ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿½ñÓÐÒ»»ìºÏÎïµÄË®ÈÜÒº£¬Ö»¿ÉÄܺ¬ÓÐÒÔÏÂÀë×ÓÖеÄÈô¸ÉÖÖ£ºK£«¡¢NH¡¢Cl£¡¢Mg2+¡¢Ba2+¡¢CO
¡¢SO
£¬ÏÖÈ¡Èý·Ý 100 mL ÈÜÒº½øÐÐÈçÏÂʵÑ飺
(1)µÚÒ»·Ý¼Ó×ãÁ¿ NaOH ÈÜÒº¼ÓÈȺó£¬ÊÕ¼¯µ½ÆøÌå 0.04 mol¡£
(2)µÚ¶þ·Ý¼Ó×ãÁ¿ BaCl2 ÈÜÒººó£¬µÃ¸ÉÔï³Áµí 6.27 g£¬¾×ãÁ¿ÑÎËáÏ´µÓ¡¢¸ÉÔïºó£¬³ÁµíÖÊÁ¿Îª 2.33 g¡£
(3)µÚÈý·Ý¼ÓÈë AgNO3 ÈÜÒºÓгÁµí²úÉú¡£¸ù¾ÝÉÏÊöʵÑ飬ÒÔÏÂÍƲⲻÕýÈ·µÄÊÇ
A.100 mL ÈÜÒºÖк¬ 0.04 mol NH
B.100 mL ÈÜÒºÖк¬ 0.01 mol SO42-ºÍ 0.02 mol CO32-
C.K£«¡¢Cl£¿ÉÄÜ´æÔÚ
D.Ò»¶¨²»´æÔÚBa2+¡¢Mg2+
¡¾´ð°¸¡¿C
¡¾½âÎö¡¿
(1)µÚÒ»·Ý¼Ó×ãÁ¿ NaOH ÈÜÒº¼ÓÈȺó£¬ÊÕ¼¯µ½ÆøÌå 0.04 mol£¬ÔòÆøÌåΪNH3£¬´Ó¶øµÃ³öÔÈÜÒºÖк¬NH4+0.04mol¡£
(2)µÚ¶þ·Ý¼Ó×ãÁ¿ BaCl2 ÈÜÒººó£¬µÃ¸ÉÔï³Áµí 6.27 g£¬¾×ãÁ¿ÑÎËáÏ´µÓ¡¢¸ÉÔïºó£¬³ÁµíÖÊÁ¿Îª 2.33 g£¬ÔòBaSO4Ϊ2.33g£¬=0.01mol£¬BaCO3Ϊ6.27g-2.33g=3.94g£¬
=0.02mol¡£
(3)ÒòΪ¼ìÑéSO42-¡¢CO32-ʱ£¬¼ÓÈëÁËBaCl2£¬ÒýÈëÁËCl-£¬ËùÒÔ¼ÓÈë AgNO3 ÈÜÒºÓгÁµí²úÉú£¬²»Äܿ϶¨ÔÈÜÒºÖÐÊÇ·ñº¬ÓÐCl-¡£
A£®ÒÀÒÔÉÏ·ÖÎö£¬100 mLÈÜÒºÖк¬0.04 mol NH£¬AÕýÈ·£»
B£®ÒÀ¾ÝÉÏÃæµÄ·ÖÎöÓë¼ÆË㣬100 mLÈÜÒºÖк¬0.01 mol SO42-ºÍ0.02 mol CO32-£¬BÕýÈ·£»
C£®ÒÀ¾ÝÀë×Ó¹²´æ£¬ÈÜÒºÖк¬ÓÐNH¡¢CO
¡¢SO
£¬ÔòÒ»¶¨²»º¬ÓÐMg2+¡¢Ba2+£¬ÒÀ¾ÝµçºÉÊغ㣬ÑôÀë×ÓËù´øÕýµçºÉ×ÜÊýΪ0.04mol£¬ÒõÀë×ÓËù´ø¸ºµçºÉ×ÜÊýΪ0.06mol£¬ÔòK£«ÖÁÉÙΪ0.02mol£¬²»Äܿ϶¨Cl£ÊÇ·ñ¿ÉÄÜ£¬C²»ÕýÈ·£»
D£®ÓÉCÖзÖÎö¿ÉÖª£¬Ò»¶¨²»´æÔÚBa2+¡¢Mg2+£¬DÕýÈ·£»
¹ÊÑ¡C¡£
![](http://thumb2018.1010pic.com/images/loading.gif)