ʱ¼ä:2019-05-30 04:14:50
1¡¢Ñ¡ÔñÌâ ÏÂÁÐÅжϲ»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®pH=3µÄÑÎËáÈÜÒººÍpH=11µÄ°±Ë®ÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ
B£®pH=2µÄHAÈÜÒºÓëpH=12µÄMOHÈÜÒºÈÎÒâ±ÈÀý»ìºÏºóÓУºc£¨H+£©+c£¨M+£©¨Tc£¨OH-£©+c£¨A-£©
C£®0.1mol/LµÄNaHAÈÜÒºÆäpH=4£¬Ôòc£¨HA-£©£¾c£¨H+£©£¾c£¨H2A£©£¾c£¨A2-£©
D£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄCH3COOHºÍCH3COONaÈÜÒºµÈÌå»ý»ìºÏºóÓУºc£¨CH3COO-£©+2c£¨OH-£©¨T2c£¨H+£©+c£¨CH3COOH£©
2¡¢Ñ¡ÔñÌâ A¡¢B¡¢C¡¢D¡¢E¡¢FÁùÖÖ¶ÌÖÜÆÚÖ÷×åÔªËØ£¬ËüÃǵÄÔ×ÓÐòÊýÒÀ´ÎÔö´ó¡£ÒÑÖª£ºBÔªËØÔ×ÓµÄ×îÍâ²ãµç×ÓÊýÊÇÄÚ²ãµç×ÓÊýµÄ2±¶£¬CÔªËØµÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëÆäÇ⻯Îï·´Ó¦ÄÜÉú³ÉÑΣ¬DÓëEͬÖ÷×壬EÔªËØÔ×ÓµÄ×îÍâ²ãµç×ÓÊý±È´ÎÍâ²ãµç×ÓÊýÉÙ2¡£ÓÉÔªËØA¡¢B¡¢D×é³ÉµÄÒ»ÔªËáXΪÈÕ³£Éú»îÖеĵ÷ζ¼Á£¬ÔªËØA¡¢F×é³ÉµÄ»¯ºÏÎïΪY£¬AÓëCÐÎ³ÉµÄÆøÌ¬»¯ºÏÎïZÔÚ±ê×¼×´¿öϵÄÃܶÈΪ0.76g/L£¬ZÈÜÓÚË®ËùµÃÈÜҺΪW¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
[? ]
A.?ÔªËØCÔÚÖÜÆÚ±íÖеÄλÖÃÊǵÚÈýÖÜÆÚµÚ¢õA×å
B.?ÔÚµÈÌå»ý¡¢µÈpHµÄX¡¢YÈÜÒºÖзֱð¼ÓÈëµÈÖÊÁ¿µÄп·Û£¬³ä·Ö·´Ó¦ºó½öÓÐÒ»·ÝÈÜÒºÖдæÔÚп·Û£¬Ôò·´Ó¦¹ý³ÌÖÐÁ½ÈÜÒºÖз´Ó¦ËÙÂʵĴóС¹ØÏµÊÇ£ºX£¼Y
C.?ÓÉA¡¢B¡¢C¡¢DËÄÖÖÔªËØÐγɵϝºÏÎïÒ»¶¨Ö»Óй²¼Û¼ü
D.?25¡æÊ±£¬½«0.1?mol/LµÄWÓë0.1?mol/LµÄYÈÜÒºµÈÌå»ý»ìºÏ£¬³ä·Ö·´Ó¦ºóÈôpH=7 ÔòÈÜÒºÖÐ?c(CA4+)=c(F-)
3¡¢Ñ¡ÔñÌâ ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ
[? ]
A£®0.1 mol/L C6H5ONaÈÜÒºÖУºc(Na+)>c(C6H5O-)>c(H+)>c(OH-)
B£®Na2CO3ÈÜÒº¼ÓˮϡÊͺ󣬻ָ´ÖÁÔζȣ¬pHºÍKw¾ù¼õС
C£®pH =5µÄCH3COOHÈÜÒººÍpH =5µÄNH4ClÈÜÒºÖУ¬c(H+)²»ÏàµÈ
D£®ÔÚNa2SÈÜÒºÖмÓÈëAgCl¹ÌÌ壬ÈÜÒºÖÐc(S2-)Ͻµ
4¡¢Ñ¡ÔñÌâ ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®Ä³¶þÔªÈõËáµÄËáʽÑÎNaHAÈÜÒºÖУºc£¨H+£©+c£¨Na+£©=c£¨OH-£©+c£¨HA-£©+c£¨A2-£©
B£®Ò»¶¨Á¿µÄ£¨NH4£©2SO4ÓëNH3?H2O»ìºÏËùµÃµÄËáÐÔÈÜÒºÖУºc£¨NH4+£©£¼2c£¨SO42-£©
C£®ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.01mol/LµÄCH3COOHºÍCH3COONaµÄÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºÖÐc£¨CH3COOH£©+c£¨CH3COO-£©=0.01?mol/L
D£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄ¢ÙNH4HSO4ÈÜÒº¡¢¢ÚNH4HCO3ÈÜÒº¡¢¢ÛNH4ClÈÜÒºÖÐc£¨NH4+£©£º¢Ù£¾¢Ú£¾¢Û
5¡¢Ñ¡ÔñÌâ ÏÂÁÐÈÜÒºÖÐÓйØÎïÖʵÄÁ¿Å¨¶È¹ØÏµ»ò¼ÆËãÕýÈ·µÄÊÇ?
[? ]
A.ÈôNaOH¡¢CH3COONaºÍNaHCO3ÈýÖÖÈÜÒºµÄpHÏàµÈ£¬Ôòc(NaOH)<c(CH3COONa)<c(NaHCO3)
B.25¡æÊ±£¬0.1mol/L?Na2CO3ÈÜÒºÖÐÓÉË®µçÀë³öÀ´µÄ(OH-)´óÓÚ0.1?mol/L?NaOHÈÜÒºÖÐÓÉË®µçÀë³öÀ´µÄc(OH-) ?
C.ÒÑÖª25¡æÊ±Ksp(AgCl)=1.8¡Á10-10£¬ÔòÔÚ0.1?mol/L?AlCl3ÈÜÒºÖУ¬Ag+µÄÎïÖʵÄÁ¿Å¨¶È×î´ó¿É´ïµ½1.8¡Á10-9mol/L?
D.ÊÒÎÂÏ£¬0.1mol/L?NaHAÈÜÒºµÄpH=4£¬Ôòc(HA-)>c(H+)>c(H2A)>c(A2-)