ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿£¨1£©25¡æʱ£¬Å¨¶ÈΪ0.1 mo1¡¤L-1µÄ6ÖÖÈÜÒº£º¢ÙHCl£¬¢ÚCH3COOH ¢ÛBa(OH)2 ¢ÜNa2CO3 ¢ÝKCl ¢ÞNH4ClÈÜÒºpHÓÉСµ½´óµÄ˳ÐòΪ_______________£¨Ìîд±àºÅ£©

£¨2£©25¡æʱ£¬´×ËáµÄµçÀë³£ÊýKa=1.7¡Á10-5mol/L£¬Ôò¸ÃζÈÏÂCH3COONaµÄË®½âƽºâ³£ÊýKh=_______ mo1¡¤L-1 £¨±£Áôµ½Ð¡Êýµãºóһ룩¡£

£¨3£©25¡æʱ£¬pH=3µÄ´×ËáºÍpH=11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒº³Ê________(Ìî¡°ËáÐÔ¡±£¬¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±)£¬Çëд³öÈÜÒºÖÐÀë×ÓŨ¶È¼äµÄÒ»¸öµÈʽ£º________________________________¡£

£¨4£©25¡æʱ£¬½«mmol/LµÄ´×ËáºÍnmol/LµÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖÐc(CH3COO-)+c(CH3COOH)= _____________£¬mÓënµÄ´óС¹ØϵÊÇm_____n(Ìî¡°>¡±¡°=¡±»ò¡°<¡±)¡£

£¨5£©µ±300mL1 mo1¡¤L-1µÄNaOHÈÜÒºÎüÊÕ±ê×¼×´¿öÏÂ4.48LCO2ʱ£¬ËùµÃÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ_________________________________________¡£

¡¾´ð°¸¡¿ ¢Ù<¢Ú<¢Þ<¢Ý<¢Ü<¢Û 5.9¡Á10-10 ËáÐÔ (H+)+(Na+)=(CH3COO-)+C(OH-) mol/L > c(Na+)>c(HCO3-)>c(CO32-)>c(OH-)>c(H+)

¡¾½âÎö¡¿±¾ÌâÖ÷Òª¿¼²éÑÎÀàË®½â¡£

£¨1£©25¡æʱ£¬Å¨¶ÈΪ0.1 mo1¡¤L-1µÄ6ÖÖÈÜÒº£º¢ÙHClÍêÈ«µçÀë²úÉúH+£¬¢ÚCH3COOHС²¿·ÖµçÀë²úÉúH+£¬¢ÛBa(OH)2ÍêÈ«µçÀë²úÉúOH-£¬¢ÜNa2CO3Ë®½â²úÉúOH-£¬ÇÒË®½â³Ì¶ÈС£¬¢ÝKClÈÜÒº³ÊÖÐÐÔ£¬pH=7£¬¢ÞNH4ClË®½â²úÉúH+£¬ÇÒË®½â³Ì¶ÈСÓÚ¢ÚCH3COOHµÄµçÀë³Ì¶È£¬ËùÒÔÈÜÒºpHÓÉСµ½´óµÄ˳ÐòΪ¢Ù<¢Ú<¢Þ<¢Ý<¢Ü<¢Û¡£

£¨2£©25¡æʱ£¬´×ËáµÄµçÀë³£ÊýKa=1.7¡Á10-5mol/L£¬Ôò¸ÃζÈÏÂCH3COONaµÄË®½âƽºâ³£ÊýKh==5.9¡Á10-10mo1¡¤L-1¡£

£¨3£©25¡æʱ£¬pH=3µÄ´×ËáºÍpH=11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬Ê£Óà´óÁ¿´×ËᣬÈÜÒº³ÊËáÐÔ£¬ÈÜÒºÖеçºÉÊغ㣺(H+)+(Na+)=(CH3COO-)+C(OH-)¡£

£¨4£©c(CH3COO-)+c(CH3COOH)= mol/L£¬Èô´×ËáÓëÇâÑõ»¯ÄÆÇ¡ºÃÍêÈ«·´Ó¦Ðγɴ×ËáÄÆÈÜÒº£¬Ôò´×ËáÄÆË®½âʹÈÜÒº³Ê¼îÐÔ£¬µ±ÈÜÒºµÄpH=7ʱ£¬Ê£Óà´×Ëᣬm>n¡£

£¨5£©µ±300mL1mo1¡¤L-1µÄNaOHÈÜÒº(º¬ÓÐ0.3molNaOH)ÎüÊÕ±ê×¼×´¿öÏÂ4.48L¼´0.2molCO2ʱ£¬3OH£­+2CO2++H2O£¬ÐγɵÈŨ¶ÈµÄNaHCO3ºÍNa2CO3µÄ»ìºÏÈÜÒº£¬ ¡¢Ë®½âʹÈÜÒº³Ê¼îÐÔ£¬ËüÃǵÄË®½â³Ì¶ÈºÜС¶øÇÒË®½â³Ì¶È£º <£¬Òò´Ë£¬ËùµÃÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc(Na+)>c(HCO3-)>c(CO32-)>c(OH-)>c(H+)¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø