ Ka2

¡Ý 10-5ʱ£¬Et¿É ¡Ü_______________

4385ÏÂÁбíÊöÊÇ·ñÕýÈ·(»Ø´ðÊÇÓë·ñ)£ºÈôÓÃNa2CO3±ê¶¨HClÈÜÒº£¬ÆäÖк¬ÓÐÉÙÁ¿NaHCO3£¬Ôò²âµÃµÄHClµÄŨ¶È½«±ÈÕæÊµÖµ´ó¡£____________________________¡£

4386 ÏÂÁбíÊöÊÇ·ñÕýÈ·£ºÈô±ê¶¨HClËùÓõĻù×¼ÎïÖʺ¬Óв»ÓëHCl·´Ó¦µÄÔÓÖÊ£¬Ôò²âµÃµÄHClµÄŨ¶È½«±ÈÕæÊµÖµÐ¡¡£_______________________¡£

4387 Èô±ê¶¨HClËùÓõĻù×¼ÎïÖʺ¬Óв»ÓëHCl·´Ó¦µÄÔÓÖÊ£¬Ôò²âµÃµÄHClµÄŨ¶È½«±ÈÕæÊµÖµ___________ (´ó¡¢Ð¡»òÏàͬ)£¬ÒòΪ______________________________________________ 4388 ÈôÓÃNa2CO3±ê¶¨HClÈÜÒº£¬ÆäÖк¬ÓÐÉÙÁ¿NaHCO3£¬Ôò²âµÃµÄHClµÄŨ¶È½«±ÈÕæÊµÖµ______________ (´ó¡¢Ð¡»òÏàͬ)¡£ÊÔ¼òÊöÆäÀíÓÉ¡£

4389 ½ñÓÃNaOHµÎ¶¨Ò»ÈõËáHAÊÔÑù£¬ÎªÖкʹËHAÊÔÑù£¬¹²Ðè40.0 mL NaOHÈÜÒº£¬ÔÚ¼ÓÈë8.0 mL¼îºó£¬±»µÎ¶¨ÊÔÑùµÄpHΪ5.14¡£ÊÔ¼ÆËãHAµÄ½âÀë³£ÊýKa¡£

4390 ÏÖÓÐ0.3800 gijһ½öº¬Na2CO3ºÍNaHCO3µÄÊÔÑù£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ1:3£¬½«ÆäÈܽâÓÚË®£¬ÓÃ0.1000 mol/L HCl±ê×¼ÈÜÒºµÎ¶¨¡£ÊÔ·Ö±ð¼ÆËãµÎ¶¨ÖÁ·Ó̪ºÍ¼×»ù³ÈÖÕµãʱ£¬ËùÐèHClÈÜÒºµÄÌå»ýV·Ó̪ºÍV¼×»ù³È¡£[ÒÑÖªMr(Na2CO3)=105.99£¬Mr(NaHCO3)= 84.01 ]

4391 ÈÜÒºAº¬ÓÐ0.2650 g Na2CO3ºÍ0.1260 g NaHCO3 ,½ñÓÃ0.2000 mol/L HClµÎ¶¨¡£Èô·Ö±ðÒÔ(1) ·Ó̪ºÍ (2) ¼×»ù³ÈΪָʾ¼Á£¬¸÷ÐèÑÎËá¶àÉÙºÁÉý£¿

[ÒÑÖªMr(Na2CO3)=105.99£¬Mr(NaHCO3)=84.01]

4392 ÏÖÓÐijһ½öº¬NaOHºÍNa2CO3µÄÊÔÑù0.2400 g£¬ÆäÎïÖʵÄÁ¿Ö®±ÈΪ2:1¡£½«ÆäÈܽâÓÚË®£¬ÓÃ0.1000 mol/L HCl±ê×¼ÈÜÒºµÎ¶¨¡£ÊÔ·Ö±ð¼ÆËãµÎ¶¨ÖÁÓ÷Ó̪ºÍ¼×»ù³ÈÖÕµãʱ£¬ËùÐèHClÈÜÒºµÄÌå»ýV·Ó̪ºÍV¼×»ù³È¡£[ÒÑÖªMr(NaOH)=40.00£¬Mr(Na2CO3)=105.99]

4393 ±½¼×È©( C6H5CHO )ÊÔÑù0.8242gÓë25.00 mL 0.6 mol/LÑÎËáôǰ·ÈÜÒº»ìºÏ£¬ÊͷųöµÄHClÐèÓÃ15.12 mL 0.5020 mol/L NaOHÈÜÒºµÎ¶¨£¬ÊÔ¼ÆËãÊÔÑùÖÐC6H5CHOµÄÖÊÁ¿·ÖÊý¡£ ÒÑÖª Mr(C6H5CHO)=106.12£¬·´Ó¦Ê½ÈçÏ£º C6H5CHO £« H2NOH¡¤HCl = C6H5CH==NOH + H2O + HCl

HCl + NaOH = NaCl + H2O

4394 ÄñàÑßÊ(2 -°±»ù- 6 -ôÇ»ù-àÑßÊ) [ M (C5H5N5O ) = 151.13 g / mol ]²»ÈÜÓÚË®£¬µ«ÈÜÓÚËá¡£ÊÔ¸ù¾ÝÏÂÁÐʵÑéÊý¾Ý¼ÆËãÊÔÑùÖÐÄñàÑßʵÄÖÊÁ¿·ÖÊý£º³ÆÈ¡0.1650 gÊÔÑù£¬ÈÜÓÚ25.00 mL 0.1000 mol/L HClÈÜÒºÖУ¬¹ýÁ¿µÄËáÐèÓÃ15.32mL 0.1000 mol/L NaOHÈÜÒºµÎ¶¨£¬ÒÑÖªÄñàÑßÊÓëHCl·´Ó¦¼ÆÁ¿±ÈΪ1:1¡£

4395 ³ÆÈ¡0.3228 g »¯Ñ§´¿Ä³¶þÔªÓлúËᣬÐè42.67 mL 0.1008 mol/L NaOHÈÜÒº²ÅÄÜÍêÈ«Öкͣ¬ÊÔ¼ÆËã¸ÃËáµÄĦ¶ûÖÊÁ¿¡£

4396 ÒÔ¼×»ù³ÈΪָʾ¼Á£¬ÓÃNaOHµÎ¶¨HClʱ£¬ÖÕµãÑÕÉ«±ä»¯ÊÇ ___________________£¬ÓÃHClµÎ¶¨NaOHʱ£¬ÖÕµãÑÕÉ«±ä»¯ÊÇ __________________________________¡£

4397 ÏÂÁбíÊöÊÇ·ñÕýÈ·(»Ø´ðÊÇÓë·ñ)£º½ñÓñê×¼ÈÜÒºAµÎ¶¨Ä³ÈÜÒºB£¬ÁîVºÍV0·Ö±ðΪµÎ¶¨ºÍ¿Õ°×µÎ¶¨Ê±ÈÜÒºAµÄÌå»ý¡£Èç¹ûÎóÓÃV´úÌæÐ£ÕýÖµ( V - V0 )À´¼ÆËãÈÜÒºBµÄŨ¶È£¬ÄÇôËãµÃµÄŨ¶È½«´óÓÚÕæÊµÖµ¡£____________________________¡£

4398 Ò»ÇâÑõ»¯ÄÆÈÜÒºÒòÎüÊÕÁË´óÆøÖеÄCO2±»Na2CO3ÎÛȾ£¬µ±ÒÔ¼×»ù³ÈΪָʾ¼Áʱ²âµÃÆäŨ¶ÈΪ0.1000 mol/L£¬ÒÔ·Ó̪Ϊָʾ¼Áʱ²âµÃÆäŨ¶ÈΪ0.0990 mol/L¡£ÊÔÎÊÔÚ1 L¸ÃÈÜÒºÖÐÓжàÉÙ¿ËNaOHºÍNa2CO3¡£[ÒÑÖª Mr(NaOH)=40.01£¬ Mr(Na2CO3)=105.99] 4401

È¡25.0gij¹ÌÌåÊÔÑùÖÃÓÚÒ»ÃܱյIJ£Á§ÖÓÕÖÏ£¬Í¬Ê±ÖÃÒ»ÄÚÊ¢100.0 mL NaOHµÄСÃó

ÒÔÎüÊÕCO2£¬48 hºó£¬È¡25.00 mLNaOH£¬ÒÔ·Ó̪×÷ָʾ¼Á£¬ÐèÓÃ12.17 mL 0.1250 mol / L HClÈÜÒºµÎ¶¨¡£Áí×÷Ò»·Ý¿Õ°×(²»·ÅÖøùÌÌå)²â¶¨£¬ÐèHClÈÜÒº23.31 mL¡£ÊÔ¼ÆËãÔÚϸ¾ú×÷ÓÃÏ£¬¸Ã¹ÌÌå»Ó·¢CO2µÄËÙÂÊ¡£[ÒÔÿ¿ËÿСʱ¶àÉÙºÁ¿ËCO2¼Æ£¬ÒÑÖªMr(CO2)=44.01 ]

4402 ÊÔ¸ù¾ÝÏÂÁÐʵÑéÊý¾Ý¼ÆËãÊÔÑùÖб½·Ó[Mr(C6H5OH)= 94.11 ]µÄÖÊÁ¿£ºº¬±½·ÓÊÔÑùÈÜÓÚÒ»¶èÐÔÈܼÁÖУ¬ÓùýÁ¿µÄÒÒôûÒÒõ£»¯£¬È»ºó¼Óˮʹ¹ýÁ¿µÄÒÒôûË®½â£¬ÈÜÒºÓÃ8.12 mL 0.5040 mol/L NaOHÒÒ´¼ÈÜÒºµÎ¶¨¡£ÏàͬÌå»ýµÄÒÒôûÐè18.37 mL NaOHÈÜÒº×÷¿Õ°×²â¶¨£¬·´Ó¦Ê½ÈçÏ£º (CH3CO)2O + C6H5OH = CH3COOC6H5 + CH3COOH

(CH3CO)2O + H2O = 2CH3COOH

CH3COOH + OH- = CH3COO- + H2O

4403 ijһԪÓлúËáµÄKaΪ6.7 ? 10-4£¬µ±½«100g¸ÃËáÈÜÓÚ1 LË®ÖÐʱÓÐ3.5%½âÀë¡£¸ÃËáµÄĦ¶ûÖÊÁ¿ÊǶàÉÙ£¿

4404 È¡100.0mLÌìȻˮÑù£¬ÓÃ0.0250 mol/LHCl±ê×¼ÈÜÒºµÎ¶¨ÖÁ·Ó̪ÖÕµãʱ£¬ÐèHClÈÜÒº2.80mL£¬µÎ¶¨ÖÁ¼×»ù³ÈÖÕµãʱ£¬ÐèHClÈÜÒº29.60 mL¡£ÊÔ¼ÆËãÌìȻˮÑùÖÐCO32-ºÍHCO3-µÄÖÊÁ¿Å¨¶È¡£[ÒÑÖªMr(CO32-)= 60.01£¬Mr(HCO3-)= 61.02 ] 4405

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 )

4406 ¹¤³§´óÆøÖÐSO2µÄ²â¶¨¿É°´ÏÂÁз½·¨½øÐУºÒÔ10 L/minµÄËÙ¶ÈÎüÈëÒ»º¬H2O2µÄÆ¿ÖС£30 minºó²úÉúµÄH2SO4 ( SO2 + H2O2 = 2H+ + SO42- )ÐèÓÃ5.62 mL 0.01000 mol/L NaOHÈÜÒºµÎ¶¨¡£ÊÔ¼ÆËã´óÆøÖÐSO2µÄÌå»ý·ÖÊý£¬Éè? (SO2)=2.86 mg/mL¡£ÒÑÖª Mr(SO2)=64.06

4407 ÔÚÒ©ÎïµÄÖÆ±¸ÖУ¬ÅðËáµÄ²â¶¨¿É°´ÏÂÁз½·¨½øÐУº½«ÊÔÑù»Ò»¯£¬³ýÈ¥ÓлúÎÔÙÈܽâÓÚH2SO4£¬Öк͸ÃÈÜÒº£¬¼ÓÈëCH3OHºÍH2SO4£¬½«Éú³ÉµÄÅðËá¼×õ¥ÕôÁó½øÒ»NaOHÈÜÒº¡£È»ºó£¬³ýȥͬʱÕô³öµÄ¼×´¼£¬ÖкÍÈÜÒº£¬¼ÓÈë¸Ê¶ÌÇ´¼£¬Éú³É¸Ê¶ÌÇ´¼£­ÅðËáÅäºÏÎʹÅðËáת±äΪǿËᣬÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨¡£ÏÖijÊÔÑù0.8755 gÐè14.75 mL 0.1075 mol/L NaOHÈÜÒºµÎ¶¨¡£ÊÔ¼ÆËãÊÔÑùÖÐH3BO3µÄÖÊÁ¿·ÖÊý¡£ÒÑÖªMr(H3BO3)=61.85

4408ÎªÖÆ±¸pH = 9.30µÄ»º³åÈÜÒº£¬ÔÚ200 mL 0.600 mol/L NH3ÈÜÒºÖÐÐè¼ÓÈë¶àÉÙ¿ËNH4Cl£¿

[ÒÑÖªpKa(NH4+)=9.26£¬Mr(NH4Cl)=53.49]

4409 ÊÔ¼ÆËã0.10 mol/LÈýÂÈÒÒËáÈÜÒºµÄpH£¬ÈôÈÜÒºÖл¹º¬ÓÐ0.10 mol/LÈýÂÈÒÒËáÄÆ£¬pHÓÖÊǶàÉÙ£¿ [ÒÑÖªKa(Cl3CCOOH)=0.60]

4410 ΪÅäÖÆ100 mL pH=5.00µÄ»º³åÈÜÒº£¬Ó¦°´Ê²Ã´Ìå»ý(mL)½«0.100 mol/LijÈõËáHAÈÜÒººÍ0.0500 mol/L NaOHÈÜÒº»ìºÏ£¿ÉèKa(HA)=5.0 ? 10-5¡£

4411 ·Ö±ð´ÓËáHA¼°Æä¹²éî¼îNaAÅäÖÆ³ÉpHΪ5.00ºÍ6.00Á½ÖÖ»º³åÈÜÒºXºÍY£¬ÉèÁ½ÖÖÈÜÒºÖÐHAµÄŨ¶È¾ùΪ0.500 mol/L¡£Èô½«ÕâÁ½ÖÖÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpHÊǶàÉÙ£¿ÉèKa(HA)=1.00 ? 10-5¡£

4412ÏÖÓÐ0.2000 gijһ½öº¬Na2CO3ºÍNaHCO3µÄÊÔÑù£¬ÓÃ0.1000 mol/L HCl±ê×¼ÈÜÒºµÎ¶¨ÖÁ¼×»ù³ÈÖÕµãʱ£¬ÐèHClÈÜÒº24.25 mL£¬ÊÔ¼ÆËãNa2CO3ºÍNaHCO3µÄÖÊÁ¿·ÖÊý¡£

[ÒÑÖªMr(Na2CO3)= 105.99£¬Mr(NaHCO3)=84.01]

4413 ÔÚ1 L 0.100 mol/L NH3ÈÜÒºÖУ¬¼ÓÈë¶àÉÙºÁÉý2.00 mol/L HClÈÜÒº·½ÄÜÊ¹ÖÆ³ÉµÄ»º³åÈÜÒºµÄpHΪ9.00£¿ ÒÑÖªpKa(NH4+)=9.26¡£

4414 ij»º³åÈÜÒºXÿÉýº¬ÓÐ1.00 mol HAcºÍ1.00 mol NaAc£¬ÎªÊ¹500 mL XÈÜÒºµÄpHÌá¸ß1µ¥Î»£¬Ó¦ÔÚÆäÖмÓÈë¶àÉÙ¿ËNaOH£¿ [ÒÑÖªKa(HAc)=1.8 ? 10-5£¬Mr(NaOH)=40.00]

4415 ½ñÓÃNaOH´¦ÀíÒ»(NH4)2SO4ÈÜÒº£¬ÊÔÎʸÃÈÜÒºµÄpHӦΪ _________________ʱ£¬²ÅÄÜʹ99.95% NH4+ת±äΪNH3£¿ ÒÑÖªpKa(NH4+)=9.26¡£(д³öËãʽ)

4416 Ïò100mL 0.100 mol/L HAc-0.100 mol/L NaAc»º³åÈÜÒºÖУ¬¼ÓÈë0.010 mol¹ÌÌåNaOH¡£ÊÔ¼ÆËãÈÜÒºµÄ£ÛH+£ÝºÍpH¸Ä±ä¶àÉÙ£¿ [ÒÑÖªpKa(HAc)=4.74]

4417 ÏÖÓÐÒ»HAc-NaAc»º³åÈÜÒº£¬Æä[ H+

]Ϊ9.0 ? 10-6

mol/L£¬½ñ½«10.0 mmol HCl¼ÓÈë1 L¸ÃÈÜÒºÖÐʱ£¬[H+]±äΪ1.0 ? 10-5mol/L£¬ÊÔ¼ÆËãÔ­»º³åÈÜÒºÖÐHAcºÍNaAcµÄŨ¶È¡£

[ÒÑÖªKa(HAc)=1.8 ? 10 ¨C5 ]

4418 ÏÖÓÉijÈõËáHX ( pKa = 5.30 )¼°Æä¹²éî¼îNaXÅäÖÆpHΪ5.00µÄ»º³åÈÜÒº¡£ÈôÔÚ100 mL»º³åÈÜÒºÖмÓÈë10 mmol HCl£¬ÊÔÏò×î³õ»º³åÈÜÒºµÄŨ¶ÈÐè¶à´óʱ²ÅÄÜÒòËáµÄ¼ÓÈë¶øÊ¹ÈÜÒºµÄpHÖ»¸Ä±ä0.20µ¥Î»£¿

4419 ΪÅäÖÆpH = 7.00µÄ»º³åÈÜÒº£¬Ó¦ÔÚ40.0 mL 0.100 mol/L H3PO4ÈÜÒºÖмÓÈë¶àÉÙºÁÉý0.500 mol/L NaOHÈÜÒº£¿(ÒÑÖªÁ×ËáµÄpKa1

=2.12£¬pKa2

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4420 Ãû´Ê½âÊÍ

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µÎ¶¨Í»Ô¾ _____________________________________________________________¡£ ·Ö²¼ÏµÊý _____________________________________________________________¡£ ±ê×¼ÈÜÒº _____________________________________________________________¡£

4420 pH = pKa(HIn) ? 1£»ÔÚ»¯Ñ§¼ÆÁ¿µãǰºó0.1%ʱ£¬ÈÜÒºµÄpH³ÊÏÖµÄͻȻ±ä»¯£»´¦ÓÚ»¯Ñ§Æ½ºâʱ£¬Ä³ÎïÖÖËùÕ¼×ÜŨ¶ÈµÄ·ÖÊý£»ÒÑ֪׼ȷŨ¶ÈµÄµÎ¶¨¼ÁÈÜÒº¡£

4421 ÓÐÁ½ÖÖ×ÜŨ¶ÈÏàµÈµÄHAc-NaAcÈÜÒº£¬ÒÑÖª¶þÕߵĻº³åÈÝÁ¿?1 > ?2£¬ÒÔϱíÊöÕýÈ·µÄÊÇ----------------------------------------------------------------------------------------------------------£¨ £© ( A ) pH1?pKa?pH2?pKa ( B ) pH1?pKa?pH2?pKa ( C )

pH1?pKa?pH2?pKa ( D ) pH1?pH2

4422 Óм×ÒÒÁ½ÖÖ×ÜŨ¶ÈÏàµÈµÄNH3-NH4Cl»º³åÈÜÒº£¬¼×£ºpH = 9.5£ºÒÒ£ºpH = 10.0£¬ÄÇô¼×µÄ»º³åÈÝÁ¿ _________ÓÚÒÒ¡£Èç¹ûÔÚÒÒÖмÓÈëNH4Cl£¬Ê¹ÆäpH½µÖÁ9.5£¬ÄÇô£¬´Ëʱ¼×µÄ»º³åÈÝÁ¿ ____________ÓÚÒÒ( NH3£ºpKb = 4.74 )¡£(´ð£º´ó»òС ) 4423 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£¿

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[ H3BO3£ºKa = 5.8 ? 10-10£»I = 0.020£¬¦Ã(H2BO3-) = 0.870 ] 4425

ÈܽâÊÊÁ¿µÄNaHSO4ºÍNa2SO4£¬µ±ÈÜÒºÖÐc(HSO4-) = c(SO42-)=0.10 mol/Lʱ£¬ÇóÈÜÒºµÄpH ¡£( H2SO4µÄKa2

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