Question

8£®°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑéÊÒÖг£ÓÃNaOHÈÜÒºÀ´½øÐÐÏ´ÆøºÍÌá´¿£¬µ±100mL 3mol/L µÄNaOHÈÜÒºÎüÊÕ±ê×¼×´¿öÏÂ4.48L CO₂ʱ£¬ËùµÃÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨Na⁺£©£¾c£¨HCO₃^-£©£¾c£¨CO₃^2-£©£¾c£¨OH^-£©£¾c£¨H⁺£©
£¨2£©³£ÎÂÏÂÏòÒ»¶¨Ìå»ýµÄ0.1mol/L´×ËáÈÜÒºÖмÓˮϡÊͺó£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇCD£®
A£®ÈÜÒºÖе¼µçÁ£×ÓµÄÊýÄ¿¼õÉÙ   B£®´×ËáµÄµçÀë³Ì¶ÈÔö´ó£¬c£¨H⁺£©Ò²Ôö´ó
C£®ÈÜÒºÖÐ$\frac{c£¨C{H}_{3}CO{O}^{-}£©}{c£¨C{H}_{3}COOH£©•c£¨O{H}^{-}£©}$²»±ä    D£®ÈÜÒºÖÐ$\frac{c£¨C{H}_{3}CO{O}^{-}£©}{c£¨{H}^{+}£©}$ ¼õС
£¨3£©¢Ù³£ÎÂϽ«0.15mol/LÏ¡ÁòËáV₁ mLÓë0.1mol/L NaOHÈÜÒºV₂ mL»ìºÏ£¬ËùµÃÈÜÒºµÄpHΪ1£¬ÔòV₁£ºV₂=1£º1£¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©£®
¢Ú³£ÎÂÏ£¬ÈôÈÜÒºÓÉpH=3µÄHAÈÜÒºV₁ mLÓëpH=11µÄNaOHÈÜÒºV₂ mL»ìºÏ¶øµÃ£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇAD£®
A£®Èô·´Ó¦ºóÈÜÒº³ÊÖÐÐÔ£¬Ôòc£¨H⁺£©+c£¨OH^-£©=2¡Á10^-7 mol/L      B£®ÈôV₁=V₂£¬·´Ó¦ºóÈÜÒºpHÒ»¶¨µÈÓÚ7
C£®Èô·´Ó¦ºóÈÜÒº³ÊËáÐÔ£¬ÔòV₁Ò»¶¨´óÓÚV₂D£®Èô·´Ó¦ºóÈÜÒº³Ê¼îÐÔ£¬ÔòV₁Ò»¶¨Ð¡ÓÚV₂
£¨4£©³£ÎÂÏ£¬Å¨¶È¾ùΪ0.1mol/LµÄÏÂÁÐÎåÖÖÈÜÒºµÄpHÈç±íËùʾ£º

ÈÜÖÊ	CH3COONa	NaHCO3	Na2CO3	NaClO	NaCN
pH	8.8	9.7	11.6	10.3	11.1

¢Ù¸ù¾Ý±íÖÐÊý¾Ý£¬½«Å¨¶È¾ùΪ0.01mol/LµÄÏÂÁÐËÄÖÖËáµÄÈÜÒº·Ö±ðÏ¡ÊÍ100±¶£¬pH±ä»¯×îСµÄÊÇA£®
A£®HCNB£®HClOC£®H₂CO₃D£®CH₃COOH
¢Ú¸ù¾ÝÒÔÉÏÊý¾Ý£¬ÅжÏÏÂÁз´Ó¦¿ÉÒÔ³ÉÁ¢µÄÊÇAB£®
A£®CH₃COOH+Na₂CO₃¨TNaHCO₃+CH₃COONa     B£®CH₃COOH+NaCN¨TCH₃COONa+HCN
C£®CO₂+H₂O+2NaClO¨TNa₂CO₃+2HClO           D£®NaHCO₃+HCN¨TNaCN+H₂O+CO₂
£¨5£©¼¸ÖÖÀë×Ó¿ªÊ¼³ÁµíʱµÄpHÈçÏÂ±í£º

Àë×Ó	Fe2+	Cu2+	Mg2+
pH	7.6	5.2	10.4

µ±Ïòº¬ÏàͬŨ¶ÈCu²⁺¡¢Mg²⁺¡¢Fe²⁺µÄÈÜÒºÖеμÓNaOHÈÜҺʱ£¬Cu²⁺£¨ÌîÀë×Ó·ûºÅ£©ÏȳÁµí£¬K_sp[Fe£¨OH£©₂]£¼K_sp[Mg£¨OH£©₂]£¨Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©£®

Answer 1

·ÖÎö £¨1£©100mL 3mol/LµÄNaOHµÄÎïÖʵÄÁ¿=3mol/L¡Á0.1L=0.3mol£¬±ê×¼×´¿öÏÂ4.48LCO₂µÄÎïÖʵÄÁ¿=$\frac{4.48L}{22.4L/mol}$=0.2mol£¬ËùÒÔÇâÑõ»¯ÄƺͶþÑõ»¯Ì¼·´Ó¦·½³ÌʽΪ2CO₂+3NaOH=Na₂CO₃+NaHCO₃+H₂O£¬¸ù¾ÝÑÎÀàµÄË®½âºÍµçºÉÊغãÀ´·ÖÎö£¬×¢ÒâÑÎÀàµÄË®½âÊÇ΢ÈõµÄ£»
£¨2£©¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룬ÈÜÒºÖд×Ëá¸ùÀë×ÓŨ¶È¡¢´×Ëá·Ö×ÓŨ¶È¡¢ÇâÀë×ÓŨ¶È¶¼¼õС£¬ÇâÑõ¸ùÀë×ÓŨ¶ÈÔö´ó£»
£¨3£©¢ÙËùµÃÈÜÒºµÄpHֵΪ1£¬ÈÜÒºÖÐÇâÀë×ÓŨ¶ÈΪ1¡Á10^-1mol•L^-1£¬ËµÃ÷ÇâÀë×Ó¹ýÁ¿£¬¸ù¾ÝÌâÖÐÏ¡ÁòËáºÍÇâÑõ»¯ÄÆÈÜÒºÌå»ýÁÐʽ¼ÆËã³öV₁£ºV₂µÄ±ÈÖµ£»
¢ÚA£®Èô·´Ó¦ºóÈÜÒº³ÊÖÐÐÔ£¬Ôòc£¨H⁺£©=c£¨OH^-£©=1¡Á10^-7mol/L£»
B£®ÈôV₁=V₂£¬·´Ó¦ºóÈÜÒºpHµÈÓÚ7»òСÓÚ7£»
C£®Èô·´Ó¦ºóÈÜÒº³ÊËáÐÔ£¬V₁¿ÉÄܵÈÓÚV₂£»
D£®Èç¹ûÈÜÒº³Ê¼îÐÔ£¬ÔòÈÜÒº¿ÉÄÜÊǼîºÍÑÎÈÜÒº£¬Ò²¿ÉÄÜÖ»ÊÇÑÎÈÜÒº£»
£¨4£©¢ÙÏàͬŨ¶ÈµÄÄÆÑÎÈÜÒº£¬ËáµÄËáÐÔÔ½Èõ£¬ÔòËá¸ùÀë×ÓË®½â³Ì¶ÈÔ½´ó£¬¸ù¾ÝÄÆÑÎÈÜÒºpHÈ·¶¨ËáµÄÇ¿Èõ£»ÏàͬŨ¶ÈµÄ²»Í¬Ëᣬ¼ÓˮϡÊÍ´Ù½øÈõËáµçÀ룬ÔòÏ¡ÊÍÏàͬµÄ±¶Êý£¬ËáµÄËáÐÔÔ½Èõ£¬ËáÈÜҺϡÊ͹ý³ÌÖÐpH±ä»¯Ô½Ð¡£»
¢ÚËá¸ùÀë×ÓË®½â³Ì¶ÈԽǿ£¬ÔòËáµÄËáÐÔÔ½Èõ£¬½áºÏÇ¿ËáÖÆÈ¡ÈõËá·ÖÎö£»
£¨5£©Éú³É³ÁµíÐèÒªµÄpHԽС£¬Ôò¸ÃÀë×ÓÏȳÁµí£»¸ù¾ÝÀë×Ó»ý³£Êý¼ÆËãÆäÈܶȻý³£Êý£®

½â´ð ½â£º£¨1£©100mL 3mol/LµÄNaOHµÄÎïÖʵÄÁ¿=3mol/L¡Á0.1L=0.3mol£¬±ê×¼×´¿öÏÂ4.48LCO₂µÄÎïÖʵÄÁ¿=$\frac{4.48L}{22.4L/mol}$=0.2mol£¬ËùÒÔÇâÑõ»¯ÄƺͶþÑõ»¯Ì¼·´Ó¦·½³ÌʽΪ2CO₂+3NaOH=Na₂CO₃+NaHCO₃+H₂O£¬ÈÜÒºÖÐ̼Ëá¸ùÀë×ÓË®½â¶øʹÈÜÒº³Ê¼îÐÔ£¬¼´c£¨OH^-£©£¾c£¨H⁺£©£¬Ì¼Ëá¸ùÀë×ÓË®½âÉú³É̼ËáÇâ¸ùÀë×Ó£¬ËùÒÔc£¨HCO₃^-£©£¾c£¨CO₃^2-£©£¬ÓÉÓÚË®½â³Ì¶È½ÏС£¬ËùÒÔc£¨CO₃^2-£©£¾c£¨OH^-£©£¬¹ÊÈÜÒºÖи÷Àë×ÓŨ¶È´óС˳ÐòÊÇc£¨Na⁺£©£¾c£¨HCO₃^-£©£¾c£¨CO₃^2-£©£¾c£¨OH^-£©£¾c£¨H⁺£©£¬
¹Ê´ð°¸Îª£ºc£¨Na⁺£©£¾c£¨HCO₃^-£©£¾c£¨CO₃^2-£©£¾c£¨OH^-£©£¾c£¨H⁺£©£»
£¨2£©A£®¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룬ÔòÈÜÒºÖе¼µçÁ£×ÓµÄÊýÄ¿Ôö¼Ó£¬¹Ê´íÎ󣻡¡¡¡ ¡¡
B£®´×ËáµÄµçÀë³Ì¶ÈÔö´ó£¬µ«ÈÜÒºÖÐc£¨H^-£©¼õС£¬¹Ê´íÎó£»
C£®ÈÜÒºÖÐ$\frac{c£¨C{H}_{3}CO{O}^{-}£©}{c£¨C{H}_{3}COOH£©•c£¨O{H}^{-}£©}$=$\frac{{K}_{a}}{{K}_{W}}$£¬Î¶Ȳ»±ä£¬ÔòµçÀëƽºâ³£Êý¡¢Ë®µÄÀë×Ó»ý³£Êý²»±ä£¬ËùÒÔÈÜÒºÖÐ$\frac{c£¨C{H}_{3}CO{O}^{-}£©}{c£¨C{H}_{3}COOH£©•c£¨O{H}^{-}£©}$²»±ä£¬¹ÊÕýÈ·£»
D£®ÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÔö´ó£¬Ôò$\frac{c£¨C{H}_{3}CO{O}^{-}£©}{c£¨{H}^{+}£©}$¼õС£¬¹ÊÕýÈ·£»
¹ÊÑ¡CD£»
£¨3£©¢ÙÈÜÒºÖÐÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Îª£ºV₂¡Á10^-3L¡Á0.1mol/L=V₂¡Á10^-4mol£¬
Ï¡ÁòËáµÄÎïÖʵÄÁ¿Îª£ºV₁¡Á10^-3L¡Á0.15mol/L¡Á2=3V₁¡Á10^-4mol£¬
Á½ÈÜÒº»ìºÏºóÈÜÒºÏÔʾËáÐÔ£¬ÇâÀë×ÓŨ¶ÈΪ1¡Á10^-1mol•L^-1£¬
ÔòÓУº3V₁¡Á10^-4mol-V₂¡Á10^-4mol=£¨V₁+V₂£©¡Á10^-3L¡Á1¡Á10^-1mol•L^-1=£¨V₁+V₂£©¡Á10^-4mol£¬
½âµÃV₁£ºV₂=1£º1£¬
¹Ê´ð°¸Îª£º1£º1£»
¢ÚA£®Èô»ìºÏÈÜÒº³ÊÖÐÐÔ£¬ÔòÈÜÒºÖÐc£¨H⁺£©=c£¨OH^-£©=2¡Á10^-7 mol•L^-1£¬ËùÒÔÈÜÒºÖÐc£¨H⁺£©+c£¨OH^-£©=2¡Á10^-7 mol•L^-1£¬¹ÊÕýÈ·£»
B£®Èç¹û¶þÕßÌå»ýÏàµÈ£¬Èç¹ûËáÊÇÇ¿ËᣬÔò»ìºÏÈÜÒº³ÊÖÐÐÔ£¬Èç¹ûËáÊÇÈõËᣬÔò»ìºÏÈÜÒº³ÊËáÐÔ£¬¹Ê´íÎó£»
C£®Èç¹ûÈÜÒº³ÊËáÐÔ£¬ÔòÈÜÒºÊÇËáºÍÑÎÈÜÒº£¬ËáÊÇÈõËᣬËáŨ¶È´óÓÚÇâÑõ»¯ÄÆŨ¶È£¬ËùÒÔV₁²»Ò»¶¨´óÓÚV₂£¬¹Ê´íÎó£»
D£®Èç¹ûÈÜÒº³Ê¼îÐÔ£¬ÔòÈÜÒº¿ÉÄÜÊǼîºÍÑÎÈÜÒº£¬Ò²¿ÉÄÜÖ»ÊÇÑÎÈÜÒº£¬ÔòV₁Ò»¶¨Ð¡ÓÚV₂£¬¹ÊÕýÈ·£»
¹ÊÑ¡AD£»
£¨4£©¢ÙÏàͬŨ¶ÈµÄÄÆÑÎÈÜÒº£¬ËáµÄËáÐÔÔ½Èõ£¬ÔòËá¸ùÀë×ÓË®½â³Ì¶ÈÔ½´ó£¬¸ù¾ÝÄÆÑÎÈÜÒºpHÈ·¶¨ËáµÄÇ¿Èõ£»ÏàͬŨ¶ÈµÄ²»Í¬Ëᣬ¼ÓˮϡÊÍ´Ù½øÈõËáµçÀ룬ÔòÏ¡ÊÍÏàͬµÄ±¶Êý£¬ËáµÄËáÐÔÔ½Èõ£¬ËáÈÜҺϡÊ͹ý³ÌÖÐpH±ä»¯Ô½Ð¡£¬¸ù¾ÝÄÆÑÎÈÜÒºµÄpHÖª£¬HCN¡¢HClO¡¢CH₃COOH¡¢H₂CO₃ËÄÖÖËáµÄËáÐÔ´óС˳ÐòÊÇCH₃COOH£¾H₂CO₃£¾HClO£¾HCN£¬ËùÒÔÈÜÒºµÄpH±ä»¯×îСµÄÊÇHCN£¬¹ÊÑ¡A£»
¢ÚHCN¡¢HClO¡¢CH₃COOH¡¢H₂CO₃ËÄÖÖËáµÄËáÐÔ´óС˳ÐòÊÇCH₃COOH£¾H₂CO₃£¾HClO£¾HCN£¬¸ù¾ÝÇ¿ËáÖÆÈ¡ÈõËáÖª£¬
A£®ËáÐÔCH₃COOH£¾H₂CO₃£¾HCO₃^-£¬ËùÒÔ¶þÕß·´Ó¦ÎªCH₃COOH+Na₂CO₃¨TNaHCO₃+CH₃COONa£¬¹ÊÕýÈ·£»  
B£®´×ËáËáÐÔ´óÓÚHCN£¬ËùÒÔ¶þÕß·´Ó¦ÎªCH₃COOH+NaCN¨TCH₃COONa+HCN£¬¹ÊÕýÈ·£»
C£®Ì¼ËáËáÐÔ´óÓÚ´ÎÂÈËᣬËùÒÔ¶þÕß·´Ó¦ÎªCO₂+H₂O+NaClO¨TNaHCO₃+NaClO£¬¹Ê´íÎó£»     
D£®ËáÐÔH₂CO₃£¾HClO£¾HCN£¬ËùÒÔNaHCO₃¡¢HCN¶þÕß²»·´Ó¦£¬¹Ê´íÎó£»
¹ÊÑ¡AB£»
£¨5£©Cu²⁺¿ªÊ¼³ÁµíµÄpH×îС£¬ËùÒÔCu²⁺ÏȳÁµí£»Mg²⁺¡¢Fe²⁺¿ªÊ¼³ÁµíµÄC£¨OH^-£©·Ö±ðΪ£º10^-3.6mol•L^-1¡¢10^-6.4mol•L^-1£¬ËùÒÔK_SP[£¨Fe£¨OH£©₂]£¼K_SP[£¨Mg£¨OH£©₂]£¬¹Ê´ð°¸Îª£ºCu²⁺£»£¼£®

µãÆÀ ±¾Ì⿼²é½Ï×ۺϣ¬Éæ¼°Àë×ÓŨ¶È´óС±È½Ï¡¢Èõµç½âÖʵĵçÀë¡¢ÄÑÈÜÎïµÄÈܽâƽºâµÈ֪ʶµã£¬¸ù¾ÝÈÜÒºÖеÄÈÜÖʼ°ÆäÐÔÖÊÈ·¶¨ÈÜÒºÖÐÀë×ÓŨ¶È´óС£¬ÖªµÀËáµÄÇ¿ÈõÓëËá¸ùÀë×ÓË®½â³Ì¶ÈµÄ¹Øϵ£¬ÖªµÀÈõµç½âÖÊÏ¡Ê͹ý³ÌÖи÷ÖÖÀë×ÓŨ¶È±ä»¯£¬ÌâÄ¿ÄѶÈÖеȣ®