ʱ¼ä:2017-02-07 17:21:37
1¡¢Ñ¡ÔñÌâ ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®³£ÎÂÏ£¬Ä³ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)£½1¡Á10-amo1¡¤L¨C1£¬Èôa£¼7ʱ£¬Ôò¸ÃÈÜÒº¿ÉÄÜΪNaHSO4ÈÜÒº
B£®³£ÎÂÏ£¬ÖкÍͬÌå»ý¡¢Í¬pHµÄÁòËá¡¢ÑÎËáºÍ´×ËáËùÐèÏàͬŨ¶ÈµÄNaOHÈÜÒºµÄÌå»ý¹ØÏµ£ºV£¨ÁòËᣩ£¾V£¨ÑÎËᣩ£½V£¨´×Ëᣩ
C£®25¡æÊ±£¬ÒÑÖªKa(CH3COOH)=1.7¡Á10-5mo1¡¤L¨C1¡¢Ka(C6H5OH) =1.0¡Á10-10mo1¡¤L¨C1¡¢ Ka1(H2CO3) = 4.2¡Á10-7mo1¡¤L¨C1 ¡¢Ka2(H2CO3) £½5.6¡Á10-11mo1¡¤L¨C1pHÏàµÈµÄ¢ÙCH3COONa ¢ÚC6H5ONa ¢ÛNaHCO3ÈÜÒºÖУ¬c(Na+)´óС¹ØÏµ£º¢Ú£¼¢Û£¼¢Ù
D£®³£ÎÂÏ£¬Ïò±¥ºÍNa2CO3ÈÜÒºÖмÓÈëÉÙÁ¿BaSO4·ÛÄ©£¬¹ýÂË£¬ÏòÏ´¾»µÄ³ÁµíÖмÓÈëÏ¡ÑÎËáÓÐÆøÅݲúÉú£¬ËµÃ÷³£ÎÂÏÂKsp£¨BaSO4£©>Ksp£¨BaCO3£©
2¡¢Ñ¡ÔñÌâ ÔÚÈÜÒºÖпÉÒÔ´óÁ¿¹²´æ£¬¼ÓOH£²úÉú³Áµí£¬¼ÓH+Éú³ÉÆøÌåµÄÒ»×éÀë×ÓÊÇ£¨?£©
A£®Na+¡¢Cu2+¡¢CO32£¡¢Cl£
B£®Na+¡¢ Cl£¡¢ HCO3£¡¢Mg2+¡¡
C£®Cl£¡¢NO3£¡¢K+¡¢ Ca2+¡¡
D£®NH4+¡¢ K+¡¢ H+¡¢ SO42£
3¡¢Ñ¡ÔñÌâ ÏÂÁи÷×éÀë×ÓÔÚÏàÓ¦µÄÌõ¼þÏÂÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ
A£®ÔÚ¼îÐÔÈÜÒºÖУºHCO3£¡¢K+¡¢C1£¡¢Na+
B£®ÓëÂÁ·Û·´Ó¦·Å³öÇâÆøµÄÎÞÉ«ÈÜÒºÖУºNO3£¡¢K+¡¢Na+¡¢SO42£
C£®ÔÚc(H+)£¯c(OH£)==1¡Á1013µÄÈÜÒºÖУºNH4+¡¢Br£¡¢C1£¡¢K+
D£®Í¨ÓÐSO2µÄÈÜÒºÖУº Ca2+¡¢C1£¡¢NO3£¡¢A13+
4¡¢Ñ¡ÔñÌâ Ҫʹº¬ÓÐBa2+¡¢Al3+¡¢Cu2+¡¢Mg2+¡¢Ag+µÈÀë×ÓµÄÈÜÒºÖеĸ÷Àë×ÓÖðÒ»ÐγɳÁµíÎö³ö£¬ÏÂÁÐËùÑ¡ÔñµÄÊÔ¼Á¼°¼ÓÈëÊÔ¼ÁµÄ˳ÐòÕýÈ·µÄÊÇ £¨ £©
A£®H2SO4¡ªHCl¡ªH2S¡ªNaOH¡ªCO2
B£®HCl¡ªH2SO4¡ªNa2S¡ªNaOH¡ªCO2
C£®NaCl¡ªNa2SO4¡ªNa2S¡ªNaOH¡ªCH3COOH
D£®Na2S¡ªNa2SO4¡ªNaCl¡ªNaOH¡ªHCl
5¡¢Ñ¡ÔñÌâ ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£º
A£®ÒÑ֪ijζÈÏ´¿Ë®ÖеÄc(H+)=2¡Ál0£7mol.L-1£¬¾Ý´ËÎÞ·¨Çó³öË®ÖÐc(OH£)
B£®ÒÑÖªMgCO3µÄKSP=6.82¡Ál0-6£¬ÔòÔÚº¬ÓйÌÌåMgCO3µÄMgCl2¡¢Na2CO3ÈÜÒºÖУ¬¶¼ÓÐc(Mg2+) =c(CO32£)£¬ÇÒc(Mg2+)¡¤c(CO32£) = 6.82¡Á10£6 mol2.L-2
C£®³£ÎÂÏÂ
µÄÈÜÒºÖдæÔÚAl3+¡¢NH4+¡¢Cl-¡¢NO3-
D£®ÒÑÖª£º
| ¹²¼Û¼ü | C¡ªC | C=C | C¡ªH | H¡ªH |
| ¼üÄÜ/kJ¡¤mol-1 | 348 | 610 | 413 | 436 ÓÉÉϱíÊý¾Ý¿ÉÒÔ¼ÆËã³ö·´Ó¦? ?µÄìʱä |