Question

14£®HAλ²»Í¬µÄÒ»ÔªËᣬÓÃHAÈÜÒººÍNaOHÈÜÒº×öÒÔÏÂʵÑ飮
£¨1£©µ±c£¨HA£©=c£¨NaOH£©=0.1mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃ»ìºÏºóÈÜÒºµÄpH£¾7£®ÔòHAΪÈõËᣨÌî¡°Ç¿Ëᡱ»ò¡°ÈõËᡱ£©£¬±íʾÈÜÒºpH£¾7µÄÔ­ÒòµÄÀë×Ó·½³ÌʽÊÇA^-+H₂O?HA+OH^-£®
£¨2£©µ±HAΪǿËᣬc£¨NaOH£©=0.1mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃÈÜÒºµÄpH=12£¬Ôò»ìºÏÇ°c£¨HA£©=0.08mol/L£®
£¨3£©µ±c£¨HA£©=0.04mol/L£¬c£¨NaOH£©=0.02mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃÈÜÒºµÄpH£¾7£¬ÔòÏÂÁÐÓйػìºÏÈÜÒºµÄ˵·¨ÖÐÕýÈ·µÄÊÇABC£®
A£®HAÒÖÖÆNaAµÄË®½â£¬NaAÒÖÖÆHAµÄµçÀë   B£®c£¨Na⁺£©£¾c£¨A^-£©  
C£®c£¨HA£©+c£¨A^-£©=0.02mol/L           D£®c£¨A^-£©£¾c£¨HA£©
£¨4£©µ±HAΪHCl£¬Óñê×¼NaOHÈÜÒºµÎ¶¨Ê±£¬ÏÂÁÐÇé¿öÄÜÒýÆðËù²âc£¨HCl£©Æ«´óµÄÊÇB£®
A£®ÓÃÕôÁóˮϴµÓ׶ÐÎÆ¿ºó£¬Î´¸ÉÔï¼´¼ÓÈë´ý²âÒº½øÐеζ¨
B£®¼îʽµÎ¶¨¹ÜÓÃˮϴµÓºó£¬Ö±½ÓÊ¢·Å±ê×¼NaOHÈÜÒºµÎ¶¨
C£®µÎ¶¨Ç°ÑöÊÓ¶ÁÊý£¬µÎ¶¨ºó¸©ÊÓ¶ÁÊý£®

Answer 1

·ÖÎö £¨1£©Ç¿ËáÈõ¼îÑÎË®½âÏÔ¼îÐÔ£»
£¨2£©Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃÈÜÒºµÄpH=12£¬Ôò»ìºÏºóÈÜÒºÖеÄÇâÑõ¸ùÀë×ÓŨ¶ÈΪc£¨OH^-£©=$\frac{c£¨NaOH£©¡ÁV-c£¨HA£©¡ÁV}{V+V}$=10^-2mol/L£¬¾Ý´Ë¼ÆË㣻
£¨3£©c£¨HA£©=0.04mol/L£¬c£¨NaOH£©=0.02mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒºÖеÄÈÜÖÊΪµÈÎïÖʵÄÁ¿µÄNaAºÍHA£¬²âµÃÈÜÒºµÄpH£¾7£¬ËµÃ÷ÑεÄË®½â³Ì¶È´óÓÚ´óÓÚËáµÄµçÀë³Ì¶È£¬½áºÏµçºÉÊغãºÍÎïÁÏÊغã·ÖÎö£»
£¨4£©¸ù¾Ýc£¨HCl£©=$\frac{c£¨NaOH£©¡ÁV£¨NaOH£©}{V£¨HCl£©}$·ÖÎö£®

½â´ð ½â£º£¨1£©µ±c£¨HA£©=c£¨NaOH£©=0.1mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬¶þÕßÇ¡ºÃ·´Ó¦Éú³ÉÑΣ¬²âµÃ»ìºÏºóÈÜÒºµÄpH£¾7£¬ËµÃ÷¸ÃÑÎΪǿËáÈõ¼îÑΣ¬Ë®½âÏÔ¼îÐÔ£¬¼´HAΪÈõËᣬÆäÑεÄË®½âÀë×Ó·½³ÌʽΪ£ºA^-+H₂O?HA+OH^-£»
¹Ê´ð°¸Îª£ºÈõË᣻A^-+H₂O?HA+OH^-£»
£¨2£©µ±HAΪǿËᣬc£¨NaOH£©=0.1mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃÈÜÒºµÄpH=12£¬Ôò»ìºÏºóÈÜÒºÖУ¬c£¨OH^-£©=10^-2mol/L£¬¼´c£¨OH^-£©=$\frac{c£¨NaOH£©¡ÁV-c£¨HA£©¡ÁV}{V+V}$=10^-2mol/L£¬½âµÃc£¨HA£©=0.08mol/L£»
¹Ê´ð°¸Îª£º0.08£»
£¨3£©c£¨HA£©=0.04mol/L£¬c£¨NaOH£©=0.02mol/L£¬Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒºÖеÄÈÜÖÊΪµÈÎïÖʵÄÁ¿µÄNaAºÍHA£¬²âµÃÈÜÒºµÄpH£¾7£¬ËµÃ÷ÑεÄË®½â³Ì¶È´óÓÚ´óÓÚËáµÄµçÀë³Ì¶È£»
A£®HAÒÖÖÆNaAµÄË®½â£¬NaAÒÖÖÆHAµÄµçÀ룬¶þÕßÏ໥ÒÖÖÆ£¬¹ÊAÕýÈ·£»
B£®ÈÜÒºÖеĵçºÉÊغãΪc£¨Na⁺£©+c£¨H⁺£©=c£¨A^-£©+c£¨OH^-£©£¬ÒÑÖª pH£¾7£¬¼´c£¨H⁺£©£¼c£¨OH^-£©£¬ËùÒÔc£¨Na⁺£©£¾c£¨A^-£©£¬¹ÊBÕýÈ·£»
C£®Á½ÈÜÒºµÈÌå»ý»ìºÏ£¬HAºÍNaAµÄ×ÜŨ¶È±äΪԭÀ´µÄÒ»°ë£¬¼´c£¨HA£©+c£¨A^-£©=0.02mol/L£¬¹ÊCÕýÈ·£»
 D£®ÑεÄË®½â³Ì¶È´óÓÚ´óÓÚËáµÄµçÀë³Ì¶È£¬Ôòc£¨A^-£©£¼c£¨HA£©£¬¹ÊD´íÎó£»
¹Ê´ð°¸Îª£ºABC£»
£¨4£©µ±HAΪHCl£¬Óñê×¼NaOHÈÜÒºµÎ¶¨Ê±£¬c£¨HCl£©=$\frac{c£¨NaOH£©¡ÁV£¨NaOH£©}{V£¨HCl£©}$£¬
A£®ÓÃÕôÁóˮϴµÓ׶ÐÎÆ¿ºó£¬Î´¸ÉÔï¼´¼ÓÈë´ý²âÒº½øÐе樣¬´ý²âÈÜÒºµÄÎïÖʵÄÁ¿²»±ä£¬ÏûºÄµÄÇâÑõ»¯ÄƲ»±ä£¬ËùÒÔ²âµÃ½á¹û²»±ä£¬¹ÊA²»Ñ¡£»
B£®¼îʽµÎ¶¨¹ÜÓÃˮϴµÓºó£¬Ö±½ÓÊ¢·Å±ê×¼NaOHÈÜÒºµÎ¶¨£¬µ¼ÖÂÇâÑõ»¯ÄƵÄŨ¶ÈƫС£¬µÎ¶¨Ê±ÏûºÄµÄÇâÑõ»¯ÄƵÄÌå»ýÆ«´ó£¬Ôò²â¶¨½á¹ûÆ«´ó£¬¹ÊBÕýÈ·£»
C£®µÎ¶¨Ç°ÑöÊÓ¶ÁÊý£¬µÎ¶¨ºó¸©ÊÓ¶ÁÊý£¬¶Á³öµÄÇâÑõ»¯ÄƵÄÌå»ýƫС£¬ÓɼÆË㹫ʽ¿ÉÖª£¬¼ÆËãµÃµ½µÄÑÎËáµÄŨ¶ÈƫС£¬¹ÊC´íÎó£®
¹Ê´ð°¸Îª£ºB£®

µãÆÀ ±¾Ì⿼²éÁËÑεÄË®½â¡¢Èõµç½âÖʵĵçÀë¡¢ÈÜÒºÖÐÊغã¹ØϵµÄÓ¦Óá¢Ëá¼îÖк͵樵ÄÎó²î·ÖÎöµÈ£¬ÌâÄ¿×ÛºÏÐÔ½ÏÇ¿£¬ÄѶÈÖеȣ¬×¢ÖضԳ£¿¼¿¼µãµÄ×ÛºÏÓ¦ÓõĿ¼²é£®