ÌâÄ¿ÄÚÈÝ

(7·Ö)ÔÚt¡æʱ£¬Ä³NaOHÏ¡ÈÜÒºÖУ¬c(H+)=10-amol/L£¬c(OH-)=10-bmol/L£¬ÒÑÖªa+b=12,Ôò£º
£¨1£©¸ÃζÈÏ£¬Ë®µÄÀë×Ó»ý³£Êýkw=                    ¡£
£¨2£©ÔÚ¸ÃζÈÏ£¬½«100mL0.1mol/LµÄÏ¡ÁòËáÓë100mL0.4mol/LµÄNaOHÈÜÒº»ìºÏºó£¬ÈÜÒºµÄpH==             ¡£
£¨3£©¸ÃζÈÏ£¬Èô100Ìå»ýpH1=aµÄijǿËáÈÜÒºÓë1Ìå»ýpH2=bµÄijǿ¼îÈÜÒº»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°£¬¸ÃÇ¿ËáµÄpH1ÓëÇ¿¼îµÄpH2Ö®¼äÓ¦Âú×ãµÄ¹ØϵÊÇ            ¡£
£¨4£©¸ÃζÈÏ£¬pH=2µÄijËáHAÈÜÒººÍpH=10µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬»ìºÏÈÜÒºµÄpH=5¡£ÊÔ·ÖÎöÆäÔ­Òò                                                                      ¡£

£¨¹²7·Ö£¬Ã¿¿Õ2·Ö£©
£¨1£©1¡Á10-12 (1·Ö)
£¨2£©11
£¨3£©a+b=14£¨»òpH1+pH2==14£©
£¨4£©HAΪÈõËᣬHAÖкÍNaOHºó£¬»ìºÏÈÜÒºÖл¹Ê£Óà½Ï¶àµÄHA·Ö×Ó£¬¿É¼ÌÐøµçÀë³öH+£¬Ê¹ÈÜÒºµÄpH=5¡£

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø