ÌâÄ¿ÄÚÈÝ

ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH=a£¬B¼îµÄÈÜÒºpH=b¡£
(1)ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa=3£¬b=11£¬Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHΪ    ¡£
a.´óÓÚ7   b.µÈÓÚ7   c.СÓÚ7
(2)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa=4£¬b=12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ   mol/L,BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ       mol/L¡£
(3)ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa+b=14£¬Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                                             ¡£

£¨1)b   (2)10-10   10-12       (3)M++H2OMOH+H+

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÏÂͼÊÇijÑо¿ÐÔѧϰС×éÉè¼ÆµÄ¶ÔÒ»ÖַϾɺϽðµÄ¸÷³É·Ö£¨º¬ÓÐCu¡¢Fe¡¢Si ÈýÖֳɷ֣©½øÐзÖÀë¡¢»ØÊÕÔÙÀûÓõĹ¤ÒµÁ÷³Ì£¬Í¨¹ý¸ÃÁ÷³Ì½«¸÷³É·Öת»¯Îª³£Óõĵ¥Öʼ°»¯ºÏÎï¡£

ÒÑÖª£º298Kʱ£¬Ksp[Cu(OH)2]=2.2¡Á10-20£¬Ksp[Fe(OH)3]=4.0¡Á10-38£¬ Ksp[Mn(OH)2] =1.9¡Á10-13£¬
¸ù¾ÝÉÏÃæÁ÷³Ì»Ø´ðÓйØÎÊÌ⣺
£¨1£©²Ù×÷¢ñ¡¢¢ò¡¢¢óÖ¸µÄÊÇ                                     ¡£
£¨2£©¼ÓÈë¹ýÁ¿FeCl3ÈÜÒº¹ý³ÌÖпÉÄÜÉæ¼°µÄ»¯Ñ§·½³Ìʽ£º                    ¡£
£¨3£©¹ýÁ¿µÄ»¹Ô­¼ÁÓ¦ÊÇ                                                        ¡£
£¨4£©¢ÙÏòÈÜÒºbÖмÓÈëËáÐÔKMnO4ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ                     ¡£
¢ÚÈôÓÃX mol/L KMnO4ÈÜÒº´¦ÀíÈÜÒºb£¬µ±Ç¡ºÃ½«ÈÜÒºÖеÄÑôÀë×ÓÍêÈ«Ñõ»¯Ê±ÏûºÄKMnO4ÈÜÒºYmL£¬Ôò×îºóËùµÃºì×ØÉ«¹ÌÌåCµÄÖÊÁ¿Îª                 g£¨Óú¬X¡¢YµÄ´úÊýʽ±íʾ£©¡£
£¨5£©³£ÎÂÏÂ,ÈôÈÜÒºcÖÐËùº¬µÄ½ðÊôÑôÀë×ÓŨ¶ÈÏàµÈ£¬ÏòÈÜÒºcÖÐÖðµÎ¼ÓÈëKOHÈÜÒº£¬ÔòÈýÖÖ½ðÊôÑôÀë×Ó³ÁµíµÄÏȺó˳ÐòΪ£º             ©ƒ           ©ƒ            ¡£(Ìî½ðÊôÑôÀë×Ó)
£¨6£©×îºóÒ»²½µç½âÈôÓöèÐԵ缫µç½âÒ»¶Îʱ¼äºó£¬Îö³ö¹ÌÌåBµÄÖÊÁ¿ÎªZ g£¬Í¬Ê±²âµÃÒõÑôÁ½¼«ÊÕ¼¯µ½µÄÆøÌåÌå»ýÏàµÈ£¬Ôò±ê¿öÏÂÑô¼«Éú³ÉµÄ×îºóÒ»ÖÖÆøÌåÌå»ýΪ     L£¨Óú¬ZµÄ´úÊýʽ±íʾ£©£»¸Ãµç¼«µÄ·´Ó¦Ê½Îª                                                             .

A¡¢B¡¢C¡¢D·Ö±ðΪNH4HSO4¡¢Ba(OH)2¡¢AlCl3¡¢Na2CO3 4ÖÖÎïÖÊÖеÄ1ÖÖ£¬ÈÜÓÚË®¾ùÍêÈ«µçÀ룬ÏÖ½øÐÐÈçÏÂʵÑ飺
¢Ù×ãÁ¿AÈÜÒºÓëBÈÜÒº»ìºÏ¹²ÈÈ¿ÉÉú³É³Áµí¼×ºÍ´Ì¼¤ÐÔÆøζÆøÌ壻
¢ÚÉÙÁ¿AÈÜÒºÓëCÈÜÒº»ìºÏ¿ÉÉú³É³ÁµíÒÒ£»
¢ÛAÈÜÒºÓëBÈÜÒº¾ù¿ÉÈܽâ³ÁµíÒÒ£¬µ«¶¼²»ÄÜÈܽâ³Áµí¼×¡£
Çë»Ø´ð£º
(1)AµÄ»¯Ñ§Ê½Îª________£»ÊÒÎÂʱ£¬½«pHÏàµÈµÄAÈÜÒºÓëDÈÜÒº·Ö±ðÏ¡ÊÍ10±¶£¬pH·Ö±ð±äΪaºÍb£¬Ôòa________b(Ìî¡°>¡±¡°£½¡±»ò¡°<¡±)¡£
(2)¼ÓÈÈÕô¸ÉCÈÜÒº²¢×ÆÉÕ£¬×îºóËùµÃ¹ÌÌåΪ________(Ìѧʽ)¡£
(3)CÈÜÒºÓëDÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________
(4)ÏòBÈÜÒºÖÐÖðµÎ¼ÓÈëµÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº£¬µÎ¼Ó¹ý³ÌÖÐË®µÄµçÀëƽºâ½«________(Ìî¡°ÕýÏò¡±¡°²»¡±»ò¡°ÄæÏò¡±)Òƶ¯£»×îÖÕËùµÃÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ________________________________________________________________________¡£
(5)ÒÑÖª³Áµí¼×µÄKsp£½x¡£½«0.03 mol¡¤L£­1µÄAÈÜÒºÓë0.01 mol¡¤L£­1µÄBÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÈÜÒºÖÐËá¸ùÀë×ÓµÄŨ¶ÈΪ________(Óú¬xµÄ´úÊýʽ±íʾ£¬»ìºÏºóÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ)¡£

ϱíÊǼ¸ÖÖÈõµç½âÖʵĵçÀëƽºâ³£Êý¡¢ÄÑÈܵç½âÖʵÄÈܶȻýKsp£¨25¡æ£©¡£

µç½âÖÊ
µçÀë·½³Ìʽ
µçÀë³£ÊýK
Ksp
H2CO3
H2CO3HCO3£­£«H£«
HCO3£­CO32£­£«H£«
K1£½4.31¡Á10£­7
K2£½5.61¡Á10£­11
£­
C6H5OH
C6H5OHC6H5O£­£«H£«
1.1¡Á10£­10
£­
H3PO4
H3PO4H2PO4£­£«H£«
H2PO4£­HPO42£­£«H£«
HPO42£­PO43£­£«H£«
K1£½7.52¡Á10£­3
K2£½6.23¡Á10£­6
K1£½2.20¡Á10£­13
£­
NH3¡¤H2O
NH3¡¤H2OOH£­£«NH4£«
1.76¡Á10£­5
£­
BaSO4
BaSO4£¨s£©Ba2£«£«SO42£­
£­
1.07¡Á10£­10
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öC6H5OHÓëNa3PO4·´Ó¦µÄÀë×Ó·½³Ìʽ£º_________________¡£
£¨2£©25¡æʱ£¬Ïò10 mL 0. 01 mol/LC6H5OHÈÜÒºÖеμÓV mL 0.1 mol/L°±Ë®£¬»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ__________£¨ÌîÐòºÅ£©¡£
a£®Èô»ìºÏÒºpH£¾7,ÔòV¡Ý10
b£®V£½5ʱ£¬2c£¨NH3¡¤H2O£©£«2c£¨NH4£«£©£½c£¨C6H5OH£©£«c£¨C6H5O£­£©
c£®V£½10ʱ£¬»ìºÏÒºÖÐË®µÄµçÀë³Ì¶ÈСÓÚ0.01 molC6H5OHÈÜÒºÖÐË®µÄµçÀë³Ì¶È
d£®Èô»ìºÏÒºpH£¼7£¬Ôòc£¨NH4£«£©£¾c£¨C6H5O£­£©£¾c£¨H£«£©£¾c£¨OH£­£©
£¨3£©Ë®½â·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý³ÆΪˮ½â³£Êý£¨ÓÃKb±íʾ£©£¬Àà±È»¯Ñ§Æ½ºâ³£ÊýµÄ¶¨Òå¡£25¡æʱ£¬Na2CO3µÚÒ»²½Ë®½â·´Ó¦µÄË®½â³£ÊýKb£½____mol/L¡£
£¨4£©ÈçͼËùʾ£¬ÓÐT1¡¢T2²»Í¬Î¶ÈÏÂÁ½ÌõBaSO4ÔÚË®ÖеijÁµíÈܽâƽºâÇúÏߣ¨ÒÑÖªBaSO4µÄKspËæζÈÉý¸ß¶øÔö´ó£©¡£

¢ÙT2____ 25¡æ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©£»
¢ÚÌÖÂÛT1ζÈʱBaSO4µÄ³ÁµíÈܽâƽºâÇúÏߣ¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ____£¨ÌîÐòºÅ£©¡£
a£®¼ÓÈëNa2SO4²»ÄÜʹÈÜÒºÓÉaµã±äΪbµã
b£®ÔÚT1ÇúÏßÉÏ·½ÇøÓò£¨²»º¬ÇúÏߣ©ÈÎÒâÒ»µãʱ£¬¾ùÓÐBaSO4³ÁµíÉú³É
c£®Õô·¢ÈܼÁ¿ÉÄÜʹÈÜÒºÓÉdµã±äΪÇúÏßÉÏa¡¢bÖ®¼äµÄijһµã£¨²»º¬a¡¢b£©
d£®ÉýοÉʹÈÜÒºÓÉbµã±äΪdµã

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

¾«Ó¢¼Ò½ÌÍø