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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] 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 ºÍ 1.00 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

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