ÌâÄ¿ÄÚÈÝ
ij»ìºÏÎïµÄË®ÈÜÒºÖУ¬Ö»¿ÉÄܺ¬ÓÐÒÔÏÂÀë×ÓÖеÄÈô¸ÉÖÖ£ºK+¡¢Mg2+¡¢Fe3+¡¢Al3+¡¢NH4+¡¢Cl-¡¢CO32-ºÍSO42-¡£ÏÖÿ´ÎÈ¡10.00mL½øÐÐʵÑ飺
¢ÙµÚÒ»·Ý¼ÓÈëAgNO3ÈÜÒºÓгÁµí²úÉú£»
¢ÚµÚ¶þ·Ý¼ÓÈë×ãÁ¿NaOHºó¼ÓÈÈ£¬ÊÕ¼¯µ½ÆøÌå0.672L£¨±ê×¼×´¿öÏ£©
¢ÛµÚÈý·Ý¼ÓÈë×ãÁ¿BaCl2ÈÜÒººóµÃ¸ÉÔï³Áµí6.63g£¬³Áµí¾×ãÁ¿ÑÎËáÏ´µÓ£¬¸ÉÔïºóÊ£Óà4.66g¡£
Çë»Ø´ð£º
¢Å.c(CO32-)=________¡£
¢Æ.K+ÊÇ·ñ´æÔÚ£¿________£»Èô´æÔÚ£¬Å¨¶È·¶Î§ÊÇ________£¨Èô²»´æÔÚ£¬Ôò²»±Ø»Ø´ðµÚ2ÎÊ£©¡£
¢Ç.¸ù¾ÝÒÔÉÏʵÑ飬²»ÄÜÅжÏÄÄÖÖÀë×ÓÊÇ·ñ´æÔÚ£¿Èô´æÔÚ£¬ÕâÖÖÀë×ÓÈçºÎ½øÐмìÑ飿________¡¢________¡£
1mol/L ´æÔÚ ´æÔÚ ¡Ý3 mol/L Cl- È¡ÉÙÁ¿ÔÊÔÒº ¡Ý3 mol/L Cl- È¡ÉÙÁ¿ÔÊÔÒº È¡ÂËÒº¼Ó×ãÁ¿Ï¡ÏõËáºÍÏõËáÒøÈÜÒº£¬ÈôÓгÁµíÖ¤Ã÷´æÔÚÂÈÀë×Ó ·ñÔòÎÞÂÈÀë×Ó
µÚÒ»·Ý¼ÓÈëAgNO3ÈÜÒºÓгÁµí²úÉú£¬ËµÃ÷Cl£¡¢CO32£ºÍSO42£ÖÁÉÙº¬ÓÐÒ»ÖÖ£»µÚ¶þ·Ý¼ÓÈë×ãÁ¿NaOHºó¼ÓÈÈ£¬ÊÕ¼¯µ½ÆøÌå0.672L£¬ÔòÒ»¶¨º¬ÓÐNH4+£¬ÇÒÎïÖʵÄÁ¿ÊÇ0.03mol£»µÚÈý·Ý¼ÓÈë×ãÁ¿BaCl2ÈÜÒººóµÃ¸ÉÔï³Áµí6.63g£¬³Áµí¾×ãÁ¿ÑÎËáÏ´µÓ£¬¸ÉÔïºóÊ£Óà4.66g£¬ËµÃ÷Ò»¶¨º¬ÓÐCO32£ºÍSO42££¬ÇÒÎïÖʵÄÁ¿·Ö±ðÊÇ0.01molºÍ0.02mol£¬Ôòc(CO32£)=0.01molL¡Â0.01mol/L£½1mol/L¡£Òò´ËÈÜÒºÖÐÒ»¶¨Ã»ÓÐMg2£«¡¢Fe3£«¡¢Al3£«¡£ÓÉÓÚÈÜÒºÊǵçÖÐÐԵ쬶øCO32£ºÍSO42£µÄ¸ºµçºÉÊýÊÇ0.06mol£¬NH4+ÊÇ0.03mol£¬¶øÂÈÀë×ÓÊDz»ÄÜÈ·¶¨µÄ£¬ËùÒÔ¼ØÀë×ÓÒ»¶¨ÓУ¬ÇÒÆäÎïÖʵÄÁ¿´óÓÚ»òµÈÓÚ0.03mol£¬Ôò¼ØÀë×ÓŨ¶È´óÓÚ»òµÈÓÚ3 mol/L¡£¼ìÑ鼨Àë×ӵķ½·¨ÊÇÈ¡ÉÙÁ¿ÔÊÔÒº£¬¼Ó×ãÁ¿ÏõËá±µÈÜÒº¹ýÂË£¬È¡ÂËÒº¼Ó×ãÁ¿Ï¡ÏõËáºÍÏõËáÒøÈÜÒº£¬ÈôÓгÁµíÖ¤Ã÷´æÔÚÂÈÀë×Ó£¬·ñÔòÎÞÂÈÀë×Ó¡£
µÚÒ»·Ý¼ÓÈëAgNO3ÈÜÒºÓгÁµí²úÉú£¬ËµÃ÷Cl£¡¢CO32£ºÍSO42£ÖÁÉÙº¬ÓÐÒ»ÖÖ£»µÚ¶þ·Ý¼ÓÈë×ãÁ¿NaOHºó¼ÓÈÈ£¬ÊÕ¼¯µ½ÆøÌå0.672L£¬ÔòÒ»¶¨º¬ÓÐNH4+£¬ÇÒÎïÖʵÄÁ¿ÊÇ0.03mol£»µÚÈý·Ý¼ÓÈë×ãÁ¿BaCl2ÈÜÒººóµÃ¸ÉÔï³Áµí6.63g£¬³Áµí¾×ãÁ¿ÑÎËáÏ´µÓ£¬¸ÉÔïºóÊ£Óà4.66g£¬ËµÃ÷Ò»¶¨º¬ÓÐCO32£ºÍSO42££¬ÇÒÎïÖʵÄÁ¿·Ö±ðÊÇ0.01molºÍ0.02mol£¬Ôòc(CO32£)=0.01molL¡Â0.01mol/L£½1mol/L¡£Òò´ËÈÜÒºÖÐÒ»¶¨Ã»ÓÐMg2£«¡¢Fe3£«¡¢Al3£«¡£ÓÉÓÚÈÜÒºÊǵçÖÐÐԵ쬶øCO32£ºÍSO42£µÄ¸ºµçºÉÊýÊÇ0.06mol£¬NH4+ÊÇ0.03mol£¬¶øÂÈÀë×ÓÊDz»ÄÜÈ·¶¨µÄ£¬ËùÒÔ¼ØÀë×ÓÒ»¶¨ÓУ¬ÇÒÆäÎïÖʵÄÁ¿´óÓÚ»òµÈÓÚ0.03mol£¬Ôò¼ØÀë×ÓŨ¶È´óÓÚ»òµÈÓÚ3 mol/L¡£¼ìÑ鼨Àë×ӵķ½·¨ÊÇÈ¡ÉÙÁ¿ÔÊÔÒº£¬¼Ó×ãÁ¿ÏõËá±µÈÜÒº¹ýÂË£¬È¡ÂËÒº¼Ó×ãÁ¿Ï¡ÏõËáºÍÏõËáÒøÈÜÒº£¬ÈôÓгÁµíÖ¤Ã÷´æÔÚÂÈÀë×Ó£¬·ñÔòÎÞÂÈÀë×Ó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿