ÌâÄ¿ÄÚÈÝ
£¨10·Ö£©³£ÎÂÏÂÓÐŨ¶È¾ùΪ0.05mol/LµÄËÄÖÖÈÜÒº£º¢ÙNa2CO3¢ÚNaHCO3¢ÛHCl ¢ÜNH3¡¤H2O£¬»Ø´ðÏà¹ØÎÊÌ⣺
£¨1£©ÉÏÊöÈÜÒºÖУ¬¿É·¢ÉúË®½âµÄÊÇ £¨ÌîÐòºÅ£©
£¨2£©ÉÏÊöÈÜÒºÖУ¬¼ÈÄÜÓëNaOHÈÜÒº·´Ó¦£¬ÓÖÄÜÓëH2SO4ÈÜÒº·´Ó¦µÄÈÜ
ÒºÖУ¬Àë×ÓŨ¶È´óСµÄ¹Øϵ
£¨3£©Ïò¢ÜÖмÓÈëÉÙÁ¿NH4Cl¹ÌÌ壬´Ëʱc(NH4+/OH-)µÄÖµ £¨¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡± £©
£¨4£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºó£¬ÈÜҺǡºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°¢ÛµÄÌå»ý ¢ÜµÄÌå»ý£¨¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡± £©
£¨5£©È¡10mLµÄ¢ÛÈÜÒº£¬¼ÓˮϡÊ͵½500mL£¬Ôò´ËÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+) = mol/L
£¨1£©ÉÏÊöÈÜÒºÖУ¬¿É·¢ÉúË®½âµÄÊÇ £¨ÌîÐòºÅ£©
£¨2£©ÉÏÊöÈÜÒºÖУ¬¼ÈÄÜÓëNaOHÈÜÒº·´Ó¦£¬ÓÖÄÜÓëH2SO4ÈÜÒº·´Ó¦µÄÈÜ
ÒºÖУ¬Àë×ÓŨ¶È´óСµÄ¹Øϵ
£¨3£©Ïò¢ÜÖмÓÈëÉÙÁ¿NH4Cl¹ÌÌ壬´Ëʱc(NH4+/OH-)µÄÖµ £¨¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡± £©
£¨4£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºó£¬ÈÜҺǡºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°¢ÛµÄÌå»ý ¢ÜµÄÌå»ý£¨¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡± £©
£¨5£©È¡10mLµÄ¢ÛÈÜÒº£¬¼ÓˮϡÊ͵½500mL£¬Ôò´ËÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+) = mol/L
£¨10·Ö£©
£¨1£© ¢Ù¢Ú
£¨2£© c(Na+)£¾ c(HCO3¡ª) £¾ c(OH¡ª) £¾ c(H+)£¾ c(CO32¡ª)
£¨3£© Ôö´ó £¨4£© СÓÚ
£¨5£© 10-11
£¨1£© ¢Ù¢Ú
£¨2£© c(Na+)£¾ c(HCO3¡ª) £¾ c(OH¡ª) £¾ c(H+)£¾ c(CO32¡ª)
£¨3£© Ôö´ó £¨4£© СÓÚ
£¨5£© 10-11
£¨1£©ÔÚÑÎÈÜÒºÖУ¬Ö»Óк¬ÓÐÈõÀë×Ó²ÅÄÜË®½â£¬ËùÒÔ£¬ÉÏÊöÈÜÒºÖпɷ¢ÉúË®½âµÄÊÇ£ºNa2CO3¡¢NaHCO3£¬ËùÒÔÌî¢Ù¢Ú
£¨2£©¼ÈÄÜÓëNaOHÈÜÒº·´Ó¦£¬ÓÖÄÜÓëH2SO4ÈÜÒº·´Ó¦µÄÈÜÒºÖ¸µÄÊÇNaHCO3£¬NaHCO3ÈÜÒºÏÔ¼îÐÔ£¬HCO3£µÄË®½â³Ì¶È´óÓÚµçÀë³Ì¶È¡£Àë×ÓŨ¶ÈµÄ´óС¹ØϵÊÇ£ºc(Na+)£¾ c(HCO3¡ª) £¾ c(OH¡ª) £¾ c(H+)£¾ c(CO32¡ª)
£¨3£©Ïò¢ÜÖмÓÈëÉÙÁ¿NH4Cl¹ÌÌ壬¢Ü´æÔÚһˮºÏ°±µÄµçÀëƽºâ£¬°±¸ùÀë×ÓŨ¶ÈÔö´ó£¬Æ½ºâ×óÒÆ£¬´Ëʱc(NH4+/OH-)µÄÖµÔö´ó¡£
£¨4£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºó£¬»áµÃµ½ÂÈ»¯ï§£¬ÈÜҺǡºÃ³ÊÖÐÐÔ£¬°±Ë®Ó¦¸Ã¹ýÁ¿£¬ËùÒÔ£¬»ìºÏÇ°¢ÛµÄÌå»ýСÓڢܵÄÌå»ý
£¨5£©0.05mol/L10mLµÄHClÈÜÒº£¬¼ÓˮϡÊ͵½500mL£¬´ËʱÑÎËáµÄŨ¶ÈΪ0.001mol/L, c(H+) = 10-3mol/L,ÄÇôˮµçÀë³öµÄc(H+)=10-11mol/L
£¨2£©¼ÈÄÜÓëNaOHÈÜÒº·´Ó¦£¬ÓÖÄÜÓëH2SO4ÈÜÒº·´Ó¦µÄÈÜÒºÖ¸µÄÊÇNaHCO3£¬NaHCO3ÈÜÒºÏÔ¼îÐÔ£¬HCO3£µÄË®½â³Ì¶È´óÓÚµçÀë³Ì¶È¡£Àë×ÓŨ¶ÈµÄ´óС¹ØϵÊÇ£ºc(Na+)£¾ c(HCO3¡ª) £¾ c(OH¡ª) £¾ c(H+)£¾ c(CO32¡ª)
£¨3£©Ïò¢ÜÖмÓÈëÉÙÁ¿NH4Cl¹ÌÌ壬¢Ü´æÔÚһˮºÏ°±µÄµçÀëƽºâ£¬°±¸ùÀë×ÓŨ¶ÈÔö´ó£¬Æ½ºâ×óÒÆ£¬´Ëʱc(NH4+/OH-)µÄÖµÔö´ó¡£
£¨4£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºó£¬»áµÃµ½ÂÈ»¯ï§£¬ÈÜҺǡºÃ³ÊÖÐÐÔ£¬°±Ë®Ó¦¸Ã¹ýÁ¿£¬ËùÒÔ£¬»ìºÏÇ°¢ÛµÄÌå»ýСÓڢܵÄÌå»ý
£¨5£©0.05mol/L10mLµÄHClÈÜÒº£¬¼ÓˮϡÊ͵½500mL£¬´ËʱÑÎËáµÄŨ¶ÈΪ0.001mol/L, c(H+) = 10-3mol/L,ÄÇôˮµçÀë³öµÄc(H+)=10-11mol/L
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿