ÌâÄ¿ÄÚÈÝ

14£®CO2ÈÜÓÚË®Éú³É̼ËᣮÒÑÖªÏÂÁÐÊý¾Ý£º
Èôµç½âÖÊH2CO3NH3•H2O
µçÀëƽºâ³£Êý£¨25¡æ£©${K}_{{a}_{1}}$=4.30¡Á107  ${K}_{{a}_{2}}$=5.61¡Á10-11Kh=1.77¡Á10-5
ÏÖÓг£ÎÂÏÂ1mol•L-1µÄ£¨NH4£©2CO3ÈÜÒº£¬ÒÑÖªNH4+µÄË®½âƽºâ³£ÊýKh=$\frac{{K}_{w}}{{K}_{b}}$£¬CO32-µÚÒ»²½Ë®½âµÄƽºâ
³£ÊýKh=$\frac{{K}_{w}}{{K}_{{a}_{2}}}$£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£®
A£®ÓÉÊý¾Ý¿ÉÅжϸÃÈÜÒº³ÊËáÐÔ
B£®c£¨NH4+£©£¾c£¨HCO32-£©£¾c£¨CO32-£©£¾c£¨NH3•H2O£©
C£®c£¨NH4+£©+c£¨NH3•H2O£©=2c£¨CO32-£©+2c£¨HCO3-£©+2c£¨H2CO3£©
D£®c£¨NH4+£©+c£¨H+£©=c£¨HCO3-£©+c£¨OH-£©+c£¨CO32-£©

·ÖÎö A£®¸ù¾ÝCO32-µÚÒ»²½Ë®½âµÄƽºâ³£Êý´óÓÚNH4+Ë®½âµÄƽºâ³£Êý¿ÉÖªÈÜÒº³Ê¼îÐÔ£»
B£®Ì¼Ëá¸ùÀë×ÓË®½â³Ì¶È½ÏС£¬Ôòc£¨CO32-£©£¾c£¨HCO3-£©£»
C£®¸ù¾Ý̼Ëáï§ÈÜÒºÖеÄÎïÁÏÊغãÅжϣ»
D£®¸ù¾Ý̼Ëáï§ÈÜÒºÖеĵçºÉÊغãÅжϣ®

½â´ð ½â£ºA£®CO32-µÚÒ»²½Ë®½âµÄƽºâ³£Êý$\frac{{K}_{W}}{5.61¡Á1{0}^{-11}}$´óÓÚNH4+Ë®½âµÄƽºâ³£Êý$\frac{{K}_{W}}{1.77¡Á1{0}^{-5}}$£¬ÔòÈÜÒº³Ê¼îÐÔ£¬¹ÊA´íÎó£»
B£®ï§¸ùÀë×ÓµÄË®½â³Ì¶ÈÐèҪ̼Ëá¸ùÀë×Ó£¬ÔòÈÜÒºÖÐc£¨NH4+£©£¾c£¨CO32-£©¡¢c£¨HCO3-£©£¾c£¨NH3•H2O£©£¬ÓÉÓÚË®½â³Ì¶È½ÏС£¬Ôòc£¨CO32-£©£¾c£¨HCO3-£©£¬ËùÒÔÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc£¨NH4+£©£¾c£¨CO32-£©£¾c£¨HCO3-£©£¾c£¨NH3•H2O£©£¬¹ÊB´íÎó£»
C£®ÒÀ¾ÝÈÜÒºÖÐÎïÁÏÊغãµÃµ½£ºc£¨NH4+£©+c£¨NH3•H2O£©=2c£¨CO32-£©+2c£¨HCO3-£©+2c£¨H2CO3£©=2mol•L-1 £¬¹ÊCÕýÈ·£»
D£®ÈÜÒºÖдæÔÚµçºÉÊغãΪ£ºc£¨NH4+£©+c£¨H+£©=c£¨HCO3-£©+c£¨OH-£©+2c£¨CO32-£©£¬¹ÊD´íÎó£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÁËÀë×ÓŨ¶È´óС±È½Ï£¬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷ȷ̼Ëá¸ùÀë×Ó¡¢ï§¸ùÀë×ÓµÄË®½â³Ì¶È´óСΪ½â´ð¹Ø¼ü£¬×¢ÒâÕÆÎÕµçºÉÊغ㡢ÎïÁÏÊغ㼰ÑεÄË®½âÔ­ÀíÔÚÅжÏÀë×ÓŨ¶È´óСÖеÄÓ¦Ó÷½·¨£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
6£®ÂÌ·¯£¨FeSO4•7H2O£©ÔÚ»¯Ñ§ºÏ³ÉÉÏÓÃ×÷»¹Ô­¼Á¼°´ß»¯¼Á£®¹¤ÒµÉϳ£Ó÷ÏÌúмÈÜÓÚÒ»¶¨Å¨¶ÈµÄÁòËáÈÜÒºÖƱ¸ÂÌ·¯£®
£¨1£©98% 1.84g/cm3µÄŨÁòËáÔÚÏ¡Ê͹ý³ÌÖУ¬ÃܶÈϽµ£¬µ±Ï¡ÊÍÖÁ50%ʱ£¬ÃܶÈΪ1.4g/cm3£¬50%µÄÁòËáÎïÖʵÄÁ¿Å¨¶ÈΪ7.14mol•L-1£¨±£ÁôÁ½Î»Ð¡Êý£©£¬50%µÄÁòËáÓë30%µÄÁòËáµÈÌå»ý»ìºÏ£¬»ìºÏËáµÄŨ¶ÈΪ£¾£¨Ì¡¢£¼¡¢=£©40%£®
£¨2£©½«111.2gÂÌ·¯£¨FeSO4?7H2O£¬Ê½Á¿Îª278£©ÔÚ¸ßÎÂϼÓÈÈ£¬³ä·Ö·´Ó¦ºóÉú³ÉFe2O3¹ÌÌåºÍSO2¡¢SO3¡¢Ë®µÄ»ìºÏÆøÌ壬ÔòÉú³ÉFe2O3µÄÖÊÁ¿Îª32g£»SO2Ϊ0.2mol£®
ʵÑéÊÒ¿ÉÓÃÒÔÏ·½·¨ÖƱ¸Ä¦¶ûÑξ§Ìå[£¨NH4£©2SO4•FeSO4•6H2O£¬Ê½Á¿Îª392]£®
£¨3£©½«4.88gÌúм£¨º¬Fe2O3£©Óë25mL3mol/L H2SO4³ä·Ö·´Ó¦ºó£¬µÃµ½ FeSO4ºÍH2SO4µÄ»ìºÏÈÜÒº£¬Ï¡ÊÍÈÜÒºÖÁ100mL£¬²âµÃÆäpH=1£®ÌúмÖÐFe2O3µÄÖÊÁ¿·ÖÊýÊÇ0.66£¨»ò66%£©£¨±£ÁôÁ½Î»Ð¡Êý£©£®
£¨4£©ÏòÉÏÊö100mLÈÜÒºÖмÓÈëÓë¸ÃÈÜÒºÖÐFeSO4µÈÎïÖʵÄÁ¿µÄ£¨NH4£©2SO4¾§Ì壬´ý¾§ÌåÍêÈ«ÈܽâºóÕô·¢µô²¿·ÖË®£¬ÀäÈ´ÖÁt¡æ£¬Îö³öĦ¶ûÑξ§Ìå12.360g£¬Ê£ÓàÈÜÒºµÄÖÊÁ¿Îª82.560g£®t¡æʱ£¬¼ÆË㣨NH4£©2SO4•FeSO4•6H2OµÄÈܽâ¶È22.35g£®£¨±£ÁôÁ½Î»Ð¡Êý£©£®

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

¾«Ó¢¼Ò½ÌÍø