ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÊµÑéÊÒ³£ÓÃ׼ȷ³ÆÁ¿µÄÁÚ±½¶þ¼×ËáÇâ¼Ø(½á¹¹¼òʽΪ)À´×¼È·²â¶¨NaOH±ê×¼ÈÜÒºµÄŨ¶È£¬¼´·ÖÎö»¯Ñ§ÊµÑéÖг£³ÆΪ¡°±ê¶¨¡±µÄÒ»ÖÖ·½·¨¡£

ÒÑÖª£º¢ÙNaOHÈÜÒºµÄŨ¶ÈÔÚ0.1 mol¡¤L1×óÓÒ£¬µÎ¶¨ÖÕµãʱÈÜÒºµÄpHӦΪ9.1¡£

¢ÚÁÚ±½¶þ¼×ËáÇâ¼ØÏà¶Ô·Ö×ÓÖÊÁ¿Îª204

£¨1£©Ð´³öÁÚ±½¶þ¼×ËáÇâ¼ØÓëNaOH·´Ó¦µÄÀë×Ó·½³Ìʽ ¡£

£¨2£©½«ÓÃÍÐÅÌÌìƽ³ÆºÃµÄÁÚ±½¶þ¼×ËáÇâ¼Ø·ÅÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÊÊÁ¿µÄË®Èܽ⣬ÈÜÒº³ÊÎÞÉ«£¬ÔÙ¼ÓÈëָʾ¼Á·Ó̪£¬ÓÃNaOHÈÜÒºµÎ¶¨µ½ÖÕµãʱ£¬ÏÖÏóÊÇ ¡£

£¨3£©Ä³Ñ§Éú½øÐÐÁËËÄ´ÎʵÑ飬ʵÑéÊý¾ÝÈçÏÂ±í£º

ʵÑé±àºÅ

ÁÚ±½¶þ¼×ËáÇâ

¼ØµÄÖÊÁ¿/g

´ý²âNaOHÈÜÒºµÄÌå»ý/mL

1

0.408 0

18.20

2

17.10

3

16.90

4

17.00

µÎ¶¨ÖÐÎó²î½Ï´óµÄÊÇµÚ ´ÎʵÑ飬Ôì³ÉÕâÖÖÎó²îµÄ¿ÉÄÜÔ­ÒòÊÇ ¡£

A£®¼îʽµÎ¶¨¹ÜÔÚװҺǰδÓôý²âNaOHÈÜÒºÈóÏ´2~3´Î

B£®µÎ¶¨¿ªÊ¼Ç°¼îʽµÎ¶¨¹Ü¼â×첿·ÖÓÐÆøÅÝ£¬Ôڵζ¨ÖÕµã¶ÁÊýʱδ·¢ÏÖÆøÅÝ

C£®Ê¢ÓÐÁÚ±½¶þ¼×ËáÇâ¼ØÈÜÒºµÄ׶ÐÎÆ¿ÖÐÓÐÉÙÁ¿Ë®

D£®´ïµ½µÎ¶¨ÖÕµãʱ£¬¸©ÊÓÈÜÒº°¼ÒºÃæ×îµÍµã¶ÁÊý

E£®µÎ¶¨¹ý³ÌÖУ¬×¶ÐÎÆ¿Õñµ´µÃÌ«¾çÁÒ£¬ÒÔÖÁÓÐЩҺµÎ·É½¦³öÀ´

£¨4£©¸ÃͬѧËù²âµÃµÄNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ mol¡¤L1(½á¹û±£ÁôÈýλСÊý)¡£

¡¾´ð°¸¡¿£¨9·Ö£¬³ý±êÃ÷Í⣬ÿ¿Õ2·Ö£©

£¨1£©+OH+H2O

£¨2£©µ±µÎÈë×îºóÒ»µÎNaOHÈÜÒººó£¬ÈÜÒºÑÕÉ«ÓÉÎÞÉ«±äΪºìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«

£¨3£©1£¨1·Ö£© AB

£¨4£©0.118

¡¾½âÎö¡¿£¨3£©AÑ¡ÏîʹÇâÑõ»¯ÄÆŨ¶È¼õС£¬ËùÓÃÇâÑõ»¯ÄÆÈÜÒºÌå»ýÆ«´ó£»CÑ¡ÏîÎÞÓ°Ï죻DÑ¡Ïîʹ¶ÁÊýÆ«µÍ£»EÑ¡ÏîÏûºÄÇâÑõ»¯ÄÆÈÜÒºÌå»ýƫС£»

£¨4£©ÉáÈ¥Îó²î´óµÄʵÑé1£¬NaOHÈÜÒºµÄÌå»ýȡƽ¾ùÖµ17.00 mL£¬n(NaOH)=n(ÁÚ±½¶þ¼×ËáÇâ¼Ø)==0.002 mol£¬c(NaOH)= =0.118 mol¡¤L1¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø