ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿´×ËáÊÇÖØÒªµÄÒ»ÔªËᣬÔÚÓлúºÍÎÞ»ú·´Ó¦Öж¼ÓÐÓ¦Óá£ÏÖÓÐ25 ¡æʱ£¬pH£½3µÄ´×Ëá¡£Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©ÈôÏò´×ËáÖмÓÈëÉÙÁ¿´×ËáÄƹÌÌ壬´ËʱÈÜÒºÖÐ________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£
£¨2£©ÈôÏò´×ËáÖмÓÈëÏ¡NaOHÈÜÒº£¬Ê¹ÆäÇ¡ºÃÍêÈ«·´Ó¦£¬ËùµÃÈÜÒºµÄpH________(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)7£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆäÔÒò_____________________________________¡£
£¨3£©ÈôÏò´×ËáÖмÓÈëpH£½11µÄNaOHÈÜÒº£¬ÇÒ¶þÕßµÄÌå»ý±ÈΪ1¡Ã1£¬ÔòËùµÃÈÜÒºÖи÷Àë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ___________________________________________¡£
£¨4£©ÈôÏò´×ËáÖмÓÈëNaOHÈÜÒºÖÁÈÜҺǡºÃ³ÊÖÐÐÔ£¬´Ëʱc(Na£«)______c(CH3COO-)(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±)¡£
£¨5£©ÈôÏò´×ËáÖмÓÈëÒ»¶¨Á¿NaOHÈÜÒº£¬ËùµÃ»ìºÏÒºpH£½6£¬Ôò´ËÈÜÒºÖÐc(CH3COO-)£C(Na£«)£½________mol/L(Ìîд׼ȷÊý¾Ý)¡£
¡¾´ð°¸¡¿ ¼õС > CH3COO££«H2OCH3COOH£«OH£ cCH3COO£)>c(Na£«)>c(H£«)>c(OH£) £½ 9.9¡Á10£7
¡¾½âÎö¡¿(1)Òò´×ËáÄƹÌÌåµçÀë²úÉúCH3COO-£¬c(CH3COO-)Ôö´ó£¬Ê¹µÄƽºâCH3COOHCH3COO-+H+ÄæÏòÒƶ¯£¬c(H+)¼õС£¬c(CH3COOH)Ôö´ó£¬ËùÒÔ¼õС£¬¹Ê´ð°¸Îª£º¼õС£»
(2)Ïò´×ËáÖмÓÈëÏ¡NaOHÈÜÒº£¬Ê¹ÆäÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³ÉÁË´×ËáÄÆ£¬´×Ëá¸ùË®½âCH3COO-+H2OCH3COOH+OH-£¬ÈÜÒº³Ê¼îÐÔ£¬¼´pH£¾7£¬¹Ê´ð°¸Îª£º£¾£»CH3COO-+H2OCH3COOH+OH-£»
(3)´×Ëá¹ýÁ¿£¬ÈÜÒºÖеijɷÖΪ´×ËáºÍ´×ËáÄÆ£¬Òò´×ËáµÄµçÀë³Ì¶È´óÓÚ´×ËáÄƵÄË®½â³Ì¶È£¬ÈÜÒº³öËáÐÔ£¬ËùÒÔc(CH3COO-)£¾c(Na+)£¾c(H+)£¾c(OH-)£¬¹Ê´ð°¸Îª£ºc(CH3COO-)£¾c(Na+)£¾c(H+)£¾c(OH-)£»
(4)ҺǡºÃ³ÊÖÐÐÔ£¬Ôòc(H+)=c(OH-)£¬¸ù¾ÝµçºÉÊغãc(Na+)+c(H+)=c(OH-)+c(CH3COO-)£¬¹Êc(Na+)=c(CH3COO-)£¬¹Ê´ð°¸Îª£º=£»
(5)pH=6ÈÜÒºÖУ¬c(H+)=10-6mol/L£¬c(OH-)=10-8mol/L£¬¸ù¾ÝµçºÉÊغãc(Na+)+c(H+) =c(OH-)+c(CH3COO-)¿ÉÖª£¬c(CH3COO-)-c(Na+)=c(H+)-c(OH-) =10-6mol/L-10-8mol/L =(10-6-10-8)mol/L=9.9¡Á10£7 mol/L£¬¹Ê´ð°¸Îª£º9.9¡Á10£7mol/L
¡¾ÌâÄ¿¡¿ÓÉCOºÍH2S·´Ó¦¿ÉÖƵÃôÊ»ùÁò£¨COS£©¡£ÔÚºãÈݵÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦²¢´ïµ½Æ½ºâ£ºCO(g)+H2S(g) COS(g)+H2(g)£¬Êý¾ÝÈçϱíËùʾ£º
ʵÑé | ζÈ/¡æ | Æðʼʱ | ƽºâʱ | |||
n(CO)/mol | n(H2S)/mol | n(COS)/mol | n(H2)/mol | n(CO)/mol | ||
1 | 150 | 10.0 | 10.0 | 0 | 0 | 7.0 |
2 | 150 | 7.0 | 8.0 | 2.0 | 4.5 | a |
3 | 400 | 20.0 | 20.0 | 0 | 0 | 16.0 |
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A. ÉÏÊö·´Ó¦ÊÇÎüÈÈ·´Ó¦
B. ʵÑé1´ïƽºâʱ£¬COµÄת»¯ÂÊΪ70%
C. ʵÑé2´ïƽºâʱ£¬a<7.0
D. ʵÑé3´ïƽƽºâºó£¬ÔÙ³äÈë1.0molH2£¬Æ½ºâÄæÏòÒƶ¯£¬Æ½ºâ³£ÊýÖµÔö´ó