ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÑÖªAg2SO4µÄKspΪ2.0¡Á10£­5 mol3¡¤L£­3£¬½«ÊÊÁ¿Ag2SO4¹ÌÌåÈÜÓÚ100 mLË®ÖÐÖÁ¸ÕºÃ±¥ºÍ£¬¸Ã¹ý³ÌÖÐAg£« ºÍSO42£­Å¨¶ÈËæʱ¼ä±ä»¯¹ØϵÈçÓÒͼ[±¥ºÍAg2SO4ÈÜÒºÖÐc(Ag£«)=0.034 molL£­1]¡£Èôt1ʱ¿ÌÔÚÉÏÊöÌåϵÖмÓÈë100 mL 0.020 molL£­1 Na2SO4ÈÜÒº£¬ÏÂÁÐʾÒâͼÖУ¬ÄÜÕýÈ·±íʾt1ʱ¿ÌºóAg£« ºÍSO42£­Å¨¶ÈËæʱ¼ä±ä»¯¹ØϵµÄÊÇ

A. A B. B C. C D. D

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿

Ag2SO4¸ÕºÃΪ100mlµÄ±¥ºÍÈÜÒº£¬ÒòΪc(Ag+)=0.034molL-1£¬Ôòc(SO42-)=0.017molL-1£»µ±¼ÓÈë100mL 0.020molL-1Na2SO4ÈÜÒººó£¬c(SO42-)==0.0185molL-1£¬c(Ag+)=0.017molL-1£¬ÈôûÓгÁµíÎö³ö£¬Ôò»ìºÏºóÒøÀë×ÓŨ¶ÈΪԭÀ´µÄÒ»°ë£¬ÁòËá¸ùÀë×ÓŨÂÔÓÐÔö´ó£¬ÓÉ´Ë¿ÉÒÔ¿´³öӦΪBͼÏ󣬴ËʱQ=c(SO42-)c2(Ag+)=0.0185¡Á(0.017)2=5.346¡Á10-6£¼Ksp=2.0¡Á10-3£¬Ôò¸ÃÈÜÒºÖÐûÓгÁµíÎö³ö£¬¹ÊÑ¡B¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø