ÆÕͨ»¯Ñ§Ï°ÌâÓë½â´ð£¨µÚÈýÕ£©

ceq(OH?)1.77?10?5mol?dm?3????0.009%?3c00.20mol?dm£¨3£©Í¨¹ý¼ÆËã˵Ã÷£¬Í¬Àë×ÓЧӦ¿É´ó´ó½µµÍÈõËáÔÚÈÜÒºÖеÄÀë½â¶È£¬Òò¶ø

ceq(OH?)Ͻµ¡£

12¡¢ÊÔ¼ÆËã25¡æÊ±0.10mol?dm?3H3PO4ÈÜÒºÖÐH?µÄŨ¶ÈºÍÈÜÒºµÄPH£¨Ìáʾ£º

ÔÚ0.10mol?dm?3ËáÈÜÒºÖУ¬µ±Ka£¾104ʱ£¬²»ÄÜÓ¦ÓÃÏ¡ÊͶ¨ÂɽüËÆ¼ÆËã)¡£ ½â£ºH3PO4ÊÇÖÐÇ¿ËᣬKa1?7.52?10?3?10?4£¬¹Ê²»ÄÜÓ¦ÓÃÏ¡ÊͶ¨ÂɽüËÆ¼ÆËãÆäÖÐÆ½ºâʱµÄH?Ũ¶È¡£ÓÖH3PO4ΪÈýÔªËᣬÔÚË®ÈÜÒºÖÐÖð¼¶½âÀ룬µ«

?8Ka1?6.25?10½ÏС£¬¹Êceq(H?)¿É°´Ò»¼¶½âÀëÆ½ºâ×÷½üËÆ¼ÆËã¡£

Éè0.10mol?dm?3H3PO4ÈÜÒºÖÐH?µÄƽºâŨ¶ÈΪxmol?dm?3£¬Ôò

H3PO4(aq)

Çó½âµÃ£º

x?2.4?10?2

?H?(aq)?H2PO4(aq)

ƽºâʱŨ¶È/mol?dm?3 0.10?x x x

?ceq(H?)?ceq(H2PO4)x2Ka1???7.52?10?3eqc(H3PO4)0.10?x¼´ ceq(H?)?2.4?10?2mol?dm?3

pH??lg2.4?10?2?1.6

13¡¢ÀûÓÃÊéÄ©¸½Â¼µÄÊý¾Ý£¨²»½øÐоßÌ弯Ë㣩½«ÏÂÁл¯ºÏÎïµÄ0.10mol?dm?3ÈÜ

Òº°´pHÔö´óµÄ˳ÐòÅÅÁÐÖ®¡£

£¨1£©HAc *£¨2£©NaAc £¨3£©H2SO4 £¨4£©NH3 *£¨5£©NH4Cl £¨6£©NH4Ac ½â£ºpHµÄÓÉСµ½´óµÄ˳ÐòΪ

£¨3£©H2SO4£¬£¨1£©HAc£¬£¨5£©NH4Cl£¬£¨6£©NH4Ac£¬£¨2£©NaAc£¬£¨4£©NH3

14¡¢È¡50.0cm30.100mol?dm?3ijһԪÈõËáÈÜÒº£¬Óë20.0cm30.100mol?dm?3KOHÈÜÒº»ìºÏ£¬½«»ìºÏÈÜҺϡÊÍÖÁ100cm3£¬²âµÃ´ËÈÜÒºµÄpHΪ5.25¡£Çó´ËÒ»ÔªÈõËáµÄ½âÀë³£Êý¡£

½â£º »ìºÏÈÜÒº×ÜÌå»ýΪ100cm3£¬ÔòijһԪÈõËáHAµÄŨ¶ÈΪ£º

0.100mol?dm?3?50.0cm3/100cm3?0.050mol?dm?3

KOHµÄŨ¶ÈΪ£º

0.100mol?dm?3?20.0cm3/100cm3?0.020mol?dm?3

Ò»ÔªÈõËáHAÓëKOHÖкͺó£¬HA¹ýÁ¿£¬×é³ÉHA?A?»º³åÈÜÒº£¬ÆäÖУ¬

ceq(HA)?(0.050?0.020)mol?dm?3?0.030mol?dm?3

ceq(A?)?0.020mol?dm?3

ÒÑÖªpH?5.25??lg?ceq(H?)/c??

ceq(H?)?5.62?10?6mol?dm?3

ceq(H?)?ceq(A?)?6?6Ka(HA)??5.62?10?0.020/0.030?3.7?10ceq(HA)°±µÄË®ÈÜÒº£¬Öð²½¼ÓÈë

15¡¢ÔÚÉÕ±­ÖÐÊ¢·Å20.00cm30.100mol?dm?30.100mol?dm?3HClÈÜÒº¡£ÊÔ¼ÆË㣺

£¨1£©µ±¼ÓÈë10.00cm3HClºó£¬»ìºÏÒºµÄpH£» £¨2£©µ±¼ÓÈë20.00cm3HClºó£¬»ìºÏÒºµÄpH£» £¨3£©µ±¼ÓÈë30.00cm3HClºó£¬»ìºÏÒºµÄpH£»

½â£º£¨1£©20.00cm30.100mol?dm?3NH3Óë10.00cm30.100mol?dm?3HCl»ìºÏÖкͺó£¬

?NH3¹ýÁ¿£¬·´Ó¦ÓÖÉú³ÉÁËNH4Cl£¬ËùÒÔ×é³ÉNH4?NH3»º³åÈÜÒº¡£

?ceq(NH4)0.100?10.00/30.00c(H)?Kaeq?(5.65?10?10?)mol?dm?3?5.65?10?10mol?dm?3c(NH3)0.100?10.00/30.00eq?pH??lgceq(H?)??lg5.65?10?10?9.25

£¨2£©20.00cm30.100mol?dm?3NH3Óë20.00cm30.100mol?dm?3HCl»ìºÏ£¬µÈÁ¿Öкͣ¬

?Éú³ÉNH4£¬ÆäŨ¶ÈΪ£º

?c(NH4)?0.100mol?dm?3?20.00cm3/(20.00?20.00)cm3?0.0500mol?dm?3

ceq(H?)?Ka?c?5.65?10?10?0.500mol?dm?3?5.32?10?6mol?dm?3

pH??lgceq(H?)??lg5.32?10?6?5.27

£¨3£©20.00cm30.100mol?dm?3NH3Óë30.00cm30.100mol?dm?3HCl»ìºÏÖкͺó£¬

HCl¹ýÁ¿£¬HClµÄŨ¶ÈΪ£º

(30.00?20.00)cm3?3c(HCl)?0.100mol?dm??0.0200mol?dm(30.00?20.00)cm3?3pH??lgceq(H?)??lg0.0200?1.70

16¡¢ÏÖÓÐ1.0dm3ÓÉHFºÍF?×é³ÉµÄ»º³åÈÜÒº¡£ÊÔ¼ÆË㣺

£¨1£©µ±¸Ã»º³åÈÜÒºÖк¬ÓÐ0.10molHFºÍ0.30 molNaFʱ£¬ÆäpHµÈÓÚ¶àÉÙ£¿ *£¨2£©Íù£¨1£©»º³åÈÜÒºÖмÓÈë0.40g NaOH(s)£¬²¢Ê¹ÆäÍêÈ«Èܽ⣨ûÈܽâºóÈÜÒºµÄ×ÜÌå»ýÈÔΪ1.0dm3£©¡£ÎʸÃÈÜÒºµÄpHµÈÓÚ¶àÉÙ£¿

£¨3£©µ±»º³åÈÜÒºµÄpH=3.15ʱ£¬ceq(HF)ºÍceq(F?)µÄ±ÈֵΪ¶àÉÙ£¿

½â£º £¨1£©»º³åÈÜÒºµÄpH¼ÆË㣺

ceq(H?)?Ka?ceq(HF)/ceq(F?)?(3.53?10?4?0.10/0.30)mol?dm?3?1.2?10?4mol?dm?3 pH??lgceqH(??)??4lg?1.102?3.92

£¨2£©¼ÓÈë0.40g¹ÌÌåNaOH£¬Ï൱ÓÚNaOHµÄÎïÖʵÄÁ¿£º

n(NaOH)?0.40g/40g?mol?1?0.010mol

NaOHÓëHF·´Ó¦£¬HF¹ýÁ¿£¬²¢Éú³ÉNaF£¬Á½ÕßµÄŨ¶ÈΪ£º

ceq(HF)?(0.10?0.010)mol?dm?3?0.09mol?dm?3 ceq(F?)?(0.30?0.010)mol?dm?3?0.31mol?dm?3

ceq(H?)?Ka?ceq(HF)/ceq(F?)?(3.53?10?4?0.09/0.31)mol?dm?3?1?10?4mol?dm?3pH??lgceq(H?)??lg1?10?4?4.0

£¨3£©µ±ÈÜÒºpH£½3.5ʱ£º

pH?pKa?lg?ceq(HF)/ceq(F?)?

3.15?3.45?lgceq(HF)/ceq(F?)lgceq(HF)/ceq(F?)?0.30ceq(HF)/ceq(F?)?2.017¡¢ÏÖÓÐ125 cm31.0mol?dm?3NaAcÈÜÒº£¬ÓûÅäÖÆ250cm3pHΪ5.0µÄ»º³åÈÜÒº£¬

Ðè¼ÓÈë6.0mol?dm?3HAcÈÜÒºµÄÌå»ý¶àÉÙÁ¢·½ÀåÃ×£¿

½â£º HAc?Ac?»º³åÈÜÒºÖУ¬ÒÑÖªpH£½5.0£¬pKa?4.75£¬ÉèÐè¼ÓHAcÈÜÒºµÄÌå»ýΪx¡£

5.0?4.75?lg6.0x/2501.0?125/250pH?pKa?lg?ceq(HAc)/ceq(Ac?)?

V(HAc)?x?12cm3

18¡¢ÅжÏÏÂÁз´Ó¦½øÐеķ½Ïò£¬²¢×÷¼òµ¥ËµÃ÷£¨É豸·´Ó¦ÎïÖʵÄŨ¶È¾ùΪ

1mol?dm?3£©

2?2?(1)?Cu(NH3)4??Zn2???Zn(NH3)4??Cu2?

(2)PbCO3(s)?S2??PbS(s)?CO32?2?2?

2?½â£º £¨1£©?Cu(NH3)4??Zn2???Zn(NH3)4??Cu2?ÄæÏò½øÐС£ÒòÊôͬÀàÐ͵ÄÅäÀë×Ó£¬Ki(?Cu(NH3)4?)?4.78?10?14£»Ki(?Zn(NH3)4?)?3.48?10?10£¬

2?Ki(?Cu(NH3)4?)?Ki(?Zn(NH3)4?)£¬¼´?Zn(NH3)4?¸ü²»Îȶ¨£¬Òò´Ë·´Ó¦ÄæÏò½øÐС£

2?£¨2£©PbCO3(s)?S2??PbS(s)?CO3£¬ÕýÏò½øÐС£ÒòÊôͬÀàÐ͵ÄÄÑÈܵç½âÖÊ£¬

2?2?2?Ks(PbS)?9.04?10?29£»Ks(PbCO3)?1.82?10?8£¬Ks(PbS)?Ks(PbCO3)£¬¼´PbS¸üÄÑÈÜ£¬Òò´Ë·´Ó¦ÕýÏò½øÐС£

19¡¢¸ù¾ÝPbI2µÄÈܶȻý£¬¼ÆË㣨ÔÚ25¡æÊ±£©£º

£¨1£©PbI2ÔÚË®ÖеÄÈܽâ¶È£¨mol?dm?3£©£» £¨2£©PbI2±¥ºÍÈÜÒºÖÐPb2?ºÍI?Àë×ÓµÄŨ¶È£»

£¨3£©PbI2ÔÚ0.010mol?dm?3KIµÄ±¥ºÍÈÜÒºÖÐPb2?Àë×ÓµÄŨ¶È£» £¨4£©PbI2ÔÚ0.010mol?dm?3Pb(NO3)2ÈÜÒºÖеÄÈܽâ¶È£¨mol?dm?3£©¡£ ½â£º£¨1£©ÉèPbI2ÔÚË®ÖеÄÈܽâ¶ÈΪs£¨ÒÔmol?dm?3Ϊµ¥Î»£©£¬Ôò¸ù¾Ý

PbI2(s)Pb2?(aq)?2I?(aq)

2¿ÉµÃ ceq(Pb2?)?s ceq(I?)?2s

Ks(PbI2)??ceq(Pb2?)??ceq(I?)??s?4s2?4s3

s?3Ks/4?38.49?10?9/4mol?dm?3?1.29?10?3mol?dm?3

£¨2£©ceq(Pb2?)?s?1.29?10?3mol?dm?3 ceq(I?)?2s?2.58?10?3mol?dm?3

£¨3£©ÔÚ0.010mol?dm?3KIÈÜÒºÖУ¬c(I?)?0.010mol?dm?3

Ks(PbI2)??ceq(Pb2?)??0.0102?8.49?10?9

ceq(Pb2?)?8.5?10?5mol?dm?3

£¨4£©ÔÚ0.010mol?dm?3Pb(NO3)2ÈÜÒºÖУ¬c(Pb2?)?0.010mol?dm?3

Ks(PbI2)?0.010??ceq(I?)??8.49?10?9

2ceq(I?)?9.24?10?4mol?dm?3

´ËʱPbI2µÄÈܽâ¶ÈΪ4.6?10?4mol?dm?3

20¡¢Ó¦Óñê×¼ÈÈÁ¦Ñ§Êý¾Ý¼ÆËã298.15KʱAgClµÄÈܶȻý³£Êý¡£

½â£º AgCl(s)?Ag?(aq)?Cl?(aq)

??fGm(298K)/kJ?mol?1 -109.789 77.107 -131.26

??rGm??77.107?(?131.26)?(?109.789)?kJ?mol?1?55.64kJ?mol?1

??rGm55.64?103J?mol?1lnKs??????22.45?1?1RT8.314J?mol?K?298.15K

??Ks?1.78?10?10

21¡¢½«Pb(NO3)2ÈÜÒºÓëNaClÈÜÒº»ìºÏ£¬Éè»ìºÏÒºÖÐPb(NO3)2µÄŨ¶ÈΪ

0.20mol?dm?3ÎÊ£º

£¨1£©µ±ÔÚ»ìºÏÈÜÒºÖÐCl?µÄŨ¶ÈµÈÓÚ5.0¡Á104mol?dm?3ʱ£¬ÊÇ·ñÓгÁµíÉú

ÁªÏµ¿Í·þ£º779662525#qq.com(#Ìæ»»Îª@)