ÌâÄ¿ÄÚÈÝ

ÎïÖÊÔÚË®ÖпÉÄÜ´æÔÚµçÀëƽºâ¡¢ÑεÄË®½âƽºâºÍ³ÁµíµÄÈܽâƽºâ£¬ËüÃǶ¼¿É¿´×÷»¯Ñ§Æ½ºâ¡£Çë¸ù¾ÝËùѧ֪ʶµÄ»Ø´ð£º

£¨1£©Å¨¶ÈΪ0.1mol/LµÄ8ÖÖÈÜÒº£º¢ÙHNO3 ¢ÚH2SO4 ¢ÛHCOOH ¢ÜBa(OH£©2 ¢ÝNaOH ¢ÞCH3COONa ¢ßKCl ¢àNH4Cl ÈÜÒºpHÖµÓÉСµ½´óµÄ˳ÐòÊÇ(Ìîд±àºÅ£©____________¡£

£¨2£©BΪ0.1mol•L-1NaHCO3ÈÜÒº£¬ÊµÑé²âµÃNaHCO3ÈÜÒºµÄpH£¾7£¬ÇëÓÃÀë×Ó·½³Ìʽ±íʾÆäÔ­Òò£º___________________¡£

£¨3£©ÔÚ0.10 mol•L-1ÁòËáÍ­ÈÜÒºÖмÓÈëÇâÑõ»¯ÄÆÏ¡ÈÜÒº³ä·Ö½Á°èÓÐdzÀ¶É«ÇâÑõ»¯Í­³ÁµíÉú³É£¬µ±ÈÜÒºµÄpH=8ʱ£¬c(Cu2+£©=___________(ÒÑÖªKsp[Cu(OH£© 2]=2.2x10-20£©¡£ÈôÔÚ0.10 mol•L-1ÁòËáÍ­ÈÜÒºÖÐͨÈë¹ýÁ¿H2SÆøÌ壬ʹCu2+ÍêÈ«³ÁµíΪCuS£¬´ËʱÈÜÒºÖеÄH+Ũ¶ÈÊÇ________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø