ÌâÄ¿ÄÚÈÝ

£¨9·Ö£©Ä³Ñ§ÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÁòËáÈÜÒº£¬ÊµÑéÈçÏ£ºÓÃ1.00mL´ý²âÁòËáÅäÖÆ100 mLÏ¡H2SO4ÈÜÒº£»ÒÔ0.14mol£¯LµÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO425.00mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº15.00mL¡£
(1)¸ÃѧÉúÓñê×¼0.14 mol£¯L NaOHÈÜÒºµÎ¶¨ÁòËáµÄʵÑé²Ù×÷ÈçÏ£º

A£®ÓÃËáʽµÎ¶¨¹ÜȡϡH2SO4 25.00 mL£¬×¢Èë×¶ÐÎÆ¿ÖУ¬¼ÓÈëָʾ¼Á¡£
B£®Óôý²â¶¨µÄÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü¡£
C£®ÓÃÕôÁóˮϴ¸É¾»µÎ¶¨¹Ü¡£
D£®È¡Ï¼îʽµÎ¶¨¹ÜÓñê×¼µÄNaOHÈÜÒºÈóÏ´ºó£¬½«±ê׼ҺעÈë¼îʽµÎ¶¨¹Ü¿Ì¶È¡°0¡±ÒÔÉÏ2¡«3 cm´¦£¬ÔٰѼîʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæÖÁ¿Ì¶È¡°0¡±»ò¡°0¡±¿Ì¶ÈÒÔÏ¡£ E£®¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮ¡£ F£®Áíȡ׶ÐÎÆ¿£¬ÔÙÖØ¸´²Ù×÷Ò»´Î¡£ G£®°Ñ×¶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃæ£¬Æ¿ÏµæÒ»ÕŰ×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿Ö±ÖÁµÎ¶¨Öյ㣬¼ÇÏµζ¨¹ÜÒºÃæËùÔڿ̶ȡ£
a¡¢µÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ(ÓÃÐòºÅÌîд)£º ¢Ù ¡£
b¡¢¸ÃµÎ¶¨²Ù×÷ÖÐӦѡÓõÄָʾ¼ÁÊÇ£º  ¢Ú  ¡£
c¡¢ÔÚG²Ù×÷ÖÐÈçºÎÈ·¶¨ÖÕµã?  ¢Û  ¡£
(2)¼îʽµÎ¶¨¹ÜÓÃÕôÁóË®ÈóÏ´ºó£¬Î´Óñê×¼ÒºÈóÏ´µ¼Öµζ¨½á¹û(ÌƫС¡±¡¢¡°Æ«´ó¡±»ò¡°Ç¡ºÃºÏÊÊ¡±) ¢Ü  £¬Ô­ÒòÊÇ ¢Ý ¡£
(3)¼ÆËã´ý²âÁòËá(Ï¡ÊÍǰµÄÁòËá)ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È(¼ÆËã½á¹û¾«È·µ½Ð¡Êýµãºó¶þλ)¢Þ mol/L

¢ÙECDBAGF£¨»òECBADGF£©£¨2·Ö£© ¢Ú·Ó̪£¨1·Ö£©  ¢ÛµÎÈë×îºóÒ»µÎNaOHÈÜÒº£¬ÈÜҺͻȻ±ä³ÉºìÉ«£¬°ë·ÖÖÓ²»ÍÊÉ«£¨1·Ö£©¢ÜÆ«´ó£¨1·Ö£©¢ÝµÎ¶¨¹ÜÄÚ±ÚÉϵÄˮĤ£¬½«±ê׼ҺϡÊÍ£¬Ê¹Ìå»ý¶ÁÊýÆ«´ó£¨2·Ö£©¢Þ4.20£¨2·Ö£©?

½âÎöÊÔÌâ·ÖÎö£º¼îʽµÎ¶¨¹ÜÓÃÕôÁóË®ÈóÏ´ºó£¬Î´Óñê×¼ÒºÈóÏ´£¬Ï൱ÓÚ±ê×¼Òº±»Ï¡ÊÍ£¬ËùÒÔÏûºÄµôµÄÏ¡ÁòËá¾Í»áÔö¶à£¬µÎ¶¨½á¹ûÆ«´ó¡£2NaOH+H2SO4=NaSO4+2H2O,Óɶ¨Á¿¹ØÏµ¿ÉÖª£º´ý²âÁòËá(Ï¡ÊÍǰµÄÁòËá)ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È=ÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿*2/ÁòËáµÄÌå»ý=4.20mol/L
¿¼µã£ºËá¼îÖк͵ζ¨ÊµÑé
µãÆÀ£º±¾Ì⿼²éµÄÊÇËá¼îÖк͵ζ¨ÊµÑé²Ù×÷£¬ÒªÇóѧÉúÊì¼ÇËá¼îÖк͵樵ÄʵÑé²½ÖèºÍ×¢ÒâÊÂÏ¾ÍÄÜ˳ÀûÍê³É´ËÀàÌâÄ¿¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijѧÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÁòËáÈÜÒº£¬ÊµÑéÈçÏ£ºÓÃ1.00mL´ý²âÁòËáÅäÖÆ100mLÏ¡H2SO4ÈÜÒº£»ÒÔ0.14mol/LµÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO425.00mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº15.00mL£®
£¨1£©¸ÃѧÉúÓñê×¼0.14mol/L NaOHÈÜÒºµÎ¶¨ÁòËáµÄʵÑé²Ù×÷ÈçÏ£º
A£®ÓÃËáʽµÎ¶¨¹ÜȡϡH2SO4 25.00mL£¬×¢Èë×¶ÐÎÆ¿ÖУ¬¼ÓÈëָʾ¼Á£®
B£®Óôý²â¶¨µÄÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü£® C£®ÓÃÕôÁóˮϴ¸É¾»µÎ¶¨¹Ü£®
D£®È¡Ï¼îʽµÎ¶¨¹ÜÓñê×¼µÄNaOHÈÜÒºÈóÏ´ºó£¬½«±ê׼ҺעÈë¼îʽµÎ¶¨¹Ü¿Ì¶È¡°0¡±ÒÔÉÏ2¡«3cm´¦£¬ÔٰѼîʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæÖÁ¿Ì¶È¡°0¡±»ò¡°0¡±¿Ì¶ÈÒÔÏ£® E£®¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮ£® F£®Áíȡ׶ÐÎÆ¿£¬ÔÙÖØ¸´²Ù×÷Ò»´Î£® G£®°Ñ×¶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃæ£¬Æ¿ÏµæÒ»ÕŰ×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿Ö±ÖÁµÎ¶¨Öյ㣬¼ÇÏµζ¨¹ÜÒºÃæËùÔڿ̶ȣ®
a¡¢µÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ£¨ÓÃÐòºÅÌîд£©£º
ECDBAGF£¨»òECBADGF£©
ECDBAGF£¨»òECBADGF£©
£®
b¡¢¸ÃµÎ¶¨²Ù×÷ÖÐӦѡÓõÄָʾ¼ÁÊÇ£º
·Ó̪
·Ó̪
£®
c¡¢ÔÚG²Ù×÷ÖÐÈçºÎÈ·¶¨Öյ㣿
µÎÈë×îºóÒ»µÎNaOHÈÜÒº£¬ÈÜҺͻȻ±ä³ÉºìÉ«£¬°ë·ÖÖÓ²»ÍÊÉ«
µÎÈë×îºóÒ»µÎNaOHÈÜÒº£¬ÈÜҺͻȻ±ä³ÉºìÉ«£¬°ë·ÖÖÓ²»ÍÊÉ«
£®
£¨2£©¼îʽµÎ¶¨¹ÜÓÃÕôÁóË®ÈóÏ´ºó£¬Î´Óñê×¼ÒºÈóÏ´µ¼Öµζ¨½á¹û£¨ÌƫС¡±¡¢¡°Æ«´ó¡±»ò¡°Ç¡ºÃºÏÊÊ¡±£©
Æ«´ó
Æ«´ó
£¬Ô­ÒòÊÇ
µÎ¶¨¹ÜÄÚ±ÚÉϵÄˮĤ£¬½«±ê׼ҺϡÊÍ£¬Ê¹Ìå»ý¶ÁÊýÆ«´ó
µÎ¶¨¹ÜÄÚ±ÚÉϵÄˮĤ£¬½«±ê׼ҺϡÊÍ£¬Ê¹Ìå»ý¶ÁÊýÆ«´ó
£®
£¨3£©¼ÆËã´ý²âÁòËᣨϡÊÍǰµÄÁòËᣩÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£¨¼ÆËã½á¹û¾«È·µ½Ð¡Êýµãºó¶þ룩
4.20
4.20
mol/L£®

£¨11·Ö£©Ä³Ñ§ÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÁòËáÈÜÒº£¬ÊµÑéÈçÏ£ºÒÔ0.14mol£¯LµÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO420.00mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº16.00mL¡£

£¨1£©¸ÃѧÉúÓñê×¼0.14 mol£¯LNaOHÈÜÒºµÎ¶¨ÁòËáµÄʵÑé²Ù×÷ÈçÏ£º

 A£®ÓÃËáʽµÎ¶¨¹ÜȡϡH2SO420.00 mL£¬×¢Èë×¶ÐÎÆ¿ÖУ¬¼ÓÈëָʾ¼Á¡£

 B£®Óôý²â¶¨µÄÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü¡£

 C£®ÓÃÕôÁóˮϴ¸É¾»µÎ¶¨¹Ü¡£

 D£®È¡Ï¼îʽµÎ¶¨¹ÜÓñê×¼µÄNaOHÈÜÒºÈóÏ´ºó£¬½«±ê׼ҺעÈë¼îʽµÎ¶¨¹Ü¿Ì¶È¡°0¡±ÒÔÉÏ2¡«3 cm´¦£¬ÔٰѼîʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæÖÁ¿Ì¶È¡°0¡±»ò¡°0¡±¿Ì¶ÈÒÔÏ¡£

 E£®¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮ¡£

 F£®Áíȡ׶ÐÎÆ¿£¬ÔÙÖØ¸´²Ù×÷Ò»´Î¡£

 G£®°Ñ×¶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃæ£¬Æ¿ÏµæÒ»ÕŰ×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿Ö±ÖÁµÎ¶¨Öյ㣬¼ÇÏµζ¨¹ÜÒºÃæËùÔڿ̶ȡ£

¢ÙµÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ(ÓÃÐòºÅÌîд)£º E¡ª£¨  £©¡ªD¡ª£¨  £©¡ª £¨ £©¡ª£¨  £©¡ªF  ¡£

¢Ú¸ÃµÎ¶¨²Ù×÷ÖÐӦѡÓõÄָʾ¼ÁÊÇ£º    ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡  ¡£

¢ÛÔÚG²Ù×÷ÖÐÈçºÎÈ·¶¨ÖÕµã?                                     ¡£

£¨2£©Óñê×¼µÄNaOHµÎ¶¨Î´ÖªÅ¨¶ÈµÄÑÎËᣬѡÓ÷Ó̪Ϊָʾ¼Á£¬Ôì³É²â¶¨½á¹ûÆ«¸ßµÄÔ­Òò¿ÉÄÜÊÇ_________¡£

A£®ÅäÖÆ±ê×¼ÈÜÒºµÄÇâÑõ»¯ÄÆÖлìÓÐNaClÔÓÖÊ

B£®µÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËü²Ù×÷¾ùÕýÈ·

C£®Ê¢×°Î´ÖªÒºµÄ×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´

D£®µÎ¶¨µ½ÖÕµã¶ÁÊýʱ·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº

E£®Î´Óñê×¼ÒºÈóÏ´¼îʽµÎ¶¨¹Ü

F.¿ªÊ¼ÊµÑéʱ¼îʽµÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬Ôڵζ¨¹ý³ÌÖÐÆøÅÝÏûʧ£¬Ôò²âµÃÑÎËáŨ¶È

£¨3£©¼ÆËã´ý²âÁòËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È      mol£¯L£¨¼ÆËã½á¹û±£Áô¶þλÓÐЧÊý×Ö£©

 

ijѧÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÁòËáÈÜÒº,ʵÑéÈçÏÂ:ÓÃ1.00mL´ý²âÁòËáÅäÖÆ100 mLÏ¡H2SO4ÈÜÒº¡£ÒÔ0.14mol/LµÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO4 25mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº15mL£¬Æä²Ù×÷·Ö½âΪÈçϼ¸²½£º

A£®ÒÆÈ¡Ï¡H2SO4 25.00 mL£¬×¢Èë½à¾»µÄ×¶ÐÎÆ¿ÖУ¬²¢¼ÓÈë2-3µÎָʾ¼Á¡£
B£®ÓÃÕôÁóˮϴµÓµÎ¶¨¹Ü£¬ºóÓñê×¼ÈÜÒºÈóÏ´2-3´Î¡£
C£®È¡±ê×¼NaOHÈÜҺעÈë¼îʽµÎ¶¨¹ÜÖÁ ¡°0¡±»ò¡°0¡±¿Ì¶ÈÏßÒÔÉÏ¡£
D£®µ÷½ÚÒºÃæÖÁ ¡°0¡±»ò¡°0¡±ÒÔÏÂijһ¿Ì¶È£¬¼Ç϶ÁÊý¡£
E£®°ÑÊ¢Óбê×¼ÈÜÒºµÄ¼îʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚµÎ¶¨¹Ü¼â×ìʹ֮³äÂúÈÜÒº¡£
F£®°Ñ×¶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃæ£¬Óñê×¼NaOHÈÜÒºµÎ¶¨ÖÁÖյ㣬¼ÇÏµζ¨¹ÜÒºÃæ¶ÁÊý¡£
¾Í´ËʵÑéÍê³ÉÌî¿Õ£º
£¨1£©¢ÙµÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ(ÓÃÐòºÅÌîд)                   ¡£
¢Ú¸ÃµÎ¶¨²Ù×÷ÖÐӦѡÓõÄָʾ¼ÁÊÇ             ¡£
¢ÛÔÚF²Ù×÷ÖÐÈçºÎÈ·¶¨ÖÕµã?                                 ¡£
£¨2£©¸ÃµÎ¶¨¹ý³ÌÖУ¬Ôì³É²â¶¨½á¹ûÆ«¸ßµÄÔ­Òò¿ÉÄÜÓÐÄÄЩ         ¡£
¢ÙÅäÖÆ±ê×¼ÈÜÒºµÄNaOHÖлìÓÐNa2CO3ÔÓÖÊ
¢ÚµÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËû²Ù×÷ÕýÈ· 
¢Û Ê¢×°´ý²âÒºµÄ×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´
¢ÜµÎ¶¨µ½ÖÕµã¶ÁÊýʱ£¬·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº
¢ÝµÎ¶¨Ç°¼îʽµÎ¶¨¹Ü¼â×첿·Öδ³äÂúÈÜÒº 
¢ÞµÎ¶¨Öв»É÷½«×¶ÐÎÆ¿ÄÚÒºÌåÒ¡³öÉÙÁ¿ÓÚÆ¿Íâ
£¨3£©¼ÆËã´ý²âÁòËá(Ï¡ÊÍǰµÄÁòËá)ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È         mol/L£¨±£Áô2λСÊý£©

ijѧÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÁòËáÈÜÒº,ʵÑéÈçÏÂ:ÓÃ1.00mL´ý²âÁòËáÅäÖÆ100 mLÏ¡H2SO4ÈÜÒº¡£ÒÔ0.14mol/LµÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO4 25mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº15mL£¬Æä²Ù×÷·Ö½âΪÈçϼ¸²½£º

A£®ÒÆÈ¡Ï¡H2SO4 25.00 mL£¬×¢Èë½à¾»µÄ×¶ÐÎÆ¿ÖУ¬²¢¼ÓÈë2-3µÎָʾ¼Á¡£

B£®ÓÃÕôÁóˮϴµÓµÎ¶¨¹Ü£¬ºóÓñê×¼ÈÜÒºÈóÏ´2-3´Î¡£

C£®È¡±ê×¼NaOHÈÜҺעÈë¼îʽµÎ¶¨¹ÜÖÁ ¡°0¡±»ò¡°0¡±¿Ì¶ÈÏßÒÔÉÏ¡£

D£®µ÷½ÚÒºÃæÖÁ ¡°0¡±»ò¡°0¡±ÒÔÏÂijһ¿Ì¶È£¬¼Ç϶ÁÊý¡£

E£®°ÑÊ¢Óбê×¼ÈÜÒºµÄ¼îʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚµÎ¶¨¹Ü¼â×ìʹ֮³äÂúÈÜÒº¡£

F£®°Ñ×¶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃæ£¬Óñê×¼NaOHÈÜÒºµÎ¶¨ÖÁÖյ㣬¼ÇÏµζ¨¹ÜÒºÃæ¶ÁÊý¡£

¾Í´ËʵÑéÍê³ÉÌî¿Õ£º

£¨1£©¢ÙµÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ(ÓÃÐòºÅÌîд)                   ¡£

¢Ú¸ÃµÎ¶¨²Ù×÷ÖÐӦѡÓõÄָʾ¼ÁÊÇ             ¡£

¢ÛÔÚF²Ù×÷ÖÐÈçºÎÈ·¶¨ÖÕµã?                                 ¡£

£¨2£©¸ÃµÎ¶¨¹ý³ÌÖУ¬Ôì³É²â¶¨½á¹ûÆ«¸ßµÄÔ­Òò¿ÉÄÜÓÐÄÄЩ         ¡£

¢ÙÅäÖÆ±ê×¼ÈÜÒºµÄNaOHÖлìÓÐNa2CO3ÔÓÖÊ

¢ÚµÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËû²Ù×÷ÕýÈ· 

¢Û Ê¢×°´ý²âÒºµÄ×¶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´

¢ÜµÎ¶¨µ½ÖÕµã¶ÁÊýʱ£¬·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº

¢ÝµÎ¶¨Ç°¼îʽµÎ¶¨¹Ü¼â×첿·Öδ³äÂúÈÜÒº 

¢ÞµÎ¶¨Öв»É÷½«×¶ÐÎÆ¿ÄÚÒºÌåÒ¡³öÉÙÁ¿ÓÚÆ¿Íâ

£¨3£©¼ÆËã´ý²âÁòËá(Ï¡ÊÍǰµÄÁòËá)ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È         mol/L£¨±£Áô2λСÊý£©

 

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

¾«Ó¢¼Ò½ÌÍø