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V (NaOH)/mL 0.00 20.47 40.94(»¯Ñ§¼ÆÁ¿µã) pH 2.65 4.21 8.43 ÊÔ¼ÆËã¸ÃÒ»ÔªÈõËáHBµÄĦ¶ûÖÊÁ¿ºÍpKaÖµ¡£
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pKa(HAc) = 4.74, H2SO3µÄpKa1 = 1.90, pKa2 = 7.20 0654
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±û¶þËá HOOC-CH2-COOH(¼òдΪH2M)ÓÐÏÂÁÐÁ½¸öƽºâ H2M HM-
HM-+ H+
pKa1 = 2.86
M2- + H+ pKa2 = 5.70
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ÒªÊÔ¼ÁºÍ·ÖÎö½á¹ûµÄ¼ÆË㹫ʽ) 0659
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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-] =
Ks = 10-3.1
?301.?10 »¯Ñ§¼ÆÁ¿µãºó0.1%: [SH2+] = = 10-4.3
2 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) ÀƽÐÔÈܼÁ (B) ·Ö±æÐÔÈܼÁ (C) Á½ÐÔÈܼÁ (D) ¶èÐÔÈܼÁ 2306
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(D) ´×Ëá¸ùÀë×ÓÊÇ´×ËáµÄ¹²éîËá 2307
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(D) ÿÉýº¬1.0¡Á10-7molÇ¿Ëá(ÀýÈçHCl),Ôò¸ÃÈÜÒºµÄpHΪ7.0 2308
ÏÂÁÐÎïÖÊÖÐÊôÓÚÁ½ÐÔÎïÖʵÄÊÇ-----------------------------------------------------------------( ) (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 2319
ijÈýÔªËáH3AµÄpKa1 = 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 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) 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+] = 2326
Kac D. [OH-] =
Kb c