ÌâÄ¿ÄÚÈÝ

ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÕýÈ·µÄÊÇ________(ÌîÐòºÅ)

¢ÙÓüîʽµÎ¶¨¹ÜÁ¿È¡20.00 mL¡¡0.10 mol/L¡¡KMnO4ÈÜÒº£»

¢ÚÓÃÍÐÅÌÌìƽ³ÆÈ¡10.50 g¸ÉÔïµÄNACl¹ÌÌ壻

¢Û¸÷·ÅÒ»ÕÅÖÊÁ¿ÏàͬµÄÂËÖ½ÓÚÌìƽµÄÁ½ÍÐÅÌÉÏ£¬½«NAOH¹ÌÌå·ÅÔÚ×óÅÌÖ½ÉϳÆÁ¿£»

¢ÜÅäÖÆAl2(SO4)3ÈÜҺʱ£¬¼ÓÈëÉÙÁ¿µÄÏ¡ÁòË᣻

¢Ý½«Î´¾­ÊªÈóµÄpHÊÔÖ½½þµ½Ä³ÈÜÒºÖУ¬¹ýÒ»»áÈ¡³öÓë±ê×¼±ÈÉ«¿¨±È½Ï²âµÃ¸ÃÈÜÒºµÄpH£»

¢ÞÖк͵ζ¨ÖУ¬×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºó¿ÉÖ±½ÓÍùÆäÖÐ×¢ÈëÒ»¶¨Á¿µÄ´ý²âÈÜÒº£»

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¢ÅÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÕýÈ·µÄÊÇ        £¨ÌîÐòºÅ£©¡£

¢ÙÏòÊÔ¹ÜÖеμÓÒºÌåʱ£¬Îª²»Ê¹ÒºÌåµÎµ½ÊÔ¹ÜÍâÓ¦½«½ºÍ·µÎ¹ÜÉìÈëÊÔ¹ÜÖУ»

¢ÚһС¿é½ðÊôÄƼÓÈëË®ÖкóѸËÙÈÛ³ÉСÇò£¬²»Í£µØÔÚË®ÃæÓζ¯²¢·¢³ö¡°Ö¨Ö¨¡±µÄÏìÉù£»

¢ÛÅäÖÃ100mL1.00mol/LµÄNaClÈÜҺʱ£¬¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡5.85g NaCl¹ÌÌ壻

¢ÜÏò¿ÉÄܺ¬ÓÐSO42£­¡¢SO32£­µÄÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëBa(NO3)2ÈÜÒº£¬¿É¼ìÑéSO42£­µÄ´æÔÚ£»

¢ÝÕô·¢NaClÈÜÒºÒԵõ½NaCl¹ÌÌåʱ£¬²»±Ø½«ÈÜÒºÕô¸É£»

¢ÞÏò100¡æʱµÄNaOHÏ¡ÈÜÒºÖеμӱ¥ºÍµÄFeCl3ÈÜÒº£¬ÒÔÖƱ¸Fe(OH)3½ºÌ壻

¢ßÈçͼ£¬¿É¹Û²ìµ½ÁéÃô¼ìÁ÷¼ÆµÄÖ¸Õëƫת£»

¢àÏòAlCl3ÈÜÒºÖеμÓNaOHÈÜÒººÍÏòNaOHÈÜÒºÖеμÓAlCl3ÈÜ  ÒºµÄÏÖÏóÏàͬ¡£

¢ÆÏÖÓÐ0.1mol¡¤L£­1µÄ´¿¼îÈÜÒº£¬ÓÃpHÊÔÖ½²â¶¨¸ÃÈÜÒºµÄpH£¬Æä

ÕýÈ·µÄ²Ù×÷ÊÇ                                                                           

                                                                                       

ÄãÈÏΪ¸ÃÈÜÒºpHµÄ·¶Î§Ò»¶¨½éÓÚ             Ö®¼ä¡£ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸Ö¤Ã÷´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32£­ÒýÆðµÄ£º                            ¡£

 

¢ÅÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÕýÈ·µÄÊÇ        £¨ÌîÐòºÅ£©¡£
¢ÙÏòÊÔ¹ÜÖеμÓÒºÌåʱ£¬Îª²»Ê¹ÒºÌåµÎµ½ÊÔ¹ÜÍâÓ¦½«½ºÍ·µÎ¹ÜÉìÈëÊÔ¹ÜÖУ»
¢ÚһС¿é½ðÊôÄƼÓÈëË®ÖкóѸËÙÈÛ³ÉСÇò£¬²»Í£µØÔÚË®ÃæÓζ¯²¢·¢³ö¡°Ö¨Ö¨¡±µÄÏìÉù£»
¢ÛÅäÖÃ100mL1.00mol/LµÄNaClÈÜҺʱ£¬¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡5.85g NaCl¹ÌÌ壻
¢ÜÏò¿ÉÄܺ¬ÓÐSO42£­¡¢SO32£­µÄÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëBa(NO3)2ÈÜÒº£¬¿É¼ìÑéSO42£­µÄ´æÔÚ£»
¢ÝÕô·¢NaClÈÜÒºÒԵõ½NaCl¹ÌÌåʱ£¬²»±Ø½«ÈÜÒºÕô¸É£»
¢ÞÏò100¡æʱµÄNaOHÏ¡ÈÜÒºÖеμӱ¥ºÍµÄFeCl3ÈÜÒº£¬ÒÔÖƱ¸Fe(OH)3½ºÌ壻
¢ßÈçͼ£¬¿É¹Û²ìµ½ÁéÃô¼ìÁ÷¼ÆµÄÖ¸Õëƫת£»

¢àÏòAlCl3ÈÜÒºÖеμÓNaOHÈÜÒººÍÏòNaOHÈÜÒºÖеμÓAlCl3ÈÜ ÒºµÄÏÖÏóÏàͬ¡£
¢ÆÏÖÓÐ0.1mol¡¤L£­1µÄ´¿¼îÈÜÒº£¬ÓÃpHÊÔÖ½²â¶¨¸ÃÈÜÒºµÄpH£¬Æä
ÕýÈ·µÄ²Ù×÷ÊÇ                                                                           
                                                                                       
ÄãÈÏΪ¸ÃÈÜÒºpHµÄ·¶Î§Ò»¶¨½éÓÚ             Ö®¼ä¡£ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸Ö¤Ã÷´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32£­ÒýÆðµÄ£º                            ¡£

¢ÅÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÕýÈ·µÄÊÇ         £¨ÌîÐòºÅ£©¡£

¢ÙÏòÊÔ¹ÜÖеμÓÒºÌåʱ£¬Îª²»Ê¹ÒºÌåµÎµ½ÊÔ¹ÜÍâÓ¦½«½ºÍ·µÎ¹ÜÉìÈëÊÔ¹ÜÖУ»

¢ÚһС¿é½ðÊôÄƼÓÈëË®ÖкóѸËÙÈÛ³ÉСÇò£¬²»Í£µØÔÚË®ÃæÓζ¯²¢·¢³ö¡°Ö¨Ö¨¡±µÄÏìÉù£»

¢ÛÅäÖÃ100mL1.00mol/LµÄNaClÈÜҺʱ£¬¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡5.85g NaCl¹ÌÌ壻

¢ÜÏò¿ÉÄܺ¬ÓÐSO42£­¡¢SO32£­µÄÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëBa(NO3)2ÈÜÒº£¬¿É¼ìÑéSO42£­µÄ´æÔÚ£»

¢ÝÕô·¢NaClÈÜÒºÒԵõ½NaCl¹ÌÌåʱ£¬²»±Ø½«ÈÜÒºÕô¸É£»

¢ÞÏò100¡æʱµÄNaOHÏ¡ÈÜÒºÖеμӱ¥ºÍµÄFeCl3ÈÜÒº£¬ÒÔÖƱ¸Fe(OH)3½ºÌ壻

¢ßÈçͼ£¬¿É¹Û²ìµ½ÁéÃô¼ìÁ÷¼ÆµÄÖ¸Õëƫת£»

¢àÏòAlCl3ÈÜÒºÖеμÓNaOHÈÜÒººÍÏòNaOHÈÜÒºÖеμÓAlCl3ÈÜ  ÒºµÄÏÖÏóÏàͬ¡£

¢ÆÏÖÓÐ0.1mol¡¤L£­1µÄ´¿¼îÈÜÒº£¬ÓÃpHÊÔÖ½²â¶¨¸ÃÈÜÒºµÄpH£¬Æä

ÕýÈ·µÄ²Ù×÷ÊÇ                                                                           

                                                                                       

ÄãÈÏΪ¸ÃÈÜÒºpHµÄ·¶Î§Ò»¶¨½éÓÚ              Ö®¼ä¡£ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸Ö¤Ã÷´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32£­ÒýÆðµÄ£º                             ¡£

 

(1)ÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÕýÈ·µÄÊÇ________(ÌîÐòºÅ)¡£

¢ÙÏòÊÔ¹ÜÖеμÓÒºÌåʱ£¬Îª²»Ê¹ÒºÌåµÎµ½ÊÔ¹ÜÍâÓ¦½«½ºÍ·µÎ¹ÜÉìÈëÊÔ¹ÜÖУ»

¢ÚÅäÖÃ100 mL 1.00mol/LµÄNaClÈÜҺʱ£¬¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡5.85 gNaCl¹ÌÌ壻

¢ÛÏò100¡æʱµÄNaOHÏ¡ÈÜÒºÖеμӱ¥ºÍµÄFeCl3ÈÜÒº£¬ÒÔÖƱ¸Fe(OH)3½ºÌ壻

¢ÜÅäÖÆÂÈ»¯ÌúÈÜҺʱ£¬ÏȽ«ÂÈ»¯Ìú¹ÌÌå¼ÓÈëÑÎËáÖÐÈܽ⣬Ȼºó¼ÓˮϡÊÍ£»

¢Ý·ÖҺ©¶··ÖҺʱ£¬ÏȽ«Ï²ãµÄÒºÌå´ÓÏ¿ÚÁ÷³ö£¬È»ºóÔÙ´ÓÏ¿ÚÁ÷³öÉϲãµÄÒºÌå

¢Þ²â¶¨Ä³ÈÜÒºpHʱ£¬È¡Ò»Ð¡¿é¸ÉÔïµÄÊÔÖ½·ÅÈë²£Á§Æ¬ÉÏ£¬Óò£Á§°ôպȡÈÜÒºµÎ

ÔÚÊÔÖ½µÄÖв¿£¬ÔÙÓëpHÊÔÖ½±ÈÉ«¿¨¶Ô±È(2)ÔÚÏÂͼËùʾµÄʵÑé×°ÖÃÖУ¬Ê¢ÓÐ×ãÁ¿Ë®µÄË®²ÛÀï·ÅÁ½¸öÉÕ±­£¬Ð¡ÉÕ±­Àï·ÅÓÐÉÙÁ¿Í­Æ¬ºÍ¹ýÁ¿Å¨ÏõËᣬСÉÕ±­ÍâÃæµ¹¿ÛÒ»´óÉÕ±­£¬Çå»Ø´ðÏÂÁÐÎÊÌ⣺

¢ñ.ʵÑé¹ý³ÌÖУ¬¹Û²ìµ½µÄÖ÷ÒªÏÖÏóÊÇ£º

¢ÙͭƬ±íÃæ²úÉúÆøÅÝ£¬Í­Æ¬±äС²¢Öð½¥Ïûʧ£»

¢ÚСÉÕ±­ÖÐÈÜÒºµÄÑÕÉ«Öð½¥±ä³ÉÂÌÉ«£»

¢Û________________________________________________¡£

¢Ü________________________________________________¡£

¢ò.ÓøÃ×°ÖÃ×öÍ­ÓëŨÏõËá·´Ó¦µÄʵÑ飬×îÍ»³öµÄÓŵãÊÇ______________¡£

¢ó.ÈôÒªÑéÖ¤×îÖյõ½µÄÆøÌåÉú³ÉÎ×î¼ò±ãµÄ·½·¨ÊÇ______________¡£

 ¢ÅÏÂÁÐʵÑé²Ù×÷»ò¶ÔʵÑéÊÂʵµÄÐðÊöÕýÈ·µÄÊÇ         £¨ÌîÐòºÅ£©¡£

¢ÙÏòÊÔ¹ÜÖеμÓÒºÌåʱ£¬Îª²»Ê¹ÒºÌåµÎµ½ÊÔ¹ÜÍâÓ¦½«½ºÍ·µÎ¹ÜÉìÈëÊÔ¹ÜÖУ»

¢ÚһС¿é½ðÊôÄƼÓÈëË®ÖкóѸËÙÈÛ³ÉСÇò£¬²»Í£µØÔÚË®ÃæÓζ¯²¢·¢³ö¡°Ö¨Ö¨¡±µÄÏìÉù£»

¢ÛÅäÖÃ100mL1.00mol/LµÄNaClÈÜҺʱ£¬¿ÉÓÃÍÐÅÌÌìƽ³ÆÈ¡5.85g NaCl¹ÌÌ壻

¢ÜÏò¿ÉÄܺ¬ÓÐSO42£­¡¢SO32£­µÄÈÜÒºÖмÓÈë¹ýÁ¿µÄÑÎËᣬÔÙ¼ÓÈëBa(NO3)2ÈÜÒº£¬¿É¼ìÑéSO42£­µÄ´æÔÚ£»

¢ÝÕô·¢NaClÈÜÒºÒԵõ½NaCl¹ÌÌåʱ£¬²»±Ø½«ÈÜÒºÕô¸É£»

¢ÞÏò100¡æʱµÄNaOHÏ¡ÈÜÒºÖеμӱ¥ºÍµÄFeCl3ÈÜÒº£¬ÒÔÖƱ¸Fe(OH)3½ºÌ壻

¢ßÈçͼ£¬¿É¹Û²ìµ½ÁéÃô¼ìÁ÷¼ÆµÄÖ¸Õëƫת£»

¢àÏòAlCl3ÈÜÒºÖеμÓNaOHÈÜÒººÍÏòNaOHÈÜÒºÖеμÓAlCl3ÈÜ  ÒºµÄÏÖÏóÏàͬ¡£

¢ÆÏÖÓÐ0.1mol¡¤L£­1µÄ´¿¼îÈÜÒº£¬ÓÃpHÊÔÖ½²â¶¨¸ÃÈÜÒºµÄpH£¬Æä

ÕýÈ·µÄ²Ù×÷ÊÇ                                                                           

                                                                                       

ÄãÈÏΪ¸ÃÈÜÒºpHµÄ·¶Î§Ò»¶¨½éÓÚ              Ö®¼ä¡£ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸Ö¤Ã÷´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32£­ÒýÆðµÄ£º                             ¡£

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

¾«Ó¢¼Ò½ÌÍø