12£®£¨1£©ÏàͬζÈϵÈÎïŨ¶ÈµÄÏÂÁÐÈÜÒºÖУ®
A£®NH4C1   B£®NH4HCO3   C£®NH4HSO4    D£®£¨NH4£©2SO4
¢ÙpHÖµÓÉ´óµ½Ð¡µÄ˳ÐòÊÇB£¾A£¾D£¾C£¨ÓöÔÓ¦µÄ×ÖĸÌîд£©£®
¢ÚNH4+Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇD£¾C£¾A£¾B£¨ÓöÔÓ¦µÄ×ÖĸÌîд£©£®
¢ÛÈôÉÏÊöÈÜÒºµÄpHÖµÏàͬ£¬ÔòÆäÎï³ÉµÄÁ¿Å¨¶È¹ØÏµÊÇC£¼D£¼A£¼B£¨ÓöÔÓ¦µÄ×ÖĸÌîд£©£®
£¨2£©25¡æ£®Ïò50mL 0.018mol•L-1 µÄ AgNO3ÈÜÒºÖмÓÈë50mL 0.020mol•L-lµÄÑÎËᣬÉú³É³Áµí£®¼ºÖª¸ÃζÈÏÂAgClµÄKsp=l.0¡Ál0-10£¬ºöÂÔÈÜÒºµÄÌå»ý±ä»¯£¬Çë¼ÆË㣺
¢ÙÍêÈ«³Áµíºó£¬ÈÜÒºÖÐc£¨Ag+£©=1.0¡Á10-7mol/L£®
¢ÚÍêÈ«³Áµíºó£¬ÈÜÒºµÄpH=2£®
£¨3£©ÈõËá¼°ÆäÑÎÔÚË®ÖдæÔÚ¶àÖÖÆ½ºâ¹ØÏµ£®¼ºÖªNaAË®ÈÜÒº³Ê¼îÐÔ£¬³£ÎÂϽ«0£®lmolNaAºÍ0.05molHClÈÜÓÚË®£¬µÃµ½1LÈÜÒº£®
¢Ù¼ºÖª¸Ã»ìºÏÈÜҺΪÈõËáÐÔ£¬ÔòÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨Na+£©£¾c£¨A-£©£¾c£¨Cl-£©£¾c£¨H+£©£¾c£¨OH-£©£®
¢ÚÏò¸Ã»ìºÏÎïÖÐÔÙ¼Ó0.03molNaOH£¬ÈÜÒºÖÐn£¨A-£©+n£¨OH-£©-n£¨H+£©=0.08£®

·ÖÎö £¨1£©A£®ï§¸ùÀë×ÓË®½â£¬ÈÜÒº³ÊËáÐÔ£¬B£®Ì¼ËáÇâ¸ùÀë×Ó´Ù½øï§¸ùÀë×ÓË®½â£¬ËáÐÔ½ÏÂÈ»¯ï§ÈÜÒºËáÐÔÈõ£¬C£®µçÀë³öÇâÀë×Ó£¬²¢ÒÖÖÆï§¸ùÀë×ÓË®½â£¬ÏÔÇ¿ËáÐÔ£¬D£®ï§¸ùÀë×ÓË®½â£¬Ë®½â³Ì¶È²»´ó£¬ï§¸ùÀë×ÓŨ¶È×î´ó£¬ËáÐÔ±ÈÂÈ»¯ï§ÈÜÒºµÄËáÐÔÇ¿£»
£¨2£©¢ÙÒÀ¾ÝÌâ¸ÉÊý¾Ý¼ÆËã³ÁµíºóÈÜÒºÖÐÂÈÀë×ÓŨ¶È£¬½áºÏKsp¼ÆËãÈÜÒºÖÐÒøÀë×ÓŨ¶È£»
¢Ú³ÁµíºóÈÜÒºÖÐÊ£ÓàÇâÀë×ÓŨ¶È£¬¼ÆËãpH£»
£¨3£©¢ÙNaAË®ÈÜÒº³Ê¼îÐÔ£¬ËµÃ÷A-Àë×ÓË®½â£¬HAÊÇÈõËᣬ³£ÎÂϽ«0.10molNaAºÍ0.05molHClÈÜÓÚË®£¬·¢Éú·´Ó¦£¬NaA+HCl=NaCl+HA£¬ÒÀ¾Ý¶¨Á¿¼ÆËãµÃµ½ÈÜÒºÖк¬ÓÐ0.05molNaA£¬µÃ0.05molHA£¬0.05molNaCl£»µÃµ½pH£¼7µÄÈÜÒº£¬ËµÃ÷ÈÜÒº³ÊËáÐÔ£¬Í¬Å¨¶ÈÈÜÒºÖÐHAµçÀë´óÓÚA-Àë×ÓµÄË®½â£»ÒÀ¾ÝÈÜÒºÖеçºÉÊØºã£¬µçºÉÊØºã£¬Ëá¼îÐÔ·Ö±ð¼ÆËã·ÖÎöÅжϣ»
¢Ú¸ù¾ÝµçºÉÊØºã¿ÉµÃn£¨A-£©+n£¨OH-£©-n£¨H+£©=c£¨Na+£©-c£¨Cl-£©£¬È»ºó¸ù¾ÝÄÆÀë×Ó¡¢ÂÈÀë×Ó×ÜÎïÖʵÄÁ¿½øÐмÆË㣻

½â´ð ½â£ºA£®ï§¸ùÀë×ÓË®½â£¬ÈÜÒº³ÊËáÐÔ£¬B£®Ì¼ËáÇâ¸ùÀë×Ó´Ù½øï§¸ùÀë×ÓË®½â£¬ËáÐÔ½ÏÂÈ»¯ï§ÈÜÒºËáÐÔÈõ£¬C£®µçÀë³öÇâÀë×Ó£¬²¢ÒÖÖÆï§¸ùÀë×ÓË®½â£¬ÏÔÇ¿ËáÐÔ£¬D£®ï§¸ùÀë×ÓË®½â£¬Ë®½â³Ì¶È²»´ó£¬ï§¸ùÀë×ÓŨ¶È×î´ó£¬ËáÐÔ±ÈÂÈ»¯ï§ÈÜÒºµÄËáÐÔÇ¿£®
¢ÙpHÖµÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£ºB£¾A£¾D£¾C£¬¹Ê´ð°¸Îª£ºB£¾A£¾D£¾C£»
¢ÚNH4+Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£ºD£¾C£¾A£¾B£¬¹Ê´ð°¸Îª£ºD£¾C£¾A£¾B£»
¢ÛÏàͬζÈÏ£¬ÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄÉÏÊöÈÜÒºpHÖµÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£ºB£¾A£¾D£¾C£¬ÈôÉÏÊöÈÜÒºµÄpHÖµÏàͬ£¬ÔòÆäÎï³ÉµÄÁ¿Å¨¶È¹ØÏµÓ¦ÊÇC£¼D£¼A£¼B£¬¹Ê´ð°¸Îª£ºC£¼D£¼A£¼B£»
£¨2£©¢ÙÏò50ml0.018mol/LµÄAgNO3ÈÜÒºÖмÓÈë50ml0.02mol/LµÄÑÎËᣬ»ìºÏºóÁ½ÕßŨ¶È·Ö±ðΪc£¨Ag+£©=$\frac{0.018}{2}$mol/L=0.009 mol/L¡¢c£¨Cl-£©=$\frac{0.02}{2}$mol/L=0.01mol/L£¬Ag+ÓëCl-ÊǵÈÎïÖʵÄÁ¿·´Ó¦µÄ£¬´ËʱCl-¹ýÁ¿£¬¹ýÁ¿Å¨¶Èc£¨Cl-£©=0.01mol/L-0.009mol/L=0.001mol/L£¬
Ksp=c£¨Ag+£©¡ÁC£¨Cl-£©=1.0¡Á10-10£¬Ôò£ºc£¨Ag+£©=$\frac{1.0¡Á1{0}^{-10}}{0.001}$mol/L=1.0¡Á10-7mol/L£¬
¹Ê´ð°¸Îª£º1.0¡Á10-7mol/L£»
¢Ú³ÁµíÉú³Éºó£¬ÇâÀë×ÓŨ¶Èc=$\frac{0.02mol/L}{2}$=0.01mol/L£¬ËùÒÔpH=2£¬
¹Ê´ð°¸Îª£º2£»
£¨3£©¢Ù³£ÎÂϽ«0.10mol NaAºÍ0.05mol HClÈÜÓÚË®£¬µÃµ½1LÈÜÒº£¬µÃµ½µÄÈÜÒºÖк¬ÓÐ0.05molNaA¡¢0.05molHA¡¢0.05molNaCl£»ÈÜÒºµÄpH£¼7£¬ËµÃ÷ÈÜÒº³ÊËáÐÔ£¬Í¬Å¨¶ÈÈÜÒºÖÐHAµçÀë´óÓÚA-Àë×ÓµÄË®½â£¬c£¨A-£©£¾c£¨Cl-£©¡¢c£¨H+£©£¾c£¨OH-£©£¬ÔòÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc£¨Na+£©£¾c£¨A-£©£¾c£¨Cl-£©£¾c£¨H+£©£¾c£¨OH-£©£¬
¹Ê´ð°¸Îª£ºc£¨Na+£©£¾c£¨A-£©£¾c£¨Cl-£©£¾c£¨H+£©£¾c£¨OH-£©£»
¢Ú¸ù¾Ý»ìºÏÒºÖеçºÉÊØºã¿ÉµÃ£ºn£¨A-£©+n£¨OH-£©+c£¨Cl-£©=c£¨Na+£©+n£¨H+£©£¬Ôòn£¨A-£©+n£¨OH-£©-n£¨H+£©=c£¨Na+£©-c£¨Cl-£©=0.03mol+0.10nol-0.05mol=0.08mol£¬
¹Ê´ð°¸Îª£º0.08£®

µãÆÀ ±¾±¾Ì⿼²éÁËÈõµç½âÖʵĵçÀë¼°ÆäÓ°Ïì¡¢ÈÜÒºÖÐÀë×ÓŨ¶È´óС±È½Ï£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕÓ°ÏìÈõµç½âÖʵĵçÀëÆ½ºâµÄÒòËØ£¬Äܹ»¸ù¾ÝµçºÉÊØºã¡¢ÎïÁÏÊØºã¡¢ÑεÄË®½âÅжÏÈÜÒºÖи÷Àë×ÓŨ¶È´óС£¬ÊÔÌâÅàÑøÁËѧÉúÁé»îÓ¦ÓÃËùѧ֪ʶµÄÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

2£®ÏòÒ»¶¨Á¿µÄÌúÍ­»ìºÏ·ÛÄ©ÖмÓÈëÏ¡ÏõËᣬ³ä·Ö·´Ó¦ºó£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÈôÏõËáÓÐÊ£Ó࣬ÈÜÒºÖдæÔڵĽðÊôÑôÀë×ÓÓÐFe2+¡¢Cu2+
B£®ÈôÏõËáÓÐÊ£Ó࣬ÈÜÒºÖдæÔڵĽðÊôÑôÀë×Ó¿ÉÄÜÊÇFe3+¡¢Cu2+
C£®Èô½ðÊôÓëÏõËá¾ùÎÞÊ£Ó࣬ÈÜÒºÖдæÔڵĽðÊôÑôÀë×ÓÓÐFe3+¡¢Fe2+¡¢Cu2+
D£®Èô½ðÊôÓÐÊ£Ó࣬ʣÓà½ðÊôÖÐÒ»¶¨ÓÐÌú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

3£®¶ÌÖÜÆÚÖ÷×åÔªËØM¡¢R¡¢X¡¢Y¡¢ZµÄºËµçºÉÊýÒÀ´ÎÔö´ó£®ÆäÖÐMµÄ×åÐòÊýºÍÔ­×ÓÐòÊýÏàµÈ£¬RµÄµ¥ÖÊÔÚ¿ÕÆøÖеÄÌå»ý·ÖÊý×î´ó£¬X¡¢Y¡¢ZͬÖÜÆÚ£¬ÇÒXn+Àë×ÓÔÚͬÖÜÆÚÔªËØÀë×ÓÖа뾶×îС£¬YµÄÑõ»¯ÎïÊÇÐγÉËáÓêµÄÖ÷ÒªÎïÖÊ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÆøÌ¬Ç⻯ÎïµÄÈÈÎȶ¨ÐÔ£ºY£¾Z
B£®YÔªËØµÄÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÒ»¶¨ÊÇÇ¿Ëá
C£®MÓëR¿ÉÐγɺ¬·Ç¼«ÐÔ¼üµÄ»¯ºÏÎï
D£®¹ÌÌåX2Y3¿ÉÔÚË®ÈÜÒºÖÐÖÆÈ¡

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®ÏÂÁÐʵÑéÄÜ´ïµ½¶ÔӦʵÑéÄ¿µÄÊÇ£¨¡¡¡¡£©
A£®ÓÃ×°Öü×Íê³ÉʵÑéÊÒÖÆÈ¡ÂÈÆø
B£®ÓÃ×°ÖÃÒÒ³ýÈ¥¶þÑõ»¯Ì¼ÖеÄÉÙÁ¿ÂÈ»¯Çâ
C£®ÓÃ×°Öñû·ÖÀë¶þÑõ»¯Ã̺ÍÂÈ»¯ÃÌÈÜÒº
D£®ÓÃ×°Öö¡Õô¸É̼ËáÇâÄÆÈÜÒºÖÆNaHCO3¹ÌÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®¹ØÓÚµç½â¾«Á¶Í­£¨´ÖÍ­Öк¬ÓÐFe¡¢Zn¡¢Ni¡¢Ag¡¢AuÉÙÁ¿ÔÓÖÊ£©£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Óô¿Í­×÷Ñô¼«¡¢´ÖÍ­×÷Òõ¼«
B£®µç½âÒºµÄ³É·Ö±£³Ö²»±ä
C£®Íͽâ¹ý³ÌÖУ¬Ñô¼«ÖÊÁ¿µÄ¼õÉÙÓëÒõÃ÷¼«ÖÊÁ¿µÄÔö¼ÓÏàµÈ
D£®Òõ¼«µç¼«·´Ó¦Îª£ºCu2++2e-=Cu

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

17£®¶ÁÏÂÁÐÒ©Æ·±êÇ©£¬ÓйطÖÎö²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡ÏîABCD
ÎïÆ·±êÇ©   
·ÖÎö¸ÃÊÔ¼ÁӦװÔÚÏ𠽺ÈûµÄϸ¿ÚÆ¿ÖиÃÒ©Æ·²»ÄÜÓëÆ¤·ôÖ±½Ó½Ó´¥ÊÜÈÈÒ×·Ö½â¸ÃÒ©Æ·±êÇ©ÉÏ»¹±êÓÐ
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®ÏÂÁÐÖ¸¶¨·´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Cu ÈÜÓÚÏ¡ HNO3£ºCu+2H++NO3-¨TCu2++NO2¡ü+H2O
B£®ÂÈÆøÈÜÓÚË®£ºCl2+H2O¨T2H++Cl-+ClO-
C£®ÓÃCH3COOHÈܽâ CaCO3£ºCaCO3+2H+¨TCa2++H2O+CO2¡ü
D£®FeCl3ÈÜÒºÓëCu·´Ó¦£º2Fe3++Cu¨T2Fe2++Cu2+

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

1£®ÏòNa2CO3ÈÜÒºÖеμÓÑÎËᣬ·´Ó¦¹ý³ÌÖÐÄÜÁ¿±ä»¯ÈçͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®·´Ó¦HCO3-£¨aq£©+H+£¨aq£©=CO2£¨g£©+H2O£¨l£© Îª·ÅÈÈ·´Ó¦
B£®CO32-£¨aq£©+2H+£¨aq£©=CO2£¨g£©+H2O£¨l£©¡÷H=£¨¡÷H1+¡÷H2+¡÷H3£©
C£®¡÷H1£¾¡÷H2¡÷H2£¼¡÷H3
D£®H2CO3£¨aq£©=CO2£¨g£©+H2O£¨l£©£¬ÈôʹÓô߻¯¼Á£¬Ôò¡÷H3±äС

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®²ÝËᣨH2C2O4£©Ë®ÈÜÒºÖÐpc£¨H2C2O4£©¡¢pc£¨HC2O4-£©¡¢pc£¨C2O42-£©Ëæ×ÅÈÜÒºpHµÄ±ä»¯ÇúÏßÈçͼËùʾ£¬ÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®pH=4ʱ£¬c£¨HC2O4-£©£¾c£¨C2O42-£©
B£®c£¨H2C2O4£©+c£¨HC2O4-£©+c£¨C2O42-£©Ò»¶¨²»±ä
C£®²ÝËáµÄµçÀë³£ÊýKa1=10-1.3
D£®$\frac{c£¨{C}_{2}{{O}_{4}}^{2-}£©•c£¨{H}_{2}{C}_{2}{O}_{4}£©}{{c}^{2}£¨H{C}_{2}{{O}_{4}}^{-}£©}$=10-3

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸