ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÅðºÍÂÁλÓÚͬһÖ÷×壬ËüÃÇ¿ÉÒÔÐγÉÐí¶à×é³ÉºÍÐÔÖÊÀàËÆµÄ»¯ºÏÎï¡£Ò»ÖÖÓÃÅðþ¿ó£¨Mg2B2O5¡¤H2O£©ÖÆÈ¡µ¥ÖÊÅðµÄ¹¤ÒÕÁ÷³ÌͼÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Mg2B2O5¡¤H2OÖÐBµÄ»¯ºÏ¼ÛΪ____________¡£

£¨2£©ÈÜÒºbÖÐÈÜÖʵĻ¯Ñ§Ê½Îª______________________¡£

£¨3£©ÓÃpHÊÔÖ½²âÈÜÒºpHµÄ²Ù×÷·½·¨ÊÇ________________________¡£

£¨4£©Ð´³öNa2B4O7¡¤10H2OÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º_____________________¡£

£¨5£©ÖƵõĴÖÅðÔÚÒ»¶¨Ìõ¼þÏÂÉú³ÉBI3£¬BI3¼ÓÈÈ·Ö½â¿ÉÒԵõ½´¿¾»µÄµ¥ÖÊÅð¡£ÏÖ½«0.0200 g´ÖÅðÖÆ³ÉµÄBI3ÍêÈ«·Ö½â£¬Éú³ÉµÄI2ÓÃ0.3000 mo1¡¤L-1Na2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3ÈÜÒº15.00 mL¡££¨ÒÑÖª£ºI2+2S2O32-=2I-+S4O62-£©

¢ÙµÎ¶¨²Ù×÷ÖÐָʾ¼Áͨ³£Îª______________¡£

¢Ú¸Ã´ÖÅðÑùÆ·µÄ´¿¶ÈΪ___________________¡£

¢ÛÈôµÎ¶¨¹ÜÔÚʹÓÃǰδÓÃNa2S2O3±ê×¼ÈÜÒºÈóÏ´£¬²âµÃÑùÆ·µÄ´¿¶È½«_________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©¡£

£¨6£©Çë·ÂÕÕÉÏͼÐÎʽÉè¼Æ´Ó³ÁµíaÖлñµÃµ¥ÖÊMgµÄÁ÷³Ìͼ£¨Ìáʾ£ºÔÚ¼ýÍ·ÉÏ·½»òÏ·½±ê³öËùÓÃÊÔ¼Á»òʵÑé²Ù×÷£©¡£

¡¾´ð°¸¡¿£¨1£©+3

£¨2£©NaHCO3

£¨3£©PHÊÔÖ½·ÅÔÚ±íÃæÃóÉÏ£¬Óò£Á§°ôպȡÈÜÒº£¬µãµ½ÊÔÖ½ÖÐÑ룬Ȼºó¸ú±ê×¼±ÈÉ«¿¨¶Ô±È

£¨4£©Na2B4O7¡¤10H2O+2HCl=2NaCl+4H3BO3+5H2O

£¨5£©¢Ùµí·ÛÈÜÒº¢Ú82.5%¢ÛÆ«¸ß

£¨6£©

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©Mg2B2O5¡¤H2OÖÐMg¡¢H¡¢OµÄ»¯ºÏ¼Û·Ö±ðΪ+2¼Û¡¢+1¼ÛºÍ£­2¼Û£¬Ôò¸ù¾Ý»¯ºÏÎïÖÐÔªËØ»¯ºÏ¼ÛµÄ´úÊýºÍΪ0¿ÉÖªBµÄ»¯ºÏ¼ÛΪ£¨5¡Á2£­2¡Á2£©¡Â2£½+3¼Û£»£¨2£©NaBO2ÈÜÒºÖл¹ÓйýÁ¿µÄÇâÑõ»¯ÄÆ£¬Í¨Èë×ãÁ¿µÄCO2ºóÉú³É̼ËáÇâÄÆ£¬ÔòÈÜÒºaÖÐÈÜÖʵĻ¯Ñ§Ê½ÎªNaHCO3£»Na2B4O7¡¤10H2O¾§ÌåÓëÏ¡ÁòËá·´Ó¦Éú³ÉH3BO3¾§Ì壬Ôò¸ù¾ÝÔ­×ÓÊØºã¿ÉÖª»¹ÓÐÁòËáÄÆÉú³É£¬Òò´ËÈÜÒºbÖÐÈÜÖʵĻ¯Ñ§Ê½ÎªNa2SO4£»£¨3£©ÓÃpHÊÔÖ½²âÈÜÒºpHµÄ²Ù×÷·½·¨Êǽ«PHÊÔÖ½·ÅÔÚ±íÃæÃóÉÏ£¬Óò£Á§°ôպȡÈÜÒº£¬µãµ½ÊÔÖ½ÖÐÑ룬Ȼºó¸ú±ê×¼±ÈÉ«¿¨¶Ô±È£»£¨4£©Ð´³öMg2B2O5¡¤H2OÓëÑÎËá·´Ó¦Éú³ÉÂÈ»¯ÄÆ¡¢ÅðËᣬ»¯Ñ§·½³ÌʽΪNa2B4O7¡¤10H2O+2HCl=2NaCl+4H3BO3+5H2O£»£¨5£©¢ÙµÎ¶¨²Ù×÷ÖÐָʾ¼Áͨ³£Îªµí·ÛÈÜÒº£»¢Ú¸ù¾Ý¹ØÏµÊ½£¬ÏûºÄNa2S2O3ÈÜÒº15.00 mL 0.3000 mo1¡¤L-1Na2S2O3ÈÜÒº£¬ÐèÒªBI3µÄÎïÖʵÄÁ¿0.015L¡Á0.3000 mo1¡¤L-1¡Â3=0.0015mol£»¸Ã´ÖÅðÑùÆ·µÄ´¿¶ÈΪ0.0015mol¡Á11g/mol¡Â0.0200 g¡Á100%=82.5%£»¢ÛÈôµÎ¶¨¹ÜÔÚʹÓÃǰδÓÃNa2S2O3±ê×¼ÈÜÒºÈóÏ´£¬Na2S2O3±ê×¼ÈÜҺŨ¶ÈÆ«µÍ£¬ÏûºÄNa2S2O3±ê×¼ÈÜÒºµÄÌå»ýÆ«´ó£¬²âµÃÑùÆ·µÄ´¿¶È½«Æ«¸ß£»£¨6£©´Ó³ÁµíaÖлñµÃµ¥ÖÊMgµÄÁ÷³Ìͼ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ä³Ñо¿ÐÔѧϰС×éΪÁË̽¾¿´×ËáµÄµçÀëÇé¿ö£¬½øÐÐÁËÈçÏÂʵÑé¡£

ʵÑéÒ» ÅäÖÆ²¢±ê¶¨´×ËáÈÜÒºµÄŨ¶È

È¡±ù´×ËáÅäÖÆ250 mL 0.2 mol/L´×ËáÈÜÒº£¬ÓÃ0.2 mol/LµÄ´×ËáÈÜҺϡÊͳÉËùÐèŨ¶ÈµÄÈÜÒº£¬ÔÙÓÃNaOH±ê×¼ÈÜÒº¶ÔËùÅä´×ËáÈÜÒºµÄŨ¶È½øÐб궨¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÅäÖÆ250 mL 0.2 mol/L´×ËáÈÜҺʱÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢½ºÍ·µÎ¹ÜºÍ ¡£

£¨2£©Îª±ê¶¨Ä³´×ËáÈÜÒºµÄ׼ȷŨ¶È£¬ÓÃ0.2000 mol/L NaOHÈÜÒº¶Ô20.00 mL´×ËáÈÜÒº½øÐе樣¨·Ó̪×÷ָʾ¼Á£©£¬¼¸´ÎµÎ¶¨ÏûºÄNaOHÈÜÒºµÄÌå»ýÈçÏ£º

ʵÑéÐòºÅ

1

2

3

4

ÏûºÄNaOHÈÜÒºµÄÌå»ý£¨ml£©

20.05

20.00

18.80

19.95

Ôò¸Ã´×ËáÈÜÒºµÄ׼ȷŨ¶ÈΪ mol/L¡££¨±£ÁôСÊýµãºóËÄ룩

£¨3£©Åжϵζ¨ÖÕµãµÄ·½·¨ÊÇ ¡£

£¨4£©ÏÂÁвÙ×÷ÖпÉÄÜʹËù²â´×ËáÈÜÒºµÄŨ¶ÈÊýֵƫµÍµÄÊÇ £¨Ìî×ÖĸÐòºÅ£©¡£

A£®¼îʽµÎ¶¨¹ÜδÓñê×¼ÒºÈóÏ´¾ÍÖ±½Ó×¢Èë±ê×¼Òº

B£®µÎ¶¨Ç°Ê¢·Å´×ËáÈÜÒºµÄ×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºóûÓиÉÔï

C£®¼îʽµÎ¶¨¹ÜÔڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ

D£®Á¿È¡´×ËáÌå»ýʱ£¬¿ªÊ¼¸©ÊÓ¶ÁÊý£¬µÎ¶¨½áÊøÊ±ÑöÊÓ¶ÁÊý

ʵÑé¶þ ̽¾¿Å¨¶È¶Ô´×ËáµçÀë³Ì¶ÈµÄÓ°Ïì

ÓÃpH¼Æ²â¶¨25¡æÊ±²»Í¬Å¨¶ÈµÄ´×ËáµÄpH£¬½á¹ûÈçÏ£º

´×ËáŨ¶È£¨mol¡¤L£­1£©

0.0010

0.0100

0.0200

0.1000

pH

3.88

3.38

3.23

2.88

£¨5£©¸ù¾Ý±íÖÐÊý¾Ý£¬¿ÉÒԵóö´×ËáÊÇÈõµç½âÖʵĽáÂÛ£¬ÄãÈÏΪµÃ³ö´Ë½áÂÛµÄÒÀ¾ÝÊÇ ¡£

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

¾«Ó¢¼Ò½ÌÍø