ÌâÄ¿ÄÚÈÝ
6£®Ç¦µÄÒ±Á¶¡¢¼Ó¹¤Í¬Ñù»áʹˮÌåÖÐÖØ½ðÊôǦµÄº¬Á¿Ôö´óÔì³ÉÑÏÖØÎÛȾ£®Ë®ÈÜÒºÖÐǦµÄ´æÔÚÐÎ̬Ö÷ÒªÓÐPb2+¡¢Pb£¨OH£©+¡¢Pb£¨OH£©2¡¢Pb£¨OH£©3-¡¢Pb£¨OH£©42-£®¸÷ÐÎ̬µÄŨ¶È·ÖÊý¦ÁËæÈÜÒºpH ±ä»¯µÄ¹ØÏµÈçͼËùʾ£º£¨1£©Íùº¬Pb2+µÄÈÜÒºÖеÎÏ¡NaOHÈÜÒº£¬pH=8ʱÈÜÒºÖдæÔÚµÄÑôÀë×Ó³ýH+¡¢Na+Í⻹ÓÐPb2+¡¢Pb£¨OH£©+£¬pH=9ʱ£¬Ö÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪPb2++2OH-¨TPb£¨OH£©2¡ý£®
£¨2£©Ä³¿ÎÌâ×éÖÆ±¸ÁËÒ»ÖÖÐÂÐÍÍÑǦ¼Á£¬ÄÜÓÐЧȥ³ýË®ÖеĺÛÁ¿Ç¦£¬ÊµÑé½á¹ûÈçÏÂ±í£º
| Àë×Ó | Pb2+ | Ca2+ | Fe3+ | Mn2+ | Cl- |
| ´¦ÀíǰŨ¶È/£¨mg•L-1£© | 0.100 | 29.8 | 0.120 | 0.087 | 51.9 |
| ´¦ÀíºóŨ¶È/£¨mg•L-1£© | 0.004 | 22.6 | 0.040 | 0.053 | 49.8 |
£¨3£©Èç¹û¸ÃÍÑǦ¼Á£¨ÓÃEH±íʾ£©ÍÑǦ¹ý³ÌÖÐÖ÷Òª·¢ÉúµÄ·´Ó¦Îª£º2EH£¨s£©+Pb2+?E2Pb£¨s£©+2H+£®ÔòÍÑǦµÄ×îºÏÊÊpH ·¶Î§ÎªB£¨Ìî×Öĸ£©£®
A£®4¡«5B£®6¡«7C£®9¡«10D£®11¡«12£®
·ÖÎö £¨1£©pH=9ʱ£¬ÇúÏß2¡¢3±íʾµÄÎïÖʹ²´æ£¬ÓÉͼ¿ÉÖªPb£¨OH£©2µÄŨ¶È·ÖÊý±ÈPb£¨OH£©+µÄ´ó£¬ËùÒÔÖ÷Òª·´Ó¦ÊÇÉú³ÉPb£¨OH£©2£¬pH=12ʱ£¬Pb£¨OH£©3-¡¢Pb£¨OH£©42-¹²´æ£»
£¨2£©Óɱí¸ñÊý¾Ý¿ÉÖª£¬¼ÓÍÑǦ¼ÁÀë×ÓŨ¶È±ä»¯´óµÄ´¦ÀíЧ¹ûºÃ£»
£¨3£©¸ù¾Ý·´Ó¦Åжϣ¬²Î¼Ó·´Ó¦µÄÊÇPb2+£¬ÓÉͼÏó¿ÉÖª£¬Ñ¡ÔñPHҪʹǦȫ²¿ÒÔPb2+ÐÎʽ´æÔÚ£®
½â´ð ½â£º£¨1£©ÓÉͼÏó¿ÉÖªpH=8ʱ£¬Pb2+£¬Pb£¨OH£©+ºÍPb£¨OH£©2¹²´æ£¬ÁíÍâÈÜÒºÖл¹ÓÐH+£¬ËùÒÔÖдæÔÚµÄÑôÀë×Ó£¨Na+³ýÍ⣩ÓÐPb2+¡¢Pb£¨OH£©+¡¢H+£»pH=9ʱ£¬ÇúÏß2¡¢3±íʾµÄÎïÖʹ²´æ£¬ÓÉͼ¿ÉÖªPb£¨OH£©2µÄŨ¶È·ÖÊý±ÈPb£¨OH£©+µÄ´ó£¬ËùÒÔÖ÷Òª·´Ó¦ÊÇÉú³ÉPb£¨OH£©2£¬·´Ó¦µÄ·½³ÌʽΪPb2++2OH-¨TPb£¨OH£©2¡ý£¬
¹Ê´ð°¸Îª£ºPb2+¡¢Pb£¨OH£©+£»Pb2++2OH-¨TPb£¨OH£©2¡ý£»
£¨2£©¼ÓÈëÍÑǦ¼Á£¬Pb2+Ũ¶Èת»¯ÂÊΪ$\frac{0.1-0.004}{0.1}$¡Á100%=96%£¬Fe3+Ũ¶Èת»¯ÂÊΪ$\frac{0.12-0.04}{0.12}$¡Á100%=67%£¬¶øÆäËüÁ½ÖÖÀë×Óת»¯ÂʽÏС£¬Ôò³ýPb2+Í⣬¸ÃÍÑǦ¼Á¶ÔÆäËüÀë×ÓµÄÈ¥³ýЧ¹û×îºÃµÄÊÇFe3+£¬
¹Ê´ð°¸Îª£ºFe3+£»
£¨3£©·´Ó¦Îª2EH£¨s£©+Pb2+?E2Pb£¨s£©+2H+£¬²Î¼Ó·´Ó¦µÄÊÇPb2+£¬ÓÉͼÏó¿ÉÖª£¬Ñ¡ÔñPHҪʹǦȫ²¿ÒÔPb2+ÐÎʽ´æÔÚ£¬Ó¦Îª6¡«7Ö®¼ä£¬¹Ê´ð°¸Îª£ºB£®
µãÆÀ ±¾Ì⿼²éÎïÖÊ·ÖÀëÌá´¿£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕϰÌâÖÐͼÏó¡¢±í¸ñÖеÄÐÅÏ¢¼°Ó¦ÓÃΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëÇ¨ÒÆÓ¦ÓÃÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®
| A£® | MgBr2ºÍKCl | B£® | Cl2ºÍHCl | C£® | HIºÍNaI | D£® | Na2O2ºÍH2O2 |
| A£® | º¬ÑõËáµÄËáÐÔ£ºX¶ÔÓ¦µÄÇ¿ÓÚY¶ÔÓ¦µÄ | |
| B£® | ÆøÌ¬Ç⻯ÎïµÄÎȶ¨ÐÔ£ºHmXÇ¿ÓÚHnY | |
| C£® | µÚÒ»µçÀëÄÜ¿ÉÄÜY´óÓÚX | |
| D£® | XÓëYÐγɻ¯ºÏÎïʱ£¬XÏÔ¸º¼Û£¬YÏÔÕý¼Û |
¢ÙCH3OH£¨g£©+H2O£¨g£©¨TCO2£¨g£©+3H2£¨g£©£»¡÷H1=+49.0kJ•mol-1
¢ÚCH3OH£¨g£©+$\frac{1}{2}$O2£¨g£©¨TCO2£¨g£©+2H2£¨g£©£»¡÷H2=-192.9kJ•mol-1
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A£® | CH3OHµÄȼÉÕÈÈ¡÷H=-192.9 kJ•mol-1 | |
| B£® | ||
| C£® | H2ȼÉÕÄܷųö´óÁ¿µÄÈÈ£¬¹ÊCH3OHת±ä³ÉH2µÄ¹ý³Ì±ØÐëÎüÊÕÈÈÁ¿ | |
| D£® | ¸ù¾Ý¢ÚÍÆÖª£ºÔÚ25¡æ£¬101 kPaʱ£¬1 molCH3OH£¨g£©ÍêȫȼÉÕÉú³ÉCO2ºÍH2O·Å³öµÄÈÈÁ¿Ó¦´óÓÚ192.9 kJ |