ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÏÖÓг£ÎÂÏÂpH£½2µÄHClÈÜÒº¼×ºÍpH£½2µÄCH3COOHÈÜÒºÒÒ£¬Çë¸ù¾ÝÏÂÁвÙ×÷»Ø´ðÎÊÌ⣺

£¨1£©³£ÎÂÏÂ0.1mol¡¤L£­1µÄCH3COOHÈÜÒº¼ÓˮϡÊ͹ý³ÌÖУ¬ÏÂÁбí´ïʽµÄÊý¾ÝÒ»¶¨±äСµÄÊÇ_____¡£

A£®c(H£«) B£® c(H+)/c(CH3COOH) C£®c(H£«)¡¤c(OH£­)

£¨2£©È¡10mLµÄÒÒÈÜÒº£¬¼ÓÈëµÈÌå»ýµÄË®£¬CH3COOHµÄµçÀëƽºâ________(Ìî¡°Ïò×󡱡°ÏòÓÒ¡±»ò¡°²»¡±)Òƶ¯£»ÁíÈ¡10mLµÄÒÒÈÜÒº£¬¼ÓÈëÉÙÁ¿ÎÞË®´×ËáÄƹÌÌå(¼ÙÉè¼ÓÈë¹ÌÌåÇ°ºóÈÜÒºÌå»ý±£³Ö²»±ä)£¬´ý¹ÌÌåÈܽâºó£¬ÈÜÒºÖÐµÄ c(H+)/c(CH3COOH) ±ÈÖµ½«________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°ÎÞ·¨È·¶¨¡±)¡£

£¨3£©È¡µÈÌå»ýµÄ¼×¡¢ÒÒÁ½ÈÜÒº£¬·Ö±ðÓõÈŨ¶ÈµÄNaOHÏ¡ÈÜÒºÖкͣ¬ÔòÏûºÄNaOHÈÜÒºÌå»ýµÄ´óС¹ØϵΪV(¼×)______(Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±)V(ÒÒ)¡£

£¨4£©ÒÑÖª25¡æʱ£¬ÏÂÁÐËáµÄµçÀëƽºâ³£ÊýÈçÏ£º

»¯Ñ§Ê½

CH3COOH

H2CO3

HClO

H2SO3

µçÀëƽºâ³£Êý

1.8¡Á10£­5

K1£½4.3¡Á10£­7

K2£½4.7¡Á10£­11

3.0¡Á10£­8

K1£½1.54¡Á10£­2

K2£½1.02¡Á10£­7

¢ÙÏÂÁÐ΢Á£¿ÉÒÔ´óÁ¿¹²´æµÄÊÇ______Ìî×Öĸ¡£

a.CO32-¡¢HSO3- b.HCO3-¡¢HSO3-

c.SO32-¡¢HCO3- d.CO32- ¡¢H2CO3

¢Úд³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º

H2SO3£«Na2CO3(ÉÙÁ¿)£º_______________

ÊÒÎÂÏ£¬0.1 mol¡¤L£­lµÄKOHÈÜÒºµÎ10.00mL 0.10 mol¡¤L£­l H2C2O4 (¶þÔªÈõËᣩÈÜÒº£¬ËùµÃµÎ¶¨ÇúÏßÈçͼ(»ìºÏÈÜÒºµÄÌå»ý¿É¿´³É»ìºÏÇ°ÈÜÒºµÄÌå»ýÖ®ºÍ)¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨5£©µã¢ÜËùʾÈÜÒºÖУºc(K+)+c(H2C2O4)+c(HC2O4)+c(C2O42)£½_____mol/L¡£µã¢ÝËùʾµÄÈÜÒºÖи÷Àë×ÓŨ¶ÈµÄ´óС˳Ðò_____£®

£¨6£©²ÝËᾧÌå(H2C2O4¡¤2H2O)ΪÎÞÉ«£¬Ä³Í¬Ñ§Éè¼ÆʵÑé²â¶¨Æä´¿¶È¡£ÊµÑé¹ý³ÌÈçÏ£º³ÆÈ¡mg²ÝËᾧÌåÓÚ׶ÐÎÆ¿ÖÐ,¼ÓË®ÍêÈ«ÈܽâÓÃcmol¡¤L-1ËáÐÔKMnO4±ê×¼ÈÜÒº½øÐе樣¬Ôò´ïµ½µÎ¶¨ÖÕµãʱµÄÏÖÏóÊÇ_______£»¸Ã¹ý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ________£»µÎ¶¨¹ý³ÌÖÐÏûºÄVmLKMnO4±ê×¼ÈÜÒº£¬²ÝËᾧÌå´¿¶ÈΪ________¡£

¡¾´ð°¸¡¿A ÏòÓÒ ¼õС £¼ bc H2SO3£«CO32-=== HSO3-£«HCO3- 0.10 c£¨K+£©£¾c£¨C2O42-£©£¾c£¨OH-£©£¾c£¨HC2O4-£©£¾c£¨H+£© ¼ÓÈë×îºóÒ»µÎ¸ßÃÌËá¼ØÈÜÒº£¬×¶ÐÎÆ¿ÄÚÈÜÒºÓÉÎÞÉ«±äΪºìÉ«£¬ÇÒ°ë·ÖÖÓ²»ÍÊÉ« 5H2C2O4+2MnO4-+6H+=2Mn2++10CO2¡ü+8H2O 0.315cv/m

¡¾½âÎö¡¿

(1)CH3COOHÈÜÒº¼ÓˮϡÊ͹ý³Ì£¬´Ù½øµçÀ룬n(H+)Ôö´ó£¬ËáÐÔ¼õÈõc(H+)¼õС£¬c(OH-)Ôö´ó£¬Kw²»±ä£»

(2)´×ËáÊÇÈõµç½âÖÊ£¬¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룻Ïò´×ËáÖмÓÈë´×ËáÄƹÌÌ壬ÈÜÒºÖд×Ëá¸ùÀë×ÓŨ¶ÈÔö´ó£¬ÒÖÖÆ´×ËáµçÀ룻

(3)pHÏàµÈµÄ´×ËáºÍÑÎËᣬ´×ËáµÄŨ¶È´óÓÚÑÎËᣬµÈÌå»ýµÈpHµÄÁ½ÖÖËᣬ´×ËáµÄÎïÖʵÄÁ¿´óÓÚÑÎËᣬËáµÄÎïÖʵÄÁ¿Ô½¶àÐèÒªµÈŨ¶ÈµÄÇâÑõ»¯ÄÆÈÜÒºÌå»ýÔ½´ó£»

(4)¢Ù¸ù¾ÝKaÔ½´óËáÐÔԽǿ£¬¸ù¾ÝËáÐÔ½ÏÇ¿µÄÄÜÓëËáÐÔ½ÏÈõµÄËá¸ùÀë×Ó·´Ó¦£»

¢ÚH2SO3ºÍNa2CO3·´Ó¦Éú³ÉNaHSO3ºÍNaHCO3£»

(5)¸ù¾Ýµã¢ÜËùʾµÄÈÜÒºµÄÌå»ý25mL¼ÆËã³öÈÜÒºÖи÷×é·ÖµÄŨ¶È£»¸ù¾Ýµã¢ÝËùʾµÄÈÜÒºÖУ¬ÈÜÖÊÖ»ÓÐK2C2O4·ÖÎö¸÷Àë×ÓŨ¶ÈµÄ´óС£»

(6)ÓÃcmolL-1ËáÐÔKMnO4±ê×¼ÈÜÒº½øÐе樣¬ÀûÓøßÃÌËá¼ØÈÜÒºµÄÑÕɫָʾ·´Ó¦Öյ㣬¸ßÃÌËá¼ØÈÜÒºÑõ»¯²ÝËáÉú³É¶þÑõ»¯Ì¼£¬½áºÏ·´Ó¦µÄ¶¨Á¿¹Øϵ¼ÆËã²ÝËᾧÌå´¿¶È¡£

(1)A£®CH3COOHÈÜÒº¼ÓˮϡÊ͹ý³Ì£¬´Ù½øµçÀ룬c(H+)¼õС£¬¹ÊAÕýÈ·£»

B£®=¡Á=£¬K²»±ä£¬´×Ëá¸ùÀë×ÓŨ¶È¼õС£¬ÔòÏ¡Ê͹ý³ÌÖбÈÖµ±ä´ó£¬¹ÊB´íÎó£»

C£®Ï¡Ê͹ý³Ì£¬´Ù½øµçÀ룬c(H+)¼õС£¬c(OH-)Ôö´ó£¬c(H+)c(OH-)=Kw£¬Kw²»±ä£¬¹ÊC´íÎó£»

¹Ê´ð°¸ÎªA£»

(2)´×ËáÊÇÈõµç½âÖÊ£¬¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룬ËùÒÔ´×ËáµçÀëƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£»Ïò´×ËáÖмÓÈë´×ËáÄƹÌÌ壬ÈÜÒºÖд×Ëá¸ùÀë×ÓŨ¶ÈÔö´ó£¬ÒÖÖÆ´×ËáµçÀ룬ÔòÇâÀë×ÓŨ¶È¼õС£¬´×Ëá·Ö×ÓŨ¶ÈÔö´ó£¬ËùÒÔ¼õС£»

(3)pHÏàµÈµÄ´×ËáºÍÑÎËᣬ´×ËáµÄŨ¶È´óÓÚÑÎËᣬµÈÌå»ýµÈpHµÄÁ½ÖÖËᣬ´×ËáµÄÎïÖʵÄÁ¿´óÓÚÑÎËᣬËáµÄÎïÖʵÄÁ¿Ô½¶àÐèÒªµÈŨ¶ÈµÄÇâÑõ»¯ÄÆÈÜÒºÌå»ýÔ½´ó£¬ËùÒÔÏûºÄµÄNaOHÈÜÒºµÄÌå»ý´óС¹ØϵΪ£ºV(¼×)£¼V(ÒÒ)£»

(4)¢ÙÒÑÖªKaÔ½´óËáÐÔԽǿ£¬ËáÐÔÇ¿Èõ˳ÐòΪH2SO3£¾CH3COOH£¾H2CO3£¾HSO3-£¾HClO£¾HCO3-£¬ÇÒËáÐÔ½ÏÇ¿µÄÄÜÓëËáÐÔ½ÏÈõµÄËá¸ùÀë×Ó·´Ó¦£¬ÓÉÓÚHCO3-µÄËáÐÔСÓÚHSO3-µÄËáÐÔ£¬ËùÒÔHCO3-ÓëHSO3-¡¢SO32-²»·´Ó¦£¬¼´bcÄܹ²´æ£¬¹Ê´ð°¸Îªbc£»

¢ÚH2SO3ºÍNa2CO3·´Ó¦Éú³ÉNaHSO3ºÍNaHCO3£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪH2SO3£«CO32-=== HSO3-£«HCO3-£»

(5)µã¢ÜËùʾµÄÈÜÒºµÄÌå»ý25mL£¬¸ù¾ÝÎïÁÏÊغ㣺c(HC2O4-)+c(H2C2O4)+c(C2O42-)=0.10mol/L¡Á=0.04mol/L£»c(K+)=0.10mol/L¡Á=0.06mol/L£¬ËùÒÔc(HC2O4-)+c(H2C2O4)+c(C2O42-)+c(K+)=0.10molL-1£»µã¢ÝËùʾµÄÈÜÒºÖУ¬ÈÜÖÊÖ»ÓÐK2C2O4£¬Ë®½âºóÈÜÒºÏÔʾ¼îÐÔ£¬Àë×ÓŨ¶È´óС¹ØϵΪ£ºc(K+)£¾c(C2O42-)£¾c(OH-)£¾c(HC2O4-)£¾c(H+)£»

(6)³ÆÈ¡m g²ÝËᾧÌåÓÚÊÔ¹ÜÖУ¬¼ÓË®ÍêÈ«ÈܽâÓÃcmolL-1ËáÐÔKMnO4±ê×¼ÈÜÒº½øÐе樣¬Ôò´ïµ½µÎ¶¨ÖÕµãʱµÄÏÖÏóÊÇ£ºµ±µÎÈë×îºóÒ»µÎ±ê׼Һʱ£¬ÈÜÒºÓÉÎÞÉ«±ä³É×ϺìÉ«£¬ÇÒ°ë·ÖÖÓÄÚÈÜÒºÑÕÉ«²»Ôٸı䣬¸Ã¹ý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º2MnO4-+6H++5H2C2O4=2Mn2++10CO2¡ü+8H2O£¬µÎ¶¨¹ý³ÌÖÐÏûºÄV mL KMnO4±ê×¼ÈÜÒº£¬½áºÏÀë×Ó·½³Ìʽ¶¨Á¿¹Øϵ¼ÆË㣻

2MnO4-+6H++5H2C2O4=2Mn2++10CO2¡ü+8H2O
2 5

cmolL-1¡ÁV¡Á10-3L n

n=2.5cV¡Á10-3mol£¬²ÝËᾧÌåH2C2O42H2OµÄ´¿¶È=¡Á100%=%¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿ÑÇÂÈËáÄÆ£¨NaClO2£©ÊÇÒ»ÖÖÖØÒªµÄº¬ÂÈÏû¶¾¼Á£¬ÔÚË®ÖÐÈܽâ¶È½Ï´ó£¬ÓöËá·Å³öClO2£¬ÊÇÒ»ÖÖ¸ßЧµÄÑõ»¯¼ÁºÍÓÅÖÊƯ°×¼Á£¬¿ÉÓÃÓÚ¸÷ÖÖÏËάºÍijЩʳƷµÄƯ°×¡£¹ýÑõ»¯Çâ·¨ÖƱ¸NaClO2¹ÌÌåµÄʵÑé×°ÖÃÈçͼËùʾ£º

ÒÑÖª£º

¢ÙClO2µÄÈÛµãΪ-59¡æ¡¢·ÐµãΪ11¡æ£¬¼«Ò×ÈÜÓÚË®£¬ÓöÈÈË®¡¢¼û¹âÒ׷ֽ⣻ÆøÌåŨ¶È½Ï´óʱÒ×·¢Éú·Ö½â£¬ÈôÓÿÕÆø¡¢CO2¡¢µªÆøµÈÆøÌåÏ¡ÊÍʱ£¬±¬Õ¨ÐÔÔò½µµÍ¡£

¢Ú2ClO2+H2O2+2NaOH=2NaClO2+O2¡ü+2H2O

Çë»Ø´ð£º

(1)°´ÉÏͼ×é×°ºÃÒÇÆ÷ºó£¬Ê×ÏÈÓ¦¸Ã½øÐеIJÙ×÷ÊÇ____£»×°ÖÃBµÄ×÷ÓÃÊÇ___£»±ùˮԡÀäÈ´µÄÖ÷ҪĿµÄ²»°üÀ¨_£¨Ìî×Öĸ£©¡£

a£®¼õÉÙH2O2µÄ·Ö½â b£®½µµÍClO2µÄÈܽâ¶È c£®¼õÉÙClO2µÄ·Ö½â

(2)ClO2ÊǺϳÉNaClO2µÄÖØÒªÔ­ÁÏ£¬Ð´³öÈý¾±ÉÕÆ¿ÖÐÉú³ÉClO2µÄ»¯Ñ§·½³Ìʽ£º ____¡£

(3)×°ÖÃCÖмÓÈëNaOHÈÜÒºµÄÄ¿µÄ³ýÁË×÷·´Ó¦ÎïÍ⣬»¹ÒòΪ_____¡£¿ÕÆøµÄÁ÷ËÙ¹ýÂý»ò¹ý¿ì¶¼»áÓ°ÏìNaClO2µÄ²úÂÊ£¬ÊÔ·ÖÎöÔ­Òò£º________¡£

(4)¸ÃÌ××°ÖôæÔÚµÄÃ÷ÏÔȱÏÝÊÇ_________¡£

(5)Ϊ·ÀÖ¹Éú³ÉµÄNaClO2¹ÌÌå±»¼ÌÐø»¹Ô­ÎªNaCl£¬ËùÓû¹Ô­¼ÁµÄ»¹Ô­ÐÔÓ¦ÊÊÖС£³ýH2O2Í⣬»¹¿ÉÒÔÑ¡ÔñµÄ»¹Ô­¼ÁÊÇ_£¨Ìî×Öĸ£©

A£®¹ýÑõ»¯ÄÆ B£®Áò»¯ÄÆ C£®ÂÈ»¯ÑÇÌú D£®¸ßÃÌËá¼Ø

(6)Èômg NaClO3(s)×îÖÕÖƵô¿¾»µÄn g NaClO2(s)£¬ÔòNaClO2µÄ²úÂÊÊÇ_¡Á100%¡£

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

¾«Ó¢¼Ò½ÌÍø