Question

18£®¡°±ä»¯¹ÛÄîÓëƽºâ˼Ï롱ÊÇ»¯Ñ§Ñ§¿ÆµÄºËÐÄËØÑø£¬ÊÒÎÂʱ£¬0.1mol•L^-1²ÝËáÄÆÈÜÒºÖдæÔÚ¶à¸öƽºâ£¬ÆäÖÐÓйØ˵·¨ÕýÈ·µÄÊÇ[¼ºÖªÊÒÎÂʱ£¬K_sp£¨CaC₂O₄£©=2.4¡Á10^-9]£¨¡¡¡¡£©

• A£® Èô½«ÈÜÒº²»¶Ï¼ÓˮϡÊÍ£¬ÔòË®µÄµçÀë³Ì¶ÈÔö´ó
• B£® ÈÜÒºÖи÷Àë×ÓŨ¶È´óС¹Øϵ£ºc£¨Na⁺£©£¾c£¨C₂O₄^2-£©£¾c£¨OH^-£©£¾c£¨H⁺£©£¾c£¨HC₂O₄^-£©
• C£® ÈôÓÃpH¼Æ²âµÃÈÜÒºµÄpH=9£¬ÔòH₂C₂O₄µÄµÚ¶þ¼¶µçÀëƽºâ³£ÊýK_a2=10^-5
• D£® ÏòÈÜÒºÖмÓÈëµÈÌå»ýCaCl₂ÈÜÒº£¬µ±¼ÓÈëµÄCaCl₂ÈÜҺŨ¶È´óÓÚ2.4¡Á10^-8 mol•L^-1ʱ¼´ÄܲúÉú³Áµí

Answer 1

·ÖÎö A£®¼ÓÈëË®£¬ÑεÄË®½â±»´Ù½ø£¬Ë®µÄµçÀë³Ì¶È¼õС£»
B.0.1mol/L²ÝËáÄÆÈÜÒºÖвÝËá¸ùÀë×ÓË®½âÈÜÒºÏÔ¼îÐÔ£»
C£®´×ËáÄÆË®½âÈÜÒºÏÔ¼îÐÔ£¬ÒÀ¾ÝË®½âƽºâ³£ÊýºÍÀë×Ó»ý³£ÊýºÍµçÀëƽºâ³£Êý¼ÆËã·ÖÎöÅжϣ»
D£®ÒÀ¾ÝK_sp£¨CaC₂O₄£©=2.4¡Á10^-9£¬µÈÌå»ý»ìºÏc£¨C₂O₄^2-£©=0.050mol/L£¬¼ÆËãCaCl₂ÈÜÒºµÄ×îСŨ¶È£®

½â´ð ½â£ºA£®ÈôÏòÈÜÒºÖв»¶Ï¼ÓˮϡÊÍ£¬ÑÎÀàË®½â³Ì¶ÈÔö´ó£¬µ«Ë®µÄµçÀë³Ì¶È¼õС£¬¹ÊA´íÎó£»
B.0.1mol/L²ÝËáÄÆÈÜÒºÖвÝËá¸ùÀë×Ó·Ö²½Ë®½âÈÜÒºÏÔ¼îÐÔ£¬ÈÜÒºÖÐÀë×ÓŨ¶È´óСΪ£ºc£¨Na⁺£©£¾c£¨C₂O₄^2-£©£¾c£¨OH^-£©£¾c£¨HC₂O₄^-£©£¾c£¨H⁺£©£¬¹ÊB´íÎó£»
C£®ÈôÓÃpH¼Æ²âµÃÈÜÒºµÄpH=9£¬c£¨OH^-£©=c£¨HC₂O₄^-£©=10^-14+9mol/L=10^-5mol/L£¬C₂O₄^2-+H₂O?HC₂O₄^-+OH^-£¬K_h=$\frac{c£¨O{H}^{-}£©c£¨{HC}_{2}{O}_{4}^{-}£©}{c£¨{C}_{2}{O}_{4}^{2-}£©}$=$\frac{c£¨O{H}^{-}£©c£¨H{C}_{2}{O}_{4}^{-}£©}{c£¨{C}_{2}{O}_{4}^{2-}£©}•\frac{c£¨{H}^{+}£©}{c£¨{H}^{+}£©}$=$\frac{{K}_{w}}{{K}_{a2}}$=$\frac{1{0}^{-5}mol/L¡Á1{0}^{-5}mol/L}{0.1mol/L}$=10^-9mol/L£¬ÔòK_a2=10^-5mol/L£¬¹ÊCÕýÈ·£»
D£®K_sp£¨CaC₂O₄£©=c£¨Ca²⁺£©c£¨C₂O₄^2-£©=2.4¡Á10^-9£¬c£¨Ca²⁺£©=$\frac{2.4¡Á1{0}^{-9}}{0.05}$=4.8¡Á10^-8mol/L£¬¹ÊD´íÎó£®
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÁËÑÎÀàË®½â¡¢Ó°ÏìÒòËØ¡¢Ë®½âƽºâ³£Êý¡¢µçÀëƽºâ³£ÊýºÍÀë×Ó»ý³£ÊýµÄ»ìºÏ¼ÆËã¡¢ÈܶȻý³£Êý¼ÆËãµÈ֪ʶµã£®ÌâÄ¿ÄѶÈÖеȣ¬ÊÇÖеµÌ⣮