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¸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(NH3CH2COOHCl) ________________

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»´ó

ÓÚ0.15 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?

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-

Kb?c, ¹Ê 10?5.0010?14.0002500.V ??.?V10?5.006000 ½âµÃV = 40.00mL w(HA) =

02500.?4000.?8200.?100% = 51.25%

1600.?1000

2363

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2365

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Ëá¼îָʾ¼ÁµÄ½âÀëÆ½ºâ¿É±íʾΪ: HIn = H+ + In-

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