·ÖÎö £¨1£©¸ù¾ÝÈÜÒºµÄpH¼ÆËãÈÜÒºÖÐÇâÀë×ÓŨ¶È£¬¸ù¾ÝÇâÑõ»¯ÄƵÄŨ¶È¼ÆËãÇâÑõ¸ùÀë×ÓŨ¶È£¬ÔÙ½áºÏKw=c£¨H+£©£®c£¨OH-£©¼ÆËã¼´¿É£»
£¨2£©¸ù¾Ýn=$\frac{m}{M}$¼ÆËã4gNaOHµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾Ýc=$\frac{n}{V}$¼ÆËãÈÜÒºÎïÖʵÄÁ¿Å¨¶È£¬È»ºó¸ù¾Ý¼îÈÜÒºÖУ¬ÇâÀë×ÓÈ«²¿À´×ÔÓÚË®µÄµçÀëÀ´·ÖÎö£»
£¨3£©¸ù¾ÝÔÚBa£¨OH£©2ÈÜÒºÖУ¬Ba2+ºÍOH-µÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º2£¬Çó³öOH-µÄÎïÖʵÄÁ¿£¬È»ºó¸ù¾ÝC=$\frac{n}{V}$Çó³öOH-µÄÎïÖʵÄÁ¿Å¨¶È£¬È»ºó¸ù¾ÝË®µÄÀë×Ó»ýÇó³öC£¨H+£©£¬´Ó¶øÇó³öpH£»
£¨4£©Ëá¼î»ìºÏºópH=3£¬ÔòËá¹ýÁ¿£¬½áºÏ¹ýÁ¿µÄÇâÀë×ÓŨ¶È¼ÆË㣮
½â´ð ½â£º£¨1£©0.01mol•L-1µÄNaOHÈÜÒºµÄpHΪ11£¬ÔòÇâÀë×ÓŨ¶È=10-11 mol/L£¬ÇâÑõ»¯ÄÆÊÇÇ¿µç½âÖÊÍêÈ«µçÀ룬ËùÒÔÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶ÈÊÇ0.01mol/L£¬ÔòKw=c£¨H+£©£®c£¨OH-£©=0.01¡Á10-11=10-13£¬¹Ê´ð°¸Îª£º10-13£»
£¨2£©4gNaOHµÄÎïÖʵÄÁ¿=$\frac{4g}{40g/mol}$=0.1mol£¬ÈÜÓÚË®Åä³É500mLÈÜÒº£¬ËùµÃÈÜÒºÎïÖʵÄÁ¿Å¨¶È=$\frac{0.1mol}{0.5L}$=0.2mol/L£¬ÈÜÒºÖÐc£¨OH-£©=0.2mol/L£¬c£¨H+£©=$\frac{1{0}^{-14}}{0.2}$=5¡Á10-14mol/L£¬ÇÒÇâÀë×ÓÈ«²¿À´×ÔÓÚË®µÄµçÀ룬¶øË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈºÍË®µçÀë³öµÄÇâÑõ¸ùµÄŨ¶ÈÏàͬ£¬¹Ê´ËÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©Îª5¡Á10-14mol/L£¬¹Ê´ð°¸Îª£º5¡Á10-14£»
£¨3£©ÔÚBa£¨OH£©2ÈÜÒºÖУ¬Ba2+ºÍOH-µÄÎïÖʵÄÁ¿Ö®±ÈΪ1£º2£¬¹Ên£¨OH-£©=2n£¨Ba2+£©=2¡Á10-3mol£¬¹ÊÈÜÒºÖÐC£¨OH-£©=$\frac{n}{V}$=$\frac{2¡Á1{0}^{-3}mol}{0.2L}$=0.01mol/L£¬ÔòC£¨H+£©=$\frac{1{0}^{-14}}{0.01}$=10-12mol/L£¬¹ÊpH=-lgC£¨H+£©=12£¬¹Ê´ð°¸Îª£º12£»
£¨4£©½«pH=12µÄNaOHÈÜÒºVaLÓëpH=2µÄH2SO4ÈÜÒºVbL»ìºÏ£¨ºöÂÔÌå»ý±ä»¯£©£¬ÈôËùµÃ»ìºÏÒºµÄpH=3£¬ÔòËá¹ýÁ¿£¬Ôò$\frac{{V}_{b}L¡Á0.01mol/L-{V}_{a}L¡Á0.01mol/L}{{£¨V}_{a}+{V}_{b}£©L}$=0.001£¬½âµÃVa£ºVb=9£º11£¬¹Ê´ð°¸Îª£º9£º11£®
µãÆÀ ±¾Ì⿼²éÁËËá¼î·´Ó¦ºóÈÜÒºËá¼îÐÔÅжϼÆË㣬Éæ¼°ÈÜÒºpHµÄ¼ÆËã¡¢ÈÜÒºÖÐÀë×Ó»ýµÄ¼ÆËãÓ¦Óõȣ¬°ÑÎÕËá¼î·´Ó¦ºó¼î¹ýÁ¿ÊǽâÌâ¹Ø¼ü£¬×¢ÒⲻͬζÈÏÂÀë×Ó»ý²»Í¬£¬ÌâÄ¿ÄѶÈÖеȣ®
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
Ñ¡Ïî | ÏÖÏó»òÊÂʵ | ½âÊÍ |
A | ËáÐÔKMnO4ÈÜÒºÖмÓÈëH2O2ÈÜÒº£¬²úÉúÆøÌå | KMnO4´ß»¯H2O2·Ö½â²úÉúO2 |
B | ¹âÁÁµÄÌúƬ·ÅÔÚŨÏõËáÖнþÅÝÒ»¶Îʱ¼äºóÈ¡³ö£¬ÓÃÕôÁóË®ÖгåÏ´ºó·ÅÈëCuSO4ÈÜÒº£¬ÎÞÃ÷ÏÔÏÖÏó | ÌúƬÔÚŨÏõËáÖз¢Éú¶Û»¯£¬±íÃæÉú³ÉÖÂÃܵÄÑõ»¯Ä¤ |
C | Ïò·ÏFeCl3Ê´¿ÌÒºXÖмÓÈëÉÙÁ¿Ìú·Û£¬Õñµ´£¬µÃµ½³ÎÇå͸Ã÷µÄÈÜÒº | XÖÐÒ»¶¨²»º¬Cu2+ |
D | ·Ö±ðÔÚÊ¢ÓÐÏàͬŨ¶ÈZnSO4¡¢CuSO4ÈÜÒºµÄÊÔ¹ÜÖÐͨÈë×ãÁ¿H2SÆøÌ壬ºóÕßÓкÚÉ«³Áµí | KSP£¨ZnS£©£¼KSP£¨CuS£© |
A£® | A | B£® | B | C£® | C | D£® | D |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ¸÷ÎïÖʵÄŨ¶ÈÖ®±Èc£¨X£©£ºc£¨Y£©£ºc£¨Z£©=2£º3£º4 | B£® | ÏûºÄ2molX£¬Í¬Ê±ÏûºÄ4molZ | ||
C£® | »ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä | D£® | ·´Ó¦·Å³öakJÈÈÁ¿ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | ±ê×¼×´¿öÏ£¬8.0gSO3Ìå»ýΪ2.24L | |
B£® | ÎïÖʵÄÁ¿Îª1molSO2¡¢CO2µÄ»ìºÏÎïËùº¬ÑõÔ×ÓÊýΪ2NA | |
C£® | 0.1molFeCl3ÈÜÒºÖк¬ÓеÄFe3+ÊýΪ0.1NA | |
D£® | 0.5molþÔÚ¿ÕÆøÖÐÍêȫȼÉÕËùÐèÒªµÄÑõÆøµÄÎïÖʵÄÁ¿Îª0.25mol |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
A£® | 4.4g±ûÍ麬ÓÐ̼Çâ¼üµÄÊýĿΪNA | |
B£® | 3.9gNa2O2Óë×ãÁ¿CO2·´Ó¦×ªÒƵĵç×ÓÊýΪ0.1NA | |
C£® | ±ê×¼×´¿öÏ£¬11.2LH2OÖк¬ÓеķÖ×ÓÊýΪ0.5NA | |
D£® | 9.2gNO2ºÍN2O4µÄ»ìºÏÆøÌåÖк¬ÓеĵªÔ×ÓÊýΪ0.2NA |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ
Ñ¡Ïî | ʵÑé | ÏÖÏó | ÓÉÏÖÏóËùµÃ½áÂÛ |
A | ½«Cl2ͨÈëÆ·ºìÈÜÒº | Æ·ºìÍÊÉ« | Cl2¾ßÓÐƯ°×ÐÔ |
B | ·Ö±ð²âÁ¿±¥ºÍNaCO3¡¢±¥ºÍNaHCO3ÈÜÒºµÄpH | pHÖµ£ºÌ¼ËáÄÆ£¾Ì¼ËáÇâÄÆ | Ë®½âÄÜÁ¦£ºCO32-£¾HCO3- |
C | ½«CO2ͨÈëCaCl2µÄÈÜÒºÖÐ | ÎÞÃ÷ÏÔÏÖÏó | ËáÐÔ£ºÑÎË᣾̼Ëá |
D | ¹Û²ìÌú¿é¡¢ÂÁ²ÔÚ³±Êª¿ÕÆøÖеĸ¯Ê´Çé¿ö | Ìú¿éÉúÐ⣬¶øÂÁ²¼¸ºõÎޱ仯 | ÔÚ³±Êª¿ÕÆøÖÐÌú±ÈÂÁÒª»îÆà |
A£® | A | B£® | B | C£® | C | D£® | D |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com