Question

2£®ÓÃÖк͵ζ¨·¨²â¶¨ÉÕ¼îµÄ´¿¶È£¨ÉÕ¼îÖеÄÔÓÖʲ»ºÍËá·´Ó¦£©£¬Çë¸ù¾ÝʵÑé»Ø´ð£º
£¨1£©½«³ÆÁ¿ºÃµÄ4.3gÉÕ¼îÑùÆ·ÅäÖƳÉ250mL´ý²âÒº£¬ÅäÖƹý³ÌʹÓõÄÖ÷ÒªÒÇÆ÷³ý250mLÈÝÁ¿Æ¿¡¢Á¿Í²¡¢ÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÓÐÒ»ÖÖ±ØÐëʹÓõÄÒÇÆ÷ÊDz£Á§°ô£®
£¨2£©ÓüîʽµÎ¶¨¹ÜÁ¿È¡10.00mL´ý²âÒºÓÚ׶ÐÎÆ¿ÖУ¬µÎÈ뼸µÎ·Ó̪£®
£¨3£©ÓÃ0.20mol•L^-1µÄ±ê×¼ÑÎËáµÎ¶¨´ý²âÒº£¬Åжϵζ¨ÖÕµãµÄÏÖÏóÊÇ£ºµÎÈë×îºóÒ»µÎÑÎËáʱ£¬×¶ÐÎÆ¿ÄÚÈÜҺǡºÃÓɺìÉ«±äÎÞÉ«£¬²¢ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£®
£¨4£©Èç¹ûʵÑé²Ù×÷ÕýÈ·£¬´ÓµÎ¶¨¿ªÊ¼µ½½áÊø£¬ÈÜÒºÖеÄÀë×ÓŨ¶È¹Øϵ¿ÉÒÔ³öÏÖµÄÊÇB¡¢C£¨Ìî´ð°¸×ÖĸÐòºÅ£©
A£®c£¨Na⁺£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£¾c£¨OH^-£©
B£®c£¨Na⁺£©£¾c£¨OH^-£©£¾c£¨Cl^-£©£¾c£¨H⁺£©
C£®c£¨Na⁺£©+c£¨H⁺£©=c£¨OH^-£©+c£¨Cl^-£©
D£®c£¨Na⁺£©+c£¨H⁺£©£¾c£¨OH^-£©+c£¨Cl^-£©
£¨5£©¸ù¾ÝÏÂÁÐÊý¾Ý¼ÆË㣬c£¨NaOH£©=0.40mol/L£¬ÉÕ¼îµÄ´¿¶ÈΪ93%£¨±£ÁôÕûÊý£©

µÎ¶¨´ÎÊý	´ý²âÒºÌå»ý£¨mL£©	±ê×¼ÑÎËáÌå»ý£¨mL£©
		µÎ¶¨Ç°¶ÁÊý£¨mL£©	µÎ¶¨ºó¶ÁÊý£¨mL£©
µÚÒ»´Î	10.00	0.50	20.40
µÚ¶þ´Î	10.00	4.00	24.10

£¨6£©¾­¹ý·ÖÎö·¢ÏÖ£¬±¾´ÎʵÑéµÄ²â¶¨½á¹û±ÈÉÕ¼îµÄʵ¼Ê´¿¶ÈÆ«¸ß£¬Ôì³ÉÎó²îµÄ¿ÉÄÜÔ­ÒòÊÇC¡¢D£¨Ìî´ð°¸×ÖĸÐòºÅ£©
A£®µÎ¶¨Ç°Æ½ÊÓ£¬µÎ¶¨ºó¸©ÊÓ        B£®Óñê×¼ÒºÈóÏ´µÎ¶¨¹Ü
C£®Óôý²âÒºÈóϴ׶ÐÎÆ¿            D£®²»Ð¡ÐĽ«±ê×¼ÒºµÎÔÚ׶ÐÎÆ¿ÍâÃæ
E£®µÎ¶¨½Ó½üÖÕµãʱ£¬ÓÃÉÙÁ¿ÕôÁóË®³åϴ׶ÐÎÆ¿ÄÚ£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝʵÑé²Ù×÷µÄ²½ÖèÒÔ¼°Ã¿²½²Ù×÷ÐèÒªÒÇÆ÷È·¶¨·´Ó¦ËùÐèÒÇÆ÷À´½â´ð£»
£¨2£©¸ù¾Ý¼îÐÔÈÜҺʢ·ÅÔÚ¼îʽµÎ¶¨¹ÜÖУ»
£¨3£©¸ù¾ÝÈÜÒºÑÕÉ«±ä»¯ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£¬¿É˵Ã÷´ïµ½µÎ¶¨Öյ㣻
£¨4£©µÎ¶¨¹ý³ÌÖÐÈÜÒº´Ó¼îÐÔÈÜÒºÏòÖÐÐÔ¹ý¶É£¬ÈÜÒºÀïµÄÈÜÖÊ´ÓNaOHµ½NaCl¡¢NaOH¹ý¶É£¬×îºó³ÊÖÐÐÔÊÇÈÜÖÊΪNaCl£¬µ«ÈÜҺʼÖÕÏÔÖÐÐÔ£¬´æÔÚµçºÉÊغ㣺c£¨Na⁺£©+c£¨H⁺£©=c£¨OH^-£©+c£¨Cl^-£©£¬¾Ý´Ë·ÖÎöÅжϣ»
£¨5£©ÏÈ·ÖÎöËùºÄÑÎËá±ê×¼ÒºµÄÌå»ýµÄÓÐЧÐÔ£¬È»ºóÇó³öËùºÄÑÎËá±ê×¼ÒºµÄÌå»ýƽ¾ùÖµ£¬È»ºó¸ù¾Ý¹ØϵʽHCl¡«NaOHÀ´½â´ð£®
£¨6£©¸ù¾Ýc£¨´ý²â£©=$\frac{V£¨±ê×¼£©¡Ác£¨±ê×¼£©}{V£¨´ý²â£©}$·ÖÎö²»µ±²Ù×÷¶ÔV£¨±ê×¼£©µÄÓ°Ï죬ÒÔ´ËÅжÏŨ¶ÈµÄÎó²î£®

½â´ð ½â£º£¨1£©ÅäÖƲ½ÖèÓмÆËã¡¢³ÆÁ¿¡¢Èܽ⡢ÀäÈ´¡¢ÒÆÒº¡¢Ï´µÓ¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£¬Ò»°ãÓÃÍÐÅÌÌìƽ³ÆÁ¿£¬ÓÃÒ©³×È¡ÓÃÒ©Æ·£¬ÔÚÉÕ±­ÖÐÈܽ⣨¿ÉÓÃÁ¿Í²Á¿È¡£©£¬ÀäÈ´ºóתÒƵ½250mLÈÝÁ¿Æ¿ÖУ¬²¢Óò£Á§°ôÒýÁ÷£¬µ±¼ÓË®ÖÁÒºÃæ¾àÀë¿Ì¶ÈÏß1¡«2cmʱ£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼Ó£¬ËùÒÔÊÔÑé³ýÁËÐèÒª250mLÈÝÁ¿Æ¿¡¢Á¿Í²¡¢ÉÕ±­¡¢½ºÍ·µÎ¹ÜÍ⣬»¹ÓÐÒ»ÖÖ±ØÐëʹÓõÄÒÇÆ÷ÊDz£Á§°ô£»
¹Ê´ð°¸Îª£º²£Á§°ô£»
£¨2£©ÉÕ¼îÏÔ¼îÐÔ£¬¹ÊÐèÒªÓüîʽµÎ¶¨¹ÜÁ¿È¡ÉÕ¼îÈÜÒº£¬
¹Ê´ð°¸Îª£º¼î£»
£¨3£©µÎ¶¨Ç°ÈÜҺΪºìÉ«£¬µÎ¶¨ÖÕµãʱÈÜҺΪÎÞÉ«£¬ËùÒԵζ¨ÖÕµãÏÖÏóΪµÎÈë×îºóÒ»µÎÑÎËáʱ£¬×¶ÐÎÆ¿ÄÚÈÜҺǡºÃÓɺìÉ«±äÎÞÉ«£¬²¢ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£»
¹Ê´ð°¸Îª£ºµÎÈë×îºóÒ»µÎÑÎËáʱ£¬×¶ÐÎÆ¿ÄÚÈÜҺǡºÃÓɺìÉ«±äÎÞÉ«£¬²¢ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£»
£¨4£©A£®Èôc£¨Na⁺£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£¾c£¨OH^-£©³ÉÁ¢£¬Ôòc£¨Na⁺£©+c£¨H⁺£©£¾c£¨Cl^-£©+c£¨OH^-£©£¬Ã÷ÏÔÓëÈÜÒºÀï´æÔڵĵçºÉÊغãʽ²»ÎǺϣ¬¹ÊA´íÎó£»
B£®ÔÚNaOHÈÜÒºÖ»µÎ¼Ó¼¸µÎÏ¡ÑÎËᣬ´ËʱÈÜÒºÖ»Éú³ÉÉÙÁ¿NaCl£¬Ó¦¸Ã´æÔÚ£ºc£¨Na⁺£©£¾c£¨OH^-£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£¬¹ÊBÕýÈ·£»
C£®ÈÜҺʼÖÕÏÔÖÐÐÔ£¬´æÔÚµçºÉÊغãʽ£ºc£¨Na⁺£©+c£¨H⁺£©=c£¨OH^-£©+c£¨Cl^-£©£¬¹ÊCÕýÈ·£»
D£®c£¨Na⁺£©+c£¨H⁺£©£¾c£¨OH^-£©+c£¨Cl^-£©ÓëÈÜÒºÀï´æÔڵĵçºÉʽc£¨Na⁺£©+c£¨H⁺£©=c£¨OH^-£©+c£¨Cl^-£©²»·û£¬¹ÊD´íÎó£»
¹ÊÑ¡BC£»
£¨5£©ËùºÄÑÎËá±ê×¼ÒºµÄÌå»ý·Ö±ðΪ£º19.90mL£¬20.10mL£¬Á½×éÊý¾Ý¾ùÓÐЧ£¬ÑÎËá±ê×¼ÒºµÄƽ¾ùÌå»ýΪ20.00mL£»
               HCl¡«NaOH
                1      1
0.2000mol/L¡Á20.00mL     C£¨NaOH£©¡Á10mL
½âµÃ£ºC£¨NaOH£©=0.4mol/L£¬Ôò250mLµÄÈÜÒºÖк¬ÓÐNaOHµÄÖÊÁ¿Îª£º0.4mol/L¡Á0.25L¡Á40g/mol=4.0g£¬ÉÕ¼îÑùÆ·µÄ´¿¶ÈΪ$\frac{4.0g}{4.3g}$¡Á100%=93%£»
¹Ê´ð°¸Îª£º0.40£»93£»
£¨6£©A£®µÎ¶¨Ç°Æ½ÊÓ£¬µÎ¶¨ºó¸©ÊÓ£¬Ôì³ÉV£¨±ê×¼£©Æ«Ð¡£¬¸ù¾Ýc£¨´ý²â£©=$\frac{V£¨±ê×¼£©¡Ác£¨±ê×¼£©}{V£¨´ý²â£©}$·ÖÎö£¬¿ÉÖªc£¨´ý²â£©Æ«Ð¡£¬Éռ¶ÈƫС£¬¹ÊA´íÎó£»
B£®Óñê×¼ÒºÈóÏ´µÎ¶¨¹Ü£¬ÊÇÕýÈ·²Ù×÷£¬Ã»ÓÐÎó²î£¬¹ÊB´íÎó£»
C£®Óôý²âÒºÈóϴ׶ÐÎÆ¿£¬´ý²âÒºµÄÎïÖʵÄÁ¿Æ«´ó£¬Ôì³ÉV£¨±ê×¼£©Æ«´ó£¬¸ù¾Ýc£¨´ý²â£©=$\frac{V£¨±ê×¼£©¡Ác£¨±ê×¼£©}{V£¨´ý²â£©}$·ÖÎö£¬¿ÉÖªc£¨´ý²â£©Æ«´ó£¬Éռ¶ÈÆ«´ó£¬¹ÊCÕýÈ·£»
D£®²»Ð¡ÐĽ«±ê×¼ÒºµÎÔÚ׶ÐÎÆ¿ÍâÃ棬Ôì³ÉV£¨±ê×¼£©Æ«´ó£¬¸ù¾Ýc£¨´ý²â£©=$\frac{V£¨±ê×¼£©¡Ác£¨±ê×¼£©}{V£¨´ý²â£©}$·ÖÎö£¬¿ÉÖªc£¨´ý²â£©Æ«´ó£¬Éռ¶ÈÆ«´ó£¬¹ÊDÕýÈ·£»
E£®µÎ¶¨½Ó½üÖÕµãʱ£¬ÓÃÉÙÁ¿ÕôÁóË®³åϴ׶ÐÎÆ¿ÄÚ±Ú£¬´ý²âÒºµÄÎïÖʵÄÁ¿²»±ä£¬¶ÔV£¨±ê×¼£©ÎÞÓ°Ï죬¿ÉÖªc£¨´ý²â£©²»±ä£¬Éռ¶È²»±ä£¬¹ÊE´íÎó£»
¹ÊÑ¡CD£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËËá¼îÖк͵ζ¨ÊµÑé²Ù×÷¡¢¼ÆËãÓëÎó²î·ÖÎöÒÔ¼°µÎ¶¨¹ý³ÌÖÐÈÜÒºÀïÀë×ÓŨ¶È´óС±È½Ï£¬ÄѶÈÖеȣ¬ÄÜ׼ȷд³öÈÜÒºÀïµÄµçºÉÊغãʽÊǽâ´ð¹Ø¼ü£®