3£®ÆÏÌѾƳ£ÓÃNa2S2O5×ö¿¹Ñõ»¯¼Á£®
£¨1£©1.90g Na2S2O5×î¶àÄÜ»¹Ô­224mLO2£¨±ê×¼×´¿ö£©£®
£¨2£©0.5mol Na2S2O5ÈܽâÓÚË®Åä³É1LÈÜÒº£¬¸ÃÈÜÒºpH=4.5£®ÈÜÒºÖв¿·Ö΢Á£Å¨¶ÈËæÈÜÒºËá¼îÐԱ仯ÈçͼËùʾ£®Ð´³öNa2S2O5ÈܽâÓÚË®µÄ»¯Ñ§·½³ÌʽNa2S2O5+H2O=2NaHSO3£»µ±ÈÜÒºpHСÓÚ1ºó£¬ÈÜÒºÖÐH2SO3µÄŨ¶È±äС£¬ÆäÔ­Òò¿ÉÄÜÊÇÑÇÁòËá²»Îȶ¨£¬Ò×·Ö½âÉú³É¶þÑõ»¯Áò£¬»òÑÇÁòËá±»Ñõ»¯£»
ÒÑÖª£ºKsp[BaSO4]=110-10£¬Ksp[BaSO3]=510-7£®°Ñ²¿·Ö±»¿ÕÆøÑõ»¯µÄ¸ÃÈÜÒºpHµ÷Ϊ10£¬ÏòÈÜÒºÖеμÓBaCl2ʹSO42-³ÁµíÍêÈ«[c£¨SO42-£©¡Ü110-5mol•L-1]£¬´ËʱÈÜÒºÖÐc£¨SO32-£©¡Ü0.05mol•L-
£¨3£©ÆÏÌѾÆÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿²â¶¨£¨ÒÑÖª£ºSO2+I2+2H2O¨TH2SO4+2HI£©£º
׼ȷÁ¿È¡100.00mLÆÏÌѾÆÑùÆ·£¬¼ÓËáÕôÁó³ö¿¹Ñõ»¯¼Á³É·Ö£®È¡Áó·ÖÓÚ׶ÐÎÆ¿ÖУ¬µÎ¼ÓÉÙÁ¿µí·ÛÈÜÒº£¬ÓÃÎïÖʵÄÁ¿Å¨¶ÈΪ0.0225mol•L-1±ê×¼I2ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼I2ÈÜÒº16.02mL£®Öظ´ÒÔÉϲÙ×÷£¬ÏûºÄ±ê×¼I2ÈÜÒº15.98mL£®¼ÆËãÆÏÌѾÆÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿ £¨µ¥Î»£ºmg•L-1£¬ÒÔSO2¼ÆË㣬Çë¸ø³ö¼ÆËã¹ý³Ì£®£©

·ÖÎö £¨1£©ÔÚNa2S2O5ÓëO2µÄ·´Ó¦ÖУ¬Áò´Ó+4¼ÛÉýΪ+6¼Û£¬Ñõ´Ó0¼Û½µÎª-2¼Û£¬¸ù¾Ýµç×ÓµÃʧÊغã½øÐмÆË㣻
£¨2£©¸ù¾Ýͼ¿ÉÖª£¬pH=4.5ʱ£¬ÈÜÒºÖÐÖ÷ÒªÒÔÑÇÁòËáÇâ¸ùÀë×ÓÐÎʽ´æÔÚ£¬¾Ý´ËÊéдˮ½â·½³Ìʽ£»µ±ÈÜÒºpHСÓÚ1ºó£¬ÈÜÒºÖÐÖ÷ÒªÒÔÑÇÁòËáÐγɴæÔÚ£¬µ«ÑÇÁòËá²»Îȶ¨£¬Ò×·Ö½âÉú³É¶þÑõ»¯Áò£¬¸ù¾ÝKsp[BaSO4]=c£¨Ba2+£©•c£¨SO42-£©£¬¿É¼ÆËã³öÐèÒª·ÅÈÈc£¨Ba2+£©£¬½ø¶ø¼ÆËã×î´óŨ¶Èc£¨SO32-£©£¬¾Ý´Ë´ðÌ⣻
£¨3£©¸ù¾Ý·´Ó¦SO2+I2+2H2O=H2SO4+2HI£¬ÓɵâµÄÎïÖʵÄÁ¿¿É¼ÆËã³ö¶þÑõ»¯ÁòµÄÎïÖʵÄÁ¿£¬¸ù¾ÝÁòÔªËØÊغãÔÙ»»Ëã³ÉNa2S2O5£¬Óɴ˼ÆËãÆÏÌѾÆÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿£®

½â´ð ½â£º£¨1£©ÔÚNa2S2O5ÓëO2µÄ·´Ó¦ÖУ¬Áò´Ó+4¼ÛÉýΪ+6¼Û£¬Ñõ´Ó0¼Û½µÎª-2¼Û£¬1.90g Na2S2O5µÄÎïÖʵÄÁ¿Îª0.01mol£¬¸ù¾Ýµç×ÓµÃʧÊغã¿ÉÖªÄÜ»¹Ô­ÑõÆøµÄÎïÖʵÄÁ¿Îª0.01mol£¬¼´±ê×¼×´¿ö224mL£¬
¹Ê´ð°¸Îª£º224£»
£¨2£©¸ù¾Ýͼ¿ÉÖª£¬pH=4.5ʱ£¬ÈÜÒºÖÐÖ÷ÒªÒÔÑÇÁòËáÇâ¸ùÀë×ÓÐÎʽ´æÔÚ£¬ËùÒÔË®½â·½³ÌʽΪNa2S2O5+H2O=2NaHSO3£¬µ±ÈÜÒºpHСÓÚ1ºó£¬ÈÜÒºÖÐÖ÷ÒªÒÔÑÇÁòËáÐγɴæÔÚ£¬µ«ÑÇÁòËá²»Îȶ¨£¬Ò×·Ö½âÉú³É¶þÑõ»¯Áò»òÑÇÁòËá±»Ñõ»¯Ò²»áµ¼ÖÂŨ¶ÈС£¬
¸ù¾ÝKsp[BaSO4]=c£¨Ba2+£©•c£¨SO42-£©£¬¿ÉÖªÐèÒªc£¨Ba2+£©=$\frac{Ksp£¨BaS{O}_{4}£©}{c£¨S{O}_{4}^{2-}£©}$=$\frac{1¡Á1{0}^{-10}}{1¡Á1{0}^{-5}}$=10-5mol•L-1£¬ÔòÈÜÒºÖÐSO32-µÄ×î´óŨ¶Èc£¨SO32-£©=$\frac{Ksp£¨BaS{O}_{3}£©}{c£¨B{a}^{2+}£©}$=$\frac{5¡Á1{0}^{-7}}{1{0}^{-5}}$=0.05mol•L-1£¬
¹Ê´ð°¸Îª£ºNa2S2O5+H2O=2NaHSO3£»ÑÇÁòËá²»Îȶ¨£¬Ò×·Ö½âÉú³É¶þÑõ»¯Áò£¬»òÑÇÁòËá±»Ñõ»¯£»0.05£»
£¨3£©¸ù¾ÝÌâÒâ¿ÉÖª£¬ÏûºÄ±ê×¼I2ÈÜÒºµÄÌå»ýΪ$\frac{15.98mL+16.02mL}{2}$=16.0mL£¬ËùÒÔI2µÄÎïÖʵÄÁ¿Îª16.0¡Á10-3L¡Á0.0225mol•L-1=3.6¡Á10-4mol£¬¸ù¾Ý·´Ó¦SO2+I2+2H2O=H2SO4+2HI£¬¿ÉÖª¶þÑõ»¯ÁòµÄÎïÖʵÄÁ¿Îª3.6¡Á10-4mol£¬SO2µÄÖÊÁ¿Îª64g/mol¡Á3.6¡Á10-4mol=23.04mg£¬ËùÒÔÆÏÌѾÆÑùÆ·Öп¹Ñõ»¯¼ÁµÄ²ÐÁôÁ¿Îª$\frac{23.04mg}{0.1L}$=230.4mg•L-1£¬´ð£º²ÐÁôÁ¿Îª230.4mg•L-1£®

µãÆÀ ±¾Ì⿼²éͼÏó·ÖÎö¡¢ÎïÖʺ¬Á¿µÄ²â¶¨¡¢Ñõ»¯»¹Ô­·´Ó¦µÎ¶¨µÈ£¬ÄѶÈÖеȣ¬Ã÷ȷʵÑéÔ­ÀíÊǽⱾÌâ¹Ø¼ü£¬¸ù¾ÝÎïÖʵÄÐÔÖÊ·ÖÎö½â´ð£¬×¢ÒâÔªËØ»¯ºÏÎï֪ʶµÄ»ýÀÛºÍÁé»îÔËÓã®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®½«I2ÈÜÓÚKIÈÜÒºÖУ¬ÄÜÅäÖƳÉŨ¶È½Ï´óµÄµâË®£¬Ö÷ÒªÊÇ·¢ÉúÁË·´Ó¦£ºI2£¨aq£©+I-£¨aq£©?I3-£¨aq£©£¬¸ÃƽºâÌåϵÖУ¬I2µÄÎïÖʵÄÁ¿Å¨¶ÈÓëζȣ¨T£©µÄ¹ØϵÈçͼËùʾ£¨ÇúÏßÉϵÄÈκÎÒ»µã¶¼´ú±íƽºâ״̬£©£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Õý·´Ó¦ÎªÎüÈÈ·´Ó¦B£®Æ½ºâ³£Êý£ºKA£¾KB
C£®·´Ó¦ËÙÂÊ£ºvB£¾vCD£®Wµãʱ£¬vÕý£¾vÄæ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

14£®ÄøÓеç¶ÆºÍ»¯Ñ§¶ÆÁ½ÖÖ·½·¨£¬»¯Ñ§¶Æ·´Ó¦Ô­Àí£ºNi2++H2PO2-+H2O¨TNi+H2PO3-+2H+£®ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®µç¶ÆÄøʱ£¬NiΪÒõ¼«£¬¶Æ¼þΪÑô¼«
B£®µç¶ÆºÍ»¯Ñ§¶ÆÔ­Àí¶¼ÊÇÀûÓÃÑõ»¯»¹Ô­·´Ó¦
C£®»¯Ñ§¶ÆÎÞÐèͨµç£¬¶Ô¶Æ¼þµÄµ¼µçÐÔÎÞÌØÊâÒªÇó
D£®»¯Ñ§¶ÆÖÐH2PO2-ÖÐP»¯ºÏ¼ÛΪ+1£¬ÓÐÇ¿»¹Ô­ÐÔ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

11£®ÏÂͼ±íʾij¸ß·Ö×Ó»¯ºÏÎïµÄ½á¹¹Æ¬¶Ï£®¹ØÓڸø߷Ö×Ó»¯ºÏÎïµÄÍƶÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®3ÖÖµ¥Ìåͨ¹ýËõ¾Û·´Ó¦¾ÛºÏB£®Ðγɸû¯ºÏÎïµÄµ¥ÌåÖ»ÓÐ2ÖÖ
C£®ÆäÖÐÒ»ÖÖµ¥ÌåΪD£®ÆäÖÐÒ»ÖÖµ¥ÌåΪ1£¬5-¶þ¼×»ù±½·Ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

18£®ÄÜÕýÈ·±íʾÏÂÁл¯Ñ§·´Ó¦µÄÀë×Ó·½³ÌʽµÄÊÇ £¨¡¡¡¡£©
A£®ÑÎËáÓëFe£¨OH£©3·´Ó¦£ºFe£¨OH£©3+3H+¨TFe3++3H2O
B£®Ï¡ÁòËáÓëÌú·Û·´Ó¦£º2Fe+6H+¨T2Fe3++3H2¡ü
C£®ÇâÑõ»¯±µÈÜÒºÓëÏ¡ÁòËá·´Ó¦£ºBa2++SO42-¨TBaSO4¡ý
D£®Ì¼Ëá¸ÆÓëÑÎËá·´Ó¦£ºCO32-+2H+¨TH2O+CO2¡ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

8£®Ä³¶þÔªËᣨH2A£©ÔÚË®ÖеĵçÀë·½³ÌʽΪH2A¨TH++HA-£¬HA-?H++A2-£¨Ka2=1.0¡Á10-2£©£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®½«Í¬Å¨¶ÈµÄNaHAºÍNa2AÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpHÒ»¶¨Ð¡ÓÚ7
B£®0.1 mol•L-1µÄNaHAÈÜÒºÖÐÀë×ÓŨ¶ÈΪ£ºc£¨Na+£©£©£¾c£¨HA-£©£¾c£¨H+£¾c£¨A2-£©£¾c£¨OH-£©
C£®ÔÚ0.1 mol•L-1µÄH2AÈÜÒºÖУ¬c£¨H+£©£¾0.12 mol•L-1
D£®ÔÚ0.1 mol•L-1µÄNa2AÈÜÒºÖУ¬c£¨A2-£©+c£¨HA-£©+c£¨Na+£©=0.3mol•L-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

15£®·´Ó¦A£¨g£©+B£¨g£©?C£¨g£©+D£¨g£©¹ý³ÌÖеÄÄÜÁ¿±ä»¯ÈçͼËùʾ£¬»Ø´ðÏÂÁÐÎÊÌâ
£¨1£©¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¨Ìî¡°ÎüÈÈ¡±¡°·ÅÈÈ¡±£©£»
£¨2£©µ±·´Ó¦´ïµ½Æ½ºâʱ£¬Éý¸ßζȣ¬AµÄת»¯ÂʼõС£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±¡°²»±ä¡±£©£¬Ô­ÒòÊǸ÷´Ó¦Õý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬Éý¸ßζÈʹƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£®
£¨3£©·´Ó¦ÌåϵÖмÓÈë´ß»¯¼Á¶Ô·´Ó¦ÈÈÊÇ·ñÓÐÓ°Ï죿·ñ£¬´ß»¯¼ÁÖ»¸Ä±ä·´Ó¦»î»¯ÄÜ£¬²»¸Ä±ä·´Ó¦ÎïºÍÉúÎï³ÉµÄÄÜÁ¿£¬
£¨4£©ÔÚ·´Ó¦ÌåϵÖмÓÈë´ß»¯¼Á£¬·´Ó¦ËÙÂÊÔö´ó£¬E1µÄ±ä»¯ÊÇ£ºE1¼õС£¬£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±¡°²»±ä¡±£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÏÂÁÐÎïÖÊת»¯ÔÚ¸ø¶¨Ìõ¼þÏÂÄÜʵÏÖµÄÊÇ£¨¡¡¡¡£©
¢ÙAl2O3$¡ú_{¡÷}^{NaOH£¨aq£©}$NaAlO2£¨aq£©$\stackrel{CO_{2}}{¡ú}$Al£¨OH£©3
¢ÚS$\stackrel{O_{2}/µãȼ}{¡ú}$SO3$\stackrel{H_{2}O}{¡ú}$H2SO4
¢Û±¥ºÍNaCl£¨aq£©$\stackrel{NH_{3}£¬CO_{2}}{¡ú}$NaHCO3$\stackrel{¡÷}{¡ú}$Na2CO3
¢ÜFe2O3$\stackrel{HCl£¨aq£©}{¡ú}$FeCl3£¨aq£©$\stackrel{¡÷}{¡ú}$ÎÞË®FeCl3£®
A£®¢Ù¢ÛB£®¢Ú¢ÛC£®¢Ú¢ÜD£®¢Ù¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®ÏÂÁйØÓÚijÈÜÒºÖÐËùº¬Àë×ӵļìÑ飬ÅжÏÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¼ÓÈëAgNO3ÈÜÒººÍÏ¡ÑÎËᣬÉú³É°×É«³Áµí£¬¿ÉÈ·¶¨ÓÐCl-´æÔÚ
B£®Óýྻ²¬Ë¿ÕºÈ¡¸ÃÈÜÒºÔÚ»ðÑæÉÏ×ÆÉÕ£¬²úÉú»ÆÉ«»ðÑ棬ÔòÔ­ÈÜÒºÖÐÒ»¶¨º¬ÓÐÄÆÑÎ
C£®¼ÓÈëÑÎËᣬÉú³ÉµÄÆøÌåÄÜʹ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬ÔòÔ­ÈÜÒºÖÐÒ»¶¨º¬ÓÐCO32-
D£®¼ÓÈëKSCNÈÜÒº£¬ÈÜÒº²»±äºì£¬ÔÙ¼ÓÂÈË®ºóÈÜÒºÏÔºìÉ«£¬ÔòÔ­ÈÜÒºÒ»¶¨º¬Fe2+

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸