ÏÖÓÐ0.1mol?L-1µÄAlCl3ÈÜÒººÍ0.1mol?L-1µÄNaOHÈÜÒº£¬½øÐÐÏÂÃæµÄʵÑé¡£

(1)ÔÚÊÔ¹ÜÖÐÊ¢ÓÐÉÏÊöAlCl3ÈÜÒº10mL£¬ÏòÆäÖÐÖð½¥µÎÈëÉÏÊöµÄNaOHÈÜÒº¡£

¢Ù¼ÓÈë10mlNaOHÈÜҺʱµÄÏÖÏóÊÇ                £»¼ÓÈë30mLNaOHÈÜҺʱµÄÏÖÏóÊÇ                  £»¼ÓÈë35mlNaOHÈÜҺʱµÄÏÖÏóÊÇ                         ¡£

¢ÚÉú³É³ÁµíÖÊÁ¿×î¶àʱ£¬ÐèNaOHÈÜÒº      mL¡£

(2)ÏòÊ¢ÓÐ10mLNaOHÈÜÒºµÄÊÔ¹ÜÖеÎÈëAlCl3ÈÜÒº£¬Í¬Ê±²»Í£Ò¡¶¯ÊԹܣ¬³öÏÖµÄÏÖÏóÊÇ         £¬ÖÁ¼ÓÈë      mLAlCl3ÈÜҺʱ¿ªÊ¼³öÏÖ³Áµí£¬ÖÁ¼ÓÈë       mLAlCl3ÈÜÒº²úÉú³Áµí´ï×î´óÖµ¡£Ð´³öÉÏÊö¹ý³ÌµÄÀë×Ó·½³Ìʽ                                ¡£

(1)¢ÙÖð½¥Éú³É°×É«½º×´³Áµí£»³ÁµíÁ¿´ïµ½×î´óÖµ£»³ÁµíÈܽâÒ»°ë¡£¢Ú30

(2)¿ªÊ¼¿´²»³öÓгÁµí£»2.5£»10/3£»

   Al3++3OH-=Al(OH)3¡ý  Al3++4OH-=AlO2-+2H2

   6H2O+Al3++3AlO2-=4Al(OH)3¡ý

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐ0.1mol/L Na2CO3ÈÜÒº£¬ÊÔÓÃpHÊÔÖ½²â¶¨ÈÜÒºµÄpH£¬ÆäÕýÈ·µÄ²Ù×÷ÊÇ
°ÑС¿épHÊÔÖ½·ÅÔÚ±íÃæÃ󣨻ò²£Á§Æ¬£©ÉÏ£¬ÓÃÕºÓдý²âÈÜÒºµÄ²£Á§°ôµãÔÚÊÔÖ½µÄÖв¿£¬ÊÔÖ½±äÉ«ºó£¬Óë±ê×¼±ÈÉ«¿¨±È½ÏÀ´È·¶¨ÈÜÒºµÄpH
°ÑС¿épHÊÔÖ½·ÅÔÚ±íÃæÃ󣨻ò²£Á§Æ¬£©ÉÏ£¬ÓÃÕºÓдý²âÈÜÒºµÄ²£Á§°ôµãÔÚÊÔÖ½µÄÖв¿£¬ÊÔÖ½±äÉ«ºó£¬Óë±ê×¼±ÈÉ«¿¨±È½ÏÀ´È·¶¨ÈÜÒºµÄpH
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©ÏÖÓÐ0.1mol?L-1µÄ´¿¼îÈÜÒº£¬ÊÔÓÃpHÊÔÖ½²â¶¨ÈÜÒºµÄpH£¬ÆäÕýÈ·µÄ²Ù×÷ÊÇ
°ÑС¿épHÊÔÖ½·ÅÔÚ±íÃæÃ󣨻ò²£Á§Æ¬£©ÉÏ£¬ÓÃÕºÓдý²âÈÜÒºµÄ²£Á§°ôµãÔÚÊÔÖ½µÄÖв¿£¬ÊÔÖ½±äÉ«ºó£¬Óë±ê×¼±ÈÉ«¿¨±È½ÏÀ´È·¶¨ÈÜÒºµÄpH
°ÑС¿épHÊÔÖ½·ÅÔÚ±íÃæÃ󣨻ò²£Á§Æ¬£©ÉÏ£¬ÓÃÕºÓдý²âÈÜÒºµÄ²£Á§°ôµãÔÚÊÔÖ½µÄÖв¿£¬ÊÔÖ½±äÉ«ºó£¬Óë±ê×¼±ÈÉ«¿¨±È½ÏÀ´È·¶¨ÈÜÒºµÄpH
£®´¿¼îÈÜÒº³Ê¼îÐÔµÄÔ­Òò£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©ÊÇ
CO32-+H2O
Ë®½â
HCO3-+OH-
CO32-+H2O
Ë®½â
HCO3-+OH-
£¬¸ÃÈÜÒºpHµÄ·¶Î§Ò»¶¨½éÓÚ
7µ½13
7µ½13
Ö®¼ä£®
£¨2£©ÎªÌ½¾¿´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32-ÒýÆðµÄ£¬ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸£®
Ïò´¿¼îÈÜÒºÖеÎÈë·Ó̪ÈÜÒº£¬ÈÜÒºÏÔºìÉ«£»ÈôÔÙÏò¸ÃÈÜÒºÖеÎÈë¹ýÁ¿ÂÈ»¯¸ÆÈÜÒº£¬
²úÉú°×É«³Áµí£¬ÇÒÈÜÒºµÄºìÉ«ÍÊÈ¥£®ËµÃ÷´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32-ÒýÆðµÄ£®
Ïò´¿¼îÈÜÒºÖеÎÈë·Ó̪ÈÜÒº£¬ÈÜÒºÏÔºìÉ«£»ÈôÔÙÏò¸ÃÈÜÒºÖеÎÈë¹ýÁ¿ÂÈ»¯¸ÆÈÜÒº£¬
²úÉú°×É«³Áµí£¬ÇÒÈÜÒºµÄºìÉ«ÍÊÈ¥£®ËµÃ÷´¿¼îÈÜÒº³Ê¼îÐÔÊÇÓÉCO32-ÒýÆðµÄ£®
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨Ò»£©ÏÖÓÐ0.1mol?L-1µÄ´¿¼îÈÜÒº£¬ÊÔÓÃpHÊÔÖ½²â¶¨ÈÜÒºµÄpH£¬ÆäÕýÈ·µÄ²Ù×÷ÊÇ
 
£»´¿¼îÈÜÒº³Ê¼îÐÔµÄÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
 
£®
£¨¶þ£©Èç±íÊDz»Í¬Î¶ÈÏÂË®µÄÀë×Ó»ýÊý¾Ý£º
ζÈ/¡æ 25 t1 t2
Kw/mol2?L-2 1¡Á10-14 a 1¡Á10-12
ÊԻشðÒÔϼ¸¸öÎÊÌ⣺
£¨1£©Èô25£¼t1£¼t2£¬Ôòa
 
1¡Á10-14£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£¬ÓÉ´ËÅжϵÄÀíÓÉÊÇ£º
 
£®
£¨2£©ÔÚt1¡æÏ£¬pH=10µÄNaOHÈÜÒºÖУ¬Ë®µçÀë²úÉúµÄ[OH-]Ϊ£º
 
£®
£¨3£©ÔÚt2¡æÏ£¬½«pH=11µÄ¿ÁÐÔÄÆÈÜÒºV1LÓëpH=1µÄÁòËáÈÜÒºV2L»ìºÏ£¨Éè»ìºÏºóÈÜÒºÌå»ýΪԭÁ½ÈÜÒºÌå»ýÖ®ºÍ£©ËùµÃÈÜÒºµÄpH=2£¬ÔòV1£ºV2=
 
£¬´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÖÓÐ0.1mol/LµÄ´¿¼îÈÜÒº£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©³£ÎÂÏ£¬¼×ͬѧÓÃpHÊÔÖ½¼ìÑé¸ÃÈÜÒº£¬·¢ÏÖÆäpH£¾7£¬ÄãÈÏΪԭÒòÊÇʲô£¿
 
£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
£¨2£©³£ÎÂÏ£¬ÒÒͬѧÏò¸ÃÈÜÒºÖеÎÈ뼸µÎ·Ó̪ÈÜÒº£¬·¢ÏÖ¸ÃÈÜÒºÓöµ½·Ó̪³ÊºìÉ«£®¶Ô¸ÃÏÖÏóµÄ·ÖÎö£¬ÄãÈÏΪÄܵõ½µÄ½áÂÛÊÇ£º¸ÃÈÜÒº³Ê
 
ÐÔ£®
¶ÔΪʲôÓöµ½·Ó̪¸ÃÈÜÒº³ÊÏÖ³öºìÉ«µÄÔ­ÒòÎÊÌ⣬ͬѧÃÇÕ¹¿ªÁËÌÖÂÛ£¬ÒÒͬѧÈÏΪ£¬ÊÇÔÚÅäÖƱê׼̼ËáÄÆÈÜҺʱ»ìÈëÁËNaOHËùÖ£¬¶ø±ûͬѧÔòÈÏΪ£¬ÊÇ̼ËáÄÆÈÜҺˮ½âËùÖ£®
ÇëÄãÉè¼ÆÒ»¸ö¼òµ¥µÄʵÑé·½°¸£¬À´ÆÀÅÐÉÏÊö¹Ûµã£®£¨ÒªÇó±ØÐëд³ö²Ù×÷¡¢ÏÖÏóºÍ½áÂÛ£©
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê°²ÇðÖÐѧ¸ß¶þÊîÆÚÁ·Ï°»¯Ñ§¾í£¨ËÄ£© ÌâÐÍ£ºÊµÑéÌâ

¢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                            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸