11£®³£ÎÂÏ£¬Ïò25mL 0.1mol/L MOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2mol/L HAÈÜÒº£¬ÇúÏßÈçͼËùʾ£¨Ìå»ý±ä»¯ºöÂÔ²»¼Æ£©£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öMOHµÄµçÀë·½³Ìʽ£ºMOH¨TM++OH-
£¨2£©MOHÓëHAÇ¡ºÃÍêÈ«·´Ó¦Ê±£¬ÈÜÒº³Ê¼îÐÔ£¨Ìî¡°Ëᡱ¡¢¡°¼î¡±»ò¡°ÖС±£©£¬ÀíÓÉÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£ºA-+H2O?HA+OH -£»´Ëʱ£¬»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©£¾0.2mol/L HAÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©£®
£¨3£©Ð´³öBµã£¬»ìºÏÈÜÒºÖи÷Àë×ÓŨ¶ÈµÄ´óС¹ØϵBµãc£¨M+£©=c£¨A-£©£¾c£¨H+£©=c£¨OH-£©£®
£¨4£©Dµãʱ£¬ÈÜÒºÖÐc£¨A-£©+c£¨HA£©=2c£¨M+£©£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®

·ÖÎö £¨1£©ÓÉͼÏó¿ÉÖª0.1mol/L MOHÈÜÒºµÄpH=13£¬c£¨OH-£©=0.1mol/L£¬ËµÃ÷Ϊǿ¼î£¬ÍêÈ«µçÀ룻
£¨2£©ÓÉͼÏó¿ÉÖª£¬µ±¼ÓÈë13mL 0.2mol/L HAÈÜҺʱ£¬n£¨HA£©=0.0026mol£¬¶øn£¨MOH£©=0.0025mol£¬ËµÃ÷HA¹ýÁ¿£¬µ«ÈÜÒº³ÊÖÐÐÔ£¬ËµÃ÷HAΪÈõËᣬÈç¶þÕßÇ¡ºÃ·´Ó¦£¬Ó¦Éú³ÉÇ¿¼îÈõËáÑΣ¬´Ù½øË®µÄµçÀ룻
£¨3£©BµãÈÜÒº³ÊÖÐÐÔ£¬½áºÏµçºÉÊغãÅжϣ»
£¨4£©Dµã·´Ó¦µÃµ½µÈŨ¶ÈµÄMA¡¢HAÈÜÒº£¬½áºÏÎïÁÏÊغãÅжϣ®

½â´ð ½â£º£¨1£©ÓÉͼÏó¿ÉÖª0.1mol/L MOHÈÜÒºµÄpH=13£¬c£¨OH-£©=0.1mol/L£¬ËµÃ÷Ϊǿ¼î£¬ÍêÈ«µçÀ룬ÔòµçÀë·½³ÌʽΪMOH¨TM++OH-£¬
¹Ê´ð°¸Îª£ºMOH¨TM++OH-£»
£¨2£©ÓÉͼÏó¿ÉÖª£¬µ±¼ÓÈë13mL 0.2mol/L HAÈÜҺʱ£¬n£¨HA£©=0.0026mol£¬¶øn£¨MOH£©=0.0025mol£¬ËµÃ÷HA¹ýÁ¿£¬µ«ÈÜÒº³ÊÖÐÐÔ£¬ËµÃ÷HAΪÈõËᣬÈç¶þÕßÇ¡ºÃ·´Ó¦£¬Ó¦Éú³ÉÇ¿¼îÈõËáÑΣ¬Ë®½â³Ê¼îÐÔ£¬·¢ÉúA-+H2O?HA+OH-£¬´Ù½øË®µÄµçÀ룬Ôò»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©´óÓÚ0.2mol/L HAÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©£¬
¹Ê´ð°¸Îª£º¼î£»A-+H2O?HA+OH -£»£¾£»
£¨3£©BµãÈÜÒº³ÊÖÐÐÔ£¬Ôòc£¨H+£©=c£¨OH-£©£¬½áºÏµçºÉÊغ㣺c£¨M+£©+c£¨H+£©=c£¨A-£©+c£¨OH-£©£¬¿ÉÖª£ºc£¨M+£©=c£¨A-£©£¬ÈÜÖÊÀë×ÓŨ¶ÈÔ¶Ô¶µÍÓÚÇâÀë×ÓŨ¶È£¬ÈÜÒºÖдæÔÚc£¨M+£©=c£¨A-£©£¾c£¨H+£©=c£¨OH-£©£¬
¹Ê´ð°¸Îª£ºc£¨M+£©=c£¨A-£©£¾c£¨H+£©=c£¨OH-£©£»
£¨4£©Dµã·´Ó¦µÃµ½µÈŨ¶ÈµÄMA¡¢HAÈÜÒº£¬AÔªËØÒÔA-ºÍHAÁ½ÖÖÐÎʽ´æÔÚ£¬¶øMÈ«ÒÔM+ÐÎʽ´æÔÚ£¬ÓÉÎïÁÏÊغã¿ÉµÃ£ºc£¨A-£©+c£¨HA£©=2c£¨M+£©£¬
¹Ê´ð°¸Îª£º=£®

µãÆÀ ±¾Ì⿼²éÁËËá¼î»ìºÏÈÜÒº¶¨ÐÔÅжϡ¢ÑÎÀàË®½â¡¢Àë×ÓŨ¶È´óС±È½ÏµÈ£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦µÄ¿¼²é£¬Ã÷ȷͼÏóÖи÷¸öµãµÄº¬ÒåÊǽⱾÌâ¹Ø¼ü£¬×¢ÒâµçºÉÊغ㡢ÎïÁÏÊغãÔÚÀë×ÓŨ¶ÈµÈÁ¿¹Øϵ±È½ÏÖÐÓ¦Óã®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

13£®×©ÍßÊÇÓú¬ÌúµÈÔªËصÄð¤ÍÁ¸ô¾ø¿ÕÆøÉÕÖƶø³ÉµÄ£®µ±ÉÕÒ¤×÷ÒµÁÙ½ü½áÊøʱ£¬ÈôÓÃÁÜË®µÄ°ì·¨À´½µµÍζȣ¬Ò¤ÄÚ´¦ÓÚ»¹Ô­ÐÔÆø·Õ£¬Ôòש³ÊÇàÉ«£¬³ÆÇàש£®ÈôÓÃͱ¿ªÒ¤¶¥×ÔÈ»ÀäÈ´µÄ°ì·¨£¬Ôòש³ÊºìÉ«£¬³Æºìש£®ÓÉ´ËÅжÏש¿éÖеijɷ֣¬´íÎóµÄÊÇ£¨¡¡¡¡£©
A£®×©¿éÖеÄÖ÷Òª³É·ÖÊǹèËáÑÎ
B£®ÇàשÖеÄÌúÔªËØÖ÷ÒªÒÔÇâÑõ»¯ÑÇÌúµÄÐÎʽ´æÔÚ
C£®ºìשÖеÄÌúÔªËØÖ÷ÒªÒÔÑõ»¯ÌúµÄÐÎʽ´æÔÚ
D£®ÇàשÖеÄÌúÔªËØÖ÷ÒªÒÔÑõ»¯ÑÇÌúµÄÐÎʽ´æÔÚ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

2£®×¼È·ÒÆÈ¡25.00mLijδ֪Ũ¶ÈµÄNaOHÈÜÒºÓÚÒ»½à¾»×¶ÐÎÆ¿ÖУ¬È»ºóÓÃ0.20mol/LµÄÑÎËáÈÜÒºµÎ¶¨£¨Ö¸Ê¾¼ÁΪ¼×»ù³È£©£¬µÎ¶¨½á¹ûÈçÏ£º
HClÈÜÒºÆðʼ¶ÁÊýHClÈÜÒºÖÕµã¶ÁÊý
µÚÒ»´Î2.15mL
µÚ¶þ´Î3.10mL21.85mL
µÚÈý´Î4.20mL22.95mL
£¨1£©ÆäÖеÚÒ»´ÎµÎ¶¨´ïµ½ÖÕµãʱ£¬µÎ¶¨¹ÜÖеÄÒºÃæÈçÉÏͼËùʾ£¬ÆäÕýÈ·µÄ¶ÁÊýÊÇ22.40mL£®
£¨2£©ÅÅÈ¥ËáʽµÎ¶¨¹ÜÖÐÆøÅݵľßÌå²Ù×÷Êǽ«ËáʽµÎ¶¨¹ÜÉÔÉÔÇãб£¬Ñ¸ËÙ´ò¿ª»îÈû£¬ÆøÅÝËæÈÜÒºµÄÁ÷³ö¶ø±»Åųö£»£®
£¨3£©¸ù¾ÝÒÔÉÏÊý¾Ý¿É¼ÆËã³öNaOHÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.15mol/L£®£¨¾«È·µ½0.01£©
£¨4£©µÎ¶¨µ½ÖÕµãʱ׶ÐÎÆ¿ÖÐÈÜÒºµÄÑÕÉ«±ä»¯ÊÇÓÉ»ÆÉ«±äΪ³ÈÉ«£®
£¨5£©ÏÂÁвÙ×÷»áµ¼Ö´ý²âNaOHÈÜҺŨ¶ÈÆ«¸ßµÄÊÇAC£¨ÌîÐòºÅ£©£®
A£®µÎ¶¨¹ÜÓÃÕôÁóˮϴºóδÓñê×¼ËáÒºÈóÏ´£¬Ö±½Ó×°Èë±ê×¼ËáÒº
B£®µÎ¶¨¹Ü¶ÁÊýʱ£¬µÎ¶¨Ç°ÑöÊӵζ¨ºó¸©ÊÓ
C£®µÎ¶¨Ç°µÎ¶¨¹Ü¼â¶ËÓÐÆøÅÝ£¬µÎºóÆøÅÝÏûʧ
D£®µÎ¶¨µ½Öյ㸽½üʱ£¬ÓÃÉÙÁ¿ÕôÁóË®³åϴ׶ÐÎÆ¿ÄÚ±ÚÉÏÕ´¸½µÄÈÜÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

19£®ÓÃO2½«HClת»¯ÎªCl2£¬¿ÉÌá¸ßЧÒ棬¼õÉÙÎÛȾ£¬
£¨1£©´«Í³ÉϸÃת»¯Í¨¹ýÈçͼËùʾµÄ´ß»¯¼ÁÑ­»·ÊµÏÖ£¬ÆäÖУ¬·´Ó¦¢ÙΪ£º2HCl£¨g£©+CuO£¨s£©?H2O£¨g£©+CuCl2£¨g£©¡÷H1·´Ó¦¢ÚÉú³É1molCl2£¨g£©µÄ·´Ó¦ÈÈΪ¡÷H2£¬Ôò×Ü·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ2HCl£¨g£©+$\frac{1}{2}$O2£¨g£©?H2O£¨g£©+Cl2£¨g£©£¬¡÷H=£¨¡÷H1+¡÷H2£©kJ/mol£¬£¨·´Ó¦ÈÈÓá÷H1ºÍ¡÷H2±íʾ£©£®
£¨2£©ÐÂÐÍRuO2´ß»¯¼Á¶ÔÉÏÊöHClת»¯ÎªCl2µÄ×Ü·´Ó¦¾ßÓиüºÃµÄ´ß»¯»îÐÔ£¬
¢ÙʵÑé²âµÃÔÚÒ»¶¨Ñ¹Ç¿Ï£¬×Ü·´Ó¦µÄHClƽºâת»¯ÂÊËæζȱ仯µÄaHCl-TÇúÏßÈçͼ£¬Ôò×Ü·´Ó¦µÄ¡÷H£¼0£¬£¨Ìî¡°£¾¡±¡¢¡°©„¡±»ò¡°£¼¡±£©£»A¡¢BÁ½µãµÄƽºâ³£ÊýK£¨A£©ÓëK£¨B£©ÖнϴóµÄÊÇK£¨A£©£®
¢ÚÔÚÉÏÊöʵÑéÖÐÈôѹËõÌå»ýʹѹǿÔö´ó£¬»­³öÏàÓ¦aHCl-T£¨HClµÄת»¯ÂÊÓëζȵĹØϵ£©ÇúÏßµÄʾÒâͼ£¬²¢¼òҪ˵Ã÷ÀíÓÉ£ºÔö´óѹǿ£¬Æ½ºâÓÒÒÆ£¬¨»HClÔö´ó£¬ÏàͬζÈÏ£¬HClµÄƽºâת»¯ÂʱÈ֮ǰʵÑéµÄ´ó£®
¢ÛÏÂÁдëÊ©ÖÐÓÐÀûÓÚÌá¸ßaHClµÄÓÐBD£®
A¡¢Ôö´ón£¨HCl£©       B¡¢Ôö´ón£¨O2£©   C¡¢Ê¹ÓøüºÃµÄ´ß»¯¼Á     D¡¢ÒÆÈ¥H2O
£¨3£©Ò»¶¨Ìõ¼þϲâµÃ·´Ó¦¹ý³ÌÖÐn£¨Cl2£©µÄÊý¾ÝÈçÏ£º
t£¨min£©02.04.06.08.0
n£¨Cl2£©/10-3mol01.83.75.47.2
¼ÆËã2.0¡«6.0minÄÚÒÔHClµÄÎïÖʵÄÁ¿±ä»¯±íʾµÄ·´Ó¦ËÙÂÊ£¨ÒÔmol•min-1Ϊµ¥Î»£¬Ð´³ö¼ÆËã¹ý³Ì£©£®
£¨4£©Cl2ÓÃ;¹ã·º£¬Ð´³öÓÃCl2ÖƱ¸Æ¯°×·ÛµÄ»¯Ñ§·½³Ìʽ2Cl2+2Ca£¨OH£©2=CaCl2+Ca£¨ClO£©2+2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®Ä³Î¶ÈϵÄÈÜÒºÖÐc£¨H+£©=10xmol/L£¬c£¨OH-£©=10ymol/L£®xÓëyµÄ¹ØϵÈçͼËùʾ£ºÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¸ÃζȸßÓÚ25¡æ
B£®Í¼ÖÐaµãÈÜÒº³Ê¼îÐÔ
C£®¸ÃζÈÏ£¬0.01 mol•L-1µÄHClÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄH+Ũ¶ÈΪ10-12 mol•L-1
D£®¸ÃζÈÏ£¬µÈÌå»ýŨ¶È¾ùΪ0.01 mol•L-1µÄHClÈÜÒºÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦µÄpH=7

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

16£®°Ñ0.02mol/L  CH3COOHÈÜÒººÍ0.01mol/L  NaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Ôò»ìºÏÈÜÒºÖÐ΢Á£Å¨¶È¹ØϵÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®£¨CH3COO-£©£¼c£¨Na+£©B£®c£¨CH3COOH£©£¾c£¨CH3COO-£©
C£®2c£¨H+£©=c£¨CH3COO-£©-c£¨CH3COOH£©D£®c£¨CH3COOH£©+c£¨CH3COO-£©=0.01mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®³£ÎÂÏ£¬½«Ò»¶¨Å¨¶ÈµÄHAºÍHB·Ö±ðÓë0.10mol•L-1µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ÊµÑé¼Ç¼ÈçÏÂ±í£º
»ìºÏºóµÄÈÜÒº¼ÓÈëµÄËá¡¡¡¡ËáµÄŨ¶È/£¨mol•L-1£©¡¡¡¡»ìºÏºóÈÜÒºµÄpH
¡¡¢Ù¡¡HA0.10¡¡¡¡8.7
¡¡¢ÚHB0.122
ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®HAÊÇÇ¿ËᣬHBÊÇÈõËá
B£®Éý¸ßζȣ¬ÈÜÒº¢ÚÖÐ$\frac{c£¨{B}^{-}£©}{c£¨N{a}^{+}£©}$Ôö´ó
C£®ÈÜÒº¢ÙÖÐÀë×ÓŨ¶ÈµÄ¹Øϵ£ºc£¨A-£©£¾c£¨Na+£©£¾c£¨OH-£©£¾c£¨H+£©
D£®ÈÜÒº¢ÚÖÐÀë×ÓŨ¶ÈµÄ¹Øϵ£ºc£¨Na+£©+c£¨H+£©+c£¨B-£©=0.12 mol•L-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

20£®³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHAºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºµÄpHÈçÏÂ±í£º
ʵÑé
񅧏
HAÎïÖʵÄÁ¿
Ũ¶È£¨mol•
L-1£©
NaOH땅浀
Á¿Å¨¶È£¨mol•
L-1£©
»ìºÏÈÜÒºµÄ
pH
¼×0.20.2pH=a
ÒÒc10.2pH=7
±û0.20.1pH£¾7
¶¡0.10.1pH=9
Çë»Ø´ð£º
£¨1£©²»¿¼ÂÇÆäËû×éµÄʵÑé½á¹û£¬µ¥´Ó¼××éÇé¿ö·ÖÎö£¬
Èôa=7£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬ÔòHAΪǿË᣻
Èôa£¾7£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬ÔòHAΪÈõËᣮ
£¨2£©ÔÚÒÒ×éÖлìºÏÈÜÒºÖÐÀë×ÓŨ¶Èc£¨A-£©Óëc£¨Na+£©µÄ´óС¹ØϵÊÇC
A£®Ç°Õß´ó  B£®ºóÕß´óC£®¶þÕßÏàµÈ  D£®ÎÞ·¨ÅжÏ
£¨3£©´Ó±û×éʵÑé½á¹û·ÖÎö£¬HAÊÇÈõËᣨÌî¡°Ç¿¡±»ò¡°Èõ¡±£©£®¸Ã»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇc£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©£®
£¨4£©¶¡×éʵÑéËùµÃ»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©=10-5mol•L-1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®·Ö×ÓʽΪC7H12O4£¬ÆäÖÐÖ»º¬¶þ¸ö-COOCH3»ùÍŵÄͬ·ÖÒì¹¹Ì壨²»¿¼ÂÇÊÖÐÔÒì¹¹£©¹²ÓУ¨¡¡¡¡£©
A£®4ÖÖB£®5ÖÖC£®6ÖÖD£®7ÖÖ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸