ÌâÄ¿ÄÚÈÝ

2£®»¯Ñ§ÊµÑé¿ÎÉÏ£¬ÀÏʦΪÿ×é¼äѧ·Ö±ðÌṩÁËһƿÇâÑõ»¯ÄÆÈÜÒº£¬ÈÃËûÃÇÓÃ1%µÄÏ¡ÑÎËᣨÃܶÈΪ1.1g/ml£©À´²â¶¨ÆäÈÜÖʵÄÖÊÁ¿·ÖÊý£¬¼××éͬѧµÄʵÑéÈçͼ1Ëùʾ£»ÔÚÉÕ±­ÖмÓÈë5gÇâÑõ»¯ÄÆÈÜÒº£¬µÎÈ뼸µÎ·Ó̪ÈÜÒº£¬ÓõιÜÂýÂýµÎÈë1%µÄÏ¡ÑÎËᣬ²¢²»¶Ï½Á°è£¬ÈÜÒºpHµÄ±ä»¯Èçͼ2Ëùʾ£¬£¨·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNa0H+HCl¨TNaCl+H2O£©
Çë»Ø´ð£º
£¨1£©Èçͼ2Ëùʾ£¬aµãÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇNa+£¨Àë×Ó·ûºÅ±íʾ£©£»
£¨2£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùÏûºÄÏ¡ÑÎËáµÄÈÜÖÊÖÊÁ¿0.11g£»
£¨3£©¼ÆËãÇâÑõ»¯ÄÆÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý£®£¨¾«È·µ½0.1%£©

·ÖÎö £¨1£©¸ù¾ÝaµãµÄÈÜÒº±íʾÇâÑõ»¯ÄƺÍÏ¡ÑÎÇ¡ºÃÍêÈ«·´Ó¦½øÐзÖÎö£»
£¨2£©¸ù¾ÝÇ¡ºÃÍêÈ«·´Ó¦Ê±ËùÏûºÄÏ¡ÑÎËáµÄÌå»ý£¬ÒÀ¾ÝÑÎËáÃܶȡ¢ÈÜÖÊÖÊÁ¿·ÖÊý¼ÆËãÈÜÖÊÖÊÁ¿£»
£¨3£©¸ù¾Ý»¯Ñ§·½³ÌʽºÍ²Î¼Ó·´Ó¦µÄÑÎËáÖÊÁ¿½øÐмÆË㣮

½â´ð ½â£º£¨1£©aµãµÄÈÜÒº±íʾÇâÑõ»¯ÄƺÍÏ¡ÑÎÇ¡ºÃÍêÈ«·´Ó¦£¬ËùÒÔaµãÈÜÒºÖдæÔÚµÄÑôÀë×ÓÊÇNa+£»
£¨2£©Ç¡ºÃÍêÈ«·´Ó¦Ê±ËùÏûºÄÏ¡ÑÎËáµÄÈÜÖÊÖÊÁ¿ÊÇ11mL¡Á1.1g/ml¡Á1%=0.11g£»
£¨3£©Éè²Î¼Ó·´Ó¦µÄÇâÑõ»¯ÄƵÄÖÊÁ¿Îªx
NaOH+HCl=NaCl+H2O
40  36.5
x   0.11g
$\frac{40}{x}$=$\frac{36.5}{0.11g}$
 x=0.12g
ËùÒÔÇâÑõ»¯ÄÆÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊýΪ£º$\frac{0.12g}{5g}$¡Á100%=24%£®
¹Ê´ð°¸Îª£º£¨1£©Na+£»
£¨2£©0.11g£»
£¨3£©24%£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁË»¯Ñ§·½³ÌʽµÄ¼ÆË㣬ÄѶȲ»´ó£¬×¢Òâ½âÌâµÄ¹æ·¶ÐÔºÍ׼ȷÐÔ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø