ÌâÄ¿ÄÚÈÝ

(Ò»)£¨1£©¼×ÍéÒ²ÊÇÒ»ÖÖÇå½àȼÁÏ£¬µ«²»ÍêȫȼÉÕʱÈÈЧÂʽµµÍ²¢»á²úÉúÓж¾ÆøÌåÔì³ÉÎÛȾ¡£
ÒÑÖª£º CH4(g) + 2O2(g) £½ CO2(g) + 2H2O(l)   ¦¤H1£½¨D890.3 kJ/mol
2CO (g) + O2(g) £½ 2CO2(g)            ¦¤H2£½¨D566.0 kJ/mol
Ôò¼×Íé²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮʱµÄÈÈЧÂÊÖ»ÊÇÍêȫȼÉÕʱµÄ________±¶£¨¼ÆËã½á¹û±£Áô1λСÊý£©¡£
£¨2£©¼×ÍéȼÁϵç³Ø¿ÉÒÔÌáÉýÄÜÁ¿ÀûÓÃÂÊ¡£ÏÂͼÊÇÀûÓü×ÍéȼÁϵç³Øµç½â50 mL 2 mol/LµÄÂÈ»¯Í­ÈÜÒºµÄ×°ÖÃʾÒâͼ£º

Çë»Ø´ð£º
¢Ù¼×ÍéȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½ÊÇ________¡£
¢Úµ±Ïß·ÖÐÓÐ0.1 molµç×Óͨ¹ýʱ£¬________£¨Ìî¡°a¡±»ò¡°b¡±£©¼«ÔöÖØ________g¡£
£¨¶þ£©Ï±íÊǼ¸ÖÖÈõµç½âÖʵĵçÀëƽºâ³£Êý¡¢ÄÑÈܵç½âÖʵĠ  
ÈܶȻýKsp (25¡æ)¡£

µç½âÖÊ
ƽºâ·½³Ìʽ
ƽºâ³£ÊýK
Ksp
CH3COOH
CH3COOHCH3COO-£«H+
1.76¡Á10-5
 
H2CO3
H2CO3H+£«HCO3-
HCO3-H+£«CO32-
K1£½4.31¡Á10-7
K2£½5.61¡Á10-11
 
C6H5OH
C6H5OH  C6H5O-£«H+
1.1¡Á10-10
 
H3PO4
H3PO4H+£«H2PO4-
H2PO4-H+£«HPO32-
HPO42-H+£«PO43-
K1£½7.52¡Á10-3
K2£½6.23¡Á10-8
K3£½2.20¡Á10-13
 
NH3¡¤H2O
NH3¡¤H2O NH4+£«OH-
1.76¡Á10-5
 
BaSO4
BaSO4 Ba2+£«SO42-
 
1.07¡Á10-10
BaCO3
BaCO3 Ba2+£«CO32-
 
2.58¡Á10-9
 
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÉÉϱí·ÖÎö£¬Èô¢ÙCH3COOH ¢ÚHCO3-¢ÛC6H5OH ¢ÜH2PO4- ¾ù¿É¿´×÷ËᣬÔòËüÃÇËáÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòΪ__________________________(Ìî±àºÅ)£»
(2)25¡æʱ£¬½«µÈÌå»ýµÈŨ¶ÈµÄ´×ËáºÍ°±Ë®»ìºÏ£¬»ìºÏÒºÖУºc(CH3COO-)______c(NH4+)£»(Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)
£¨3£©25¡æʱ£¬Ïò10ml 0.01mol/L±½·ÓÈÜÒºÖеμÓVml 0.01mol/L°±Ë®£¬»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ(     )£»
A£®Èô»ìºÏÒºpH£¾7£¬ÔòV¡Ý10
B£®Èô»ìºÏÒºpH£¼7£¬Ôòc((NH4+) £¾c (C6H5O-) £¾c (H+)£¾c (OH£­)
C£®V=10ʱ£¬»ìºÏÒºÖÐË®µÄµçÀë³Ì¶ÈСÓÚ10ml 0.01mol/L±½·ÓÈÜÒºÖÐË®µÄµçÀë³Ì¶È
D£®V=5ʱ£¬2c(NH3¡¤H2O)+ 2 c (NH4+)=" c" (C6H5O-)+ c (C6H5OH)
£¨4£©ÈçÏÂͼËùʾ£¬ÓÐT1¡¢T2Á½ÖÖζÈÏÂÁ½ÌõBaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏߣ¬»Ø´ðÏÂÁÐÎÊÌ⣺
ÌÖÂÛT1ζÈʱBaSO4µÄ³ÁµíÈܽâƽºâÇúÏߣ¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(   )

A£®¼ÓÈëNa2SO4¿ÉʹÈÜÒºÓÉaµã±äΪbµã
B£®ÔÚT1ÇúÏßÉÏ·½ÇøÓò(²»º¬ÇúÏß)ÈÎÒâÒ»µãʱ£¬ ¾ùÓÐBaSO4³ÁµíÉú³É
C£®Õô·¢ÈܼÁ¿ÉÄÜʹÈÜÒºÓÉdµã±äΪÇúÏßÉÏa¡¢ bÖ®¼äµÄijһµã(²»º¬a¡¢b)
D£®ÉýοÉʹÈÜÒºÓÉbµã±äΪdµã

¢å £¨1£©0.7   £¨2·Ö£©  £¨2£©¢ÙCH4-8e-+2H2O=CO2+8H+ ¢Ú  b    3.2    £¨¸÷2·Ö£© 
¢æ£¨1£©¢Ù¢Ü¢Û¢Ú  £¨2·Ö£© £¨2£©  =   £¨1·Ö£©   £¨3£© D  £¨2·Ö£© £¨4£© D £¨2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º¢å£¨1£©¼×Íé²»ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ: CH4(g) + O2(g) £½ CO(g) + 2H2O(l) ¦¤H£½¨D607.3 kJ/mol,Ôò¼×Íé²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮʱµÄÈÈЧÂÊÖ»ÊÇÍêȫȼÉÕʱµÄ=0.7±¶ £¨2£©¢Ùµç½âÖÊΪËáÐԹʸº¼«·´Ó¦Ê½Îª£ºCH4-8e-+2H2O=CO2+8H+ ¢ÚCu2+ÔÚÒõ¼«·Åµç¼´b¼«£¬Ã¿Í¨¹ý0.1molµç×Ó£¬ÓÐ0.5molCu2+·Åµç£¬¼´3.2g ¢æ£¨1£© KÖµÔ½´óËáÐÔԽǿ £¨2£©25¡æʱ´×ËáºÍ°±Ë®µÄµçÀë³Ì¶ÈÏàͬ£¬¹Ê½«µÈÌå»ýµÈŨ¶ÈµÄ´×ËáºÍ°±Ë®»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬¹Ê»ìºÏÒºÖÐc(CH3COO-)=c(NH4+)£»£¨3£©ÒòΪNH3¡¤H2OµÄµçÀë³Ì¶ÈÔ¶´óÓÚC6H5OH£¬¹ÊÏò10ml 0.01mol/L±½·ÓÈÜÒºÖеμÓVml 0.01mol/L°±Ë®£¬°±Ë®µÄÌå»ý²»ÐèÒªµ½10mLÈÜÒºµÄpH>7,A ´í£»B²»×ñÑ­µçºÉÊغ㣬´í£»V=10ʱ£¬»ìºÏҺΪ±½·Óï§ÈÜÒº£¬±½·Óï§Ë®½â´Ù½øË®µÄµçÀ룬¹Ê»ìºÏÒºÖÐË®µÄµçÀë³Ì¶È´óÓÚ10ml 0.01mol/L±½·ÓÈÜÒºÖÐË®µÄµçÀë³Ì¶È£¬C´í£¨4£©Éý¸ßζȣ¬BaSO4µÄµçÀëƽºâÏòÓÒÒƶ¯£¬c£¨SO42-£©ºÍc(Ba2+)¶¼Ôö´ó£¬D´í¡£
¿¼µã£º»¯Ñ§·´Ó¦ÓëÄÜÁ¿±ä»¯Óë˵ÈÜÒºµÄ×ۺϿ¼²é

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ϱíÊǼ¸ÖÖÈõµç½âÖʵĵçÀëƽºâ³£Êý¡¢ÄÑÈܵç½âÖʵÄÈܶȻýKsp£¨25¡æ£©¡£

µç½âÖÊ
µçÀë·½³Ìʽ
µçÀë³£ÊýK
Ksp
H2CO3
H2CO3HCO3£­£«H£«
HCO3£­CO32£­£«H£«
K1£½4.31¡Á10£­7
K2£½5.61¡Á10£­11
£­
C6H5OH
C6H5OHC6H5O£­£«H£«
1.1¡Á10£­10
£­
H3PO4
H3PO4H2PO4£­£«H£«
H2PO4£­HPO42£­£«H£«
HPO42£­PO43£­£«H£«
K1£½7.52¡Á10£­3
K2£½6.23¡Á10£­6
K1£½2.20¡Á10£­13
£­
NH3¡¤H2O
NH3¡¤H2OOH£­£«NH4£«
1.76¡Á10£­5
£­
BaSO4
BaSO4£¨s£©Ba2£«£«SO42£­
£­
1.07¡Á10£­10
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öC6H5OHÓëNa3PO4·´Ó¦µÄÀë×Ó·½³Ìʽ£º_________________¡£
£¨2£©25¡æʱ£¬Ïò10 mL 0. 01 mol/LC6H5OHÈÜÒºÖеμÓV mL 0.1 mol/L°±Ë®£¬»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ__________£¨ÌîÐòºÅ£©¡£
a£®Èô»ìºÏÒºpH£¾7,ÔòV¡Ý10
b£®V£½5ʱ£¬2c£¨NH3¡¤H2O£©£«2c£¨NH4£«£©£½c£¨C6H5OH£©£«c£¨C6H5O£­£©
c£®V£½10ʱ£¬»ìºÏÒºÖÐË®µÄµçÀë³Ì¶ÈСÓÚ0.01 molC6H5OHÈÜÒºÖÐË®µÄµçÀë³Ì¶È
d£®Èô»ìºÏÒºpH£¼7£¬Ôòc£¨NH4£«£©£¾c£¨C6H5O£­£©£¾c£¨H£«£©£¾c£¨OH£­£©
£¨3£©Ë®½â·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý³ÆΪˮ½â³£Êý£¨ÓÃKb±íʾ£©£¬Àà±È»¯Ñ§Æ½ºâ³£ÊýµÄ¶¨Òå¡£25¡æʱ£¬Na2CO3µÚÒ»²½Ë®½â·´Ó¦µÄË®½â³£ÊýKb£½____mol/L¡£
£¨4£©ÈçͼËùʾ£¬ÓÐT1¡¢T2²»Í¬Î¶ÈÏÂÁ½ÌõBaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏߣ¨ÒÑÖªBaSO4µÄKspËæζÈÉý¸ß¶øÔö´ó£©¡£

¢ÙT2____ 25¡æ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©£»
¢ÚÌÖÂÛT1ζÈʱBaSO4µÄ³ÁµíÈܽâƽºâÇúÏߣ¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ____£¨ÌîÐòºÅ£©¡£
a£®¼ÓÈëNa2SO4²»ÄÜʹÈÜÒºÓÉaµã±äΪbµã
b£®ÔÚT1ÇúÏßÉÏ·½ÇøÓò£¨²»º¬ÇúÏߣ©ÈÎÒâÒ»µãʱ£¬¾ùÓÐBaSO4³ÁµíÉú³É
c£®Õô·¢ÈܼÁ¿ÉÄÜʹÈÜÒºÓÉdµã±äΪÇúÏßÉÏa¡¢bÖ®¼äµÄijһµã£¨²»º¬a¡¢b£©
d£®ÉýοÉʹÈÜÒºÓÉbµã±äΪdµã

ÄÑÈÜÐÔÔÓ±ʯ(K2SO4¡¤MgSO4¡¤2CaSO4¡¤2H2O)ÊôÓÚ¡°´ô¿ó¡±£¬ÔÚË®ÖдæÔÚÈçÏÂÈܽâƽºâ£º
K2SO4¡¤MgSO4¡¤2CaSO4¡¤2H2O(s)2Ca2£«£«2K£«£«Mg2£«£«4SO42-£«2H2O¡£ÎªÄܳä·ÖÀûÓüØ×ÊÔ´£¬Óñ¥ºÍCa(OH)2ÈÜÒºÈܽþÔÓ±ʯÖƱ¸ÁòËá¼Ø£¬¹¤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©ÂËÔüÖ÷Òª³É·ÖÓÐ________ºÍCaSO4ÒÔ¼°Î´ÈÜÔÓ±ʯ¡£
£¨2£©Óû¯Ñ§Æ½ºâÒƶ¯Ô­Àí½âÊÍCa(OH)2ÈÜÒºÄÜÈܽâÔÓ±ʯ½þ³öK£«µÄÔ­Òò£º
                                                                                     ¡£
£¨3£©¡°³ýÔÓ¡±»·½ÚÖУ¬ÏȼÓÈë            ÈÜÒº£¬¾­½Á°èµÈ²Ù×÷ºó£¬¹ýÂË£¬ÔÙ¼ÓÈë        ÈÜÒºµ÷ÂËÒºpHÖÁÖÐÐÔ¡£
£¨4£©²»Í¬Î¶ÈÏ£¬K£«µÄ½þ³öŨ¶ÈÓëÈܽþʱ¼äµÄ¹Øϵ¼ûÓÒͼ¡£ÓÉͼ¿ÉµÃ£¬Ëæ×ÅζÈÉý¸ß£¬
¢Ù                                     £¬
¢Ú                                      £¬

¢ÛÈܽþ³öµÄK£«µÄƽºâŨ¶ÈÔö´ó¡£
£¨5£©ÓÐÈËÒÔ¿ÉÈÜÐÔ̼ËáÑÎΪÈܽþ¼Á£¬ÔòÈܽþ¹ý³ÌÖлᷢÉú£ºCaSO4(s)£«CO32-CaCO3(s)£«SO42-¡£ÒÑÖª298 Kʱ£¬Ksp(CaCO3)£½2.80¡Á10£­9£¬Ksp(CaSO4)£½4.90¡Á10£­5£¬¼ÆËã´ËζÈϸ÷´Ó¦µÄƽºâ³£Êý£¬K£½                                  ¡£

Áò¼°Æ仯ºÏÎïÓй㷺µÄÓ¦Ó㬶ÔSO2ÐÔÖʵÄÑо¿ÊǸßÖл¯Ñ§½ÌѧµÄÒ»ÏîÖØÒªÄÚÈÝ¡£
I£®¶Ô±ÈÑо¿ÊÇÒ»ÖÖÖØÒªµÄÑо¿·½·¨¡£Èô½«ÁòµÄµ¥Öʼ°²¿·Ö»¯ºÏÎï°´ÈçϱíËùʾ·Ö³É3 ×飬ÔòµÚ2×éÖÐÎïÖÊMµÄ»¯Ñ§Ê½ÊÇ        ¡£

µÚ1×é
µÚ2×é
µÚ3×é
S (µ¥ÖÊ)
SO2¡¢H2SO3¡¢M¡¢NaHSO3
SO3¡¢H2SO4¡¢Na2SO4¡¢NaHSO4
¢ò£®Ä³Ð£»¯Ñ§Ñ§Ï°Ð¡×éÓÃÏÂͼËùʾµÄʵÑé×°ÖÃÑо¿SO2ÆøÌ廹ԭFe3+¡¢Br2µÄ·´Ó¦¡£

£¨1£©ÏÂÁÐʵÑé·½°¸¿ÉÒÔÓÃÓÚÔÚʵÑéÊÒÖÆÈ¡ËùÐèSO2µÄÊÇ                        ¡£
A£®Na2SO3ÈÜÒºÓëHNO3                    B£®Na2SO3¹ÌÌåŨÁòËá
C£®¹ÌÌåÁòÔÚ´¿ÑõÖÐȼÉÕ                     D£®Í­ÓëÈȵÄŨÁòËá
£¨2£©×°ÖÃCµÄ×÷ÓÃÊdzýÈ¥¶àÓàµÄSO2£¬·ÀÖ¹ÎÛȾ¿ÕÆø¡£ÒÑÖªÔÚÓÃÇâÑõ»¯ÄÆÈÜÒºÎüÊÕSO2µÄ ¹ý³ÌÖУ¬ÍùÍùµÃµ½Na2SO3ºÍNaHSO3µÄ»ìºÏÈÜÒº£¬³£ÎÂÏ£¬ÈÜÒºpHËæn(SO32¡ª)£ºn(HSO3¡ª)±ä»¯¹ØϵÈçϱí
n(SO32¡ª)£ºn(HSO3¡ª)
91£º9
1£º1
9£º91
pH
8.2
7.2
6.2
 
µ±ÎüÊÕÒºÖÐn(SO32¡ª)£ºn(HSO3¡ª) =10£º1ʱ£¬ÈÜÒºÖÐÀë×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ                     ¡£
A£®c(Na+)+ c(H+)= 2c(SO32¡ª)+ c(HSO3¡ª)+ c(OH¡ª)
B£®c(Na+)£¾c(HSO3¡ª)£¾c(SO32¡ª)£¾c(OH¡ª)£¾c(H+)
C£®c(Na+)£¾c(SO32¡ª)£¾c(HSO3¡ª)£¾c(OH¡ª)£¾c(H+)
£¨3£©ÔÚÉÏÊö×°ÖÃÖÐͨÈë¹ýÁ¿µÄSO2£¬ÎªÁËÑéÖ¤AÖÐSO2ÓëFe3+·¢ÉúÁËÑõ»¯»¹Ô­·´Ó¦£¬ËûÃÇÈ¡AÖз´Ó¦ºóµÄÈÜÒº·Ö³ÉÈý·Ý£¬²¢Éè¼ÆÁËÈçÏÂ̽¾¿ÊµÑ飬ÇëÄãÆÀ¼Û²¢²ÎÓëËûÃǵÄ̽¾¿¹ý³Ì£¨ÏÞÑ¡ÊÔ¼Á£ºKMnO4ÈÜÒº¡¢KSCNÈÜÒº¡¢BaCl2ÈÜÒº¡¢Ï¡ÁòËᡢϡÑÎËᡢϡÏõ Ëá¡¢Ba(NO3)2ÈÜÒº¡¢ÐÂÖƵÄÂÈË®£©
ÐòºÅ
ʵÑé·½°¸
ʵÑéÏÖÏó
½áÂÛ
·½°¸¢Ù
ÍùµÚÒ»·ÝÊÔÒºÖмÓÈëKMnO4ÈÜÒºÈÜÒº
×ϺìÉ«ÍÊÈ¥
SO2ÓëFe3+·´Ó¦Éú³ÉÁËFe2+
·½°¸¢Ú
ÍùµÚ¶þ·ÝÊÔÒºÖмÓÈë
 
SO2ÓëFe3+·´Ó¦Éú³ÉÁËFe2+
·½°¸¢Û
ÍùµÚ¶þ·ÝÊÔÒºÖмÓÈë
 
SO2ÓëFe3+·´Ó¦Éú³ÉÁËSO42¡ª
 
ÉÏÊö·½°¸¢ÙµÃ³öµÄ½áÂÛÊÇ·ñºÏÀí                    £¬Ô­Òò                                ¡£
Èç¹ûËûÃÇÉè¼ÆµÄ·½°¸¢ÚÓë·½°¸¢Û¾ùºÏÀí²¢Çҵõ½ÏàÓ¦½áÂÛ£¬ÇëÄ㽫ÉÏÃæ±í¸ñ²¹³äÍêÕû¡£
£¨4£©×°ÖÃBÖÐÄܱíÃ÷Br¡ªµÄ»¹Ô­ÐÔÈõÓÚSO2µÄÏÖÏóÊÇ                           ¡£

ÎÒ¹úij´óÐ͵ç½âÍ­Éú²úÆóÒµ£¬ÆäÒ±Á¶¹¤ÒÕÖÐÍ­¡¢Áò»ØÊÕÂÊ´ïµ½97£¥¡¢87£¥¡£ÏÂͼ±íʾÆäÒ±Á¶¼Ó¹¤µÄÁ÷³Ì£º

Ò±Á¶ÖеÄÖ÷Òª·´Ó¦£ºCu2S + O2 =" 2Cu" + SO2
£¨1£©ÑÌÆøÖеÄÖ÷Òª·ÏÆøÊÇ________________£¬´ÓÌá¸ß×ÊÔ´ÀûÓÃÂʺͼõÅÅ¿¼ÂÇ£¬Æä×ÛºÏÀûÓ÷½Ê½ÊÇÖÆ___________¡£
£¨2£©µç½â·¨Á¶Í­Ê±£¬Ñô¼«ÊÇ____________£¨Ìî¡°´¿Í­°å¡±»ò¡°´ÖÍ­°å¡±£©£»´ÖÍ­Öк¬ÓеĽð¡¢ÒøÒÔµ¥ÖʵÄÐÎʽÔÚµç½â²Û_______________£¨Ìî¡°Ñô¼«¡±»ò¡°Òõ¼«¡±µÄ²Ûµ×£¬Òõ¼«µÄµç¼«·´Ó¦Ê½ÊÇ_________________________________________¡£
£¨3£©ÔÚ¾«Á¶Í­µÄ¹ý³ÌÖУ¬µç½âÖÊÈÜÒºÖÐc(Fe2+)¡¢c(Zn2+)»áÖð½¥Ôö´ó¶øÓ°Ïì½øÒ»²½µç½â¡£
¼¸ÖÖÎïÖʵÄÈܶȻý³£Êý£¨KSP£©£º

ÎïÖÊ
 
Fe(OH)2
 
Fe(OH)3
 
Zn(OH)2
 
Cu(OH)2
 
KSP
 
8.0¡Á10£­16
 
4.0¡Á10£­38
 
3.0¡Á10£­17
 
2.2¡Á10£­20
 
 
µ÷½Úµç½âÒºµÄpHÊdzýÈ¥ÔÓÖÊÀë×ӵij£Ó÷½·¨¡£¸ù¾ÝÉϱíÖÐÈܶȻýÊý¾ÝÅжϣ¬º¬ÓеÈÎïÖʵÄÁ¿Å¨¶ÈµÄFe2+¡¢Zn2+¡¢Fe3+¡¢Cu2+µÄÈÜÒº£¬ËæpHÉý¸ß×îÏȳÁµíÏÂÀ´µÄÀë×ÓÊÇ______________¡£
Ò»ÖÖ·½°¸ÊÇÏȼÓÈë¹ýÁ¿µÄH2O2£¬ÔÙµ÷½ÚpHµ½4×óÓÒ¡£¼ÓÈëH2O2ºó·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________________________________________________________________¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø