Question

²ÝËᣨH₂C₂O₄£©ÊÇÒ»ÖÖÒ×ÈÜÓÚË®µÄ¶þÔªÖÐÇ¿ËᣬÔÚË®ÖÐËüµÄ´æÔÚÐÎ̬ÓÐH₂C₂O₄¡¢HC₂O₄^-¡¢C₂O₄^2-£¬¸÷ÐÎ̬µÄ·Ö²¼ÏµÊý£¨Å¨¶È·ÖÊý£©¦ÁËæÈÜÒºpH±ä»¯µÄ¹ØϵÈçͼËùʾ£º
£¨1£©Í¼ÖÐÇúÏß1±íʾ

 

µÄ·Ö²¼ÏµÊý±ä»¯£®
£¨2£©Na₂C₂O₄ÈÜÒºÖУ¬

c(Na+)

c(C2O42-)

 

2 £¨Ìî¡°£¾¡±¡¢¡°=¡±¡¢¡°£¼¡±£©£»Íù¸ÃÈÜÒºÖеÎÈëÂÈ»¯¸ÆÈÜÒººó£¬

c(Na+)

c(C2 O 2- 4 )

Ôö¼Ó£¬¿ÉÄܵÄÔ­ÒòÊÇ

 

£®£¨ÓÃÀë×Ó·½³Ìʽ½âÊÍ£©
£¨3£©ÒÑÖªNaHC₂O₄ÈÜÒºÏÔËáÐÔ£®³£ÎÂÏ£¬Ïò10mL 0.1mol/L H₂C₂O₄ÈÜÒºÖеμÓ0.1mol/L NaOHÈÜÒº£¬Ëæ×ÅNaOHÈÜÒºÌå»ýµÄÔö¼Ó£¬µ±ÈÜÒºÖÐc£¨Na⁺£©=2c£¨C₂O₄^2-£©+c£¨HC₂O₄^-£©Ê±£¬ÈÜÒºÏÔ

 

ÐÔ£¨Ìî¡°Ëᡱ¡¢¡°¼î¡±»ò¡°ÖС±£©£¬ÇÒV£¨NaOH£©

 

10mL£¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
£¨4£©ÏÖÓÐÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol/LµÄÏÂÁÐÈÜÒº£º
a£®Na₂C₂O₄  b£®NaHC₂O₄c£®H₂C₂O₄d£®£¨NH₄£©₂C₂O₄e£®NH₄HC₂O₄
¢ÙÏÂÁйØÓÚÎåÖÖÈÜÒºµÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ

 

A£®ÈÜÒºbÖУ¬c£¨C₂O₄^2-£©£¼c£¨H₂C₂O₄£©
B£®ÈÜÒºbÖУ¬c£¨H₂C₂O₄£©+c£¨OH^-£©=2c£¨C₂O₄^2-£©+c£¨H⁺£©
C£®ÈÜÒºdºÍeÖж¼·ûºÏc£¨NH₄⁺£©+c£¨H⁺£©=c£¨HC₂O₄^-£©+2c£¨C₂O₄^2-£©+c£¨OH^-£©
D£®ÎåÖÖÈÜÒº¶¼·ûºÏc£¨H₂C₂O₄£©+c£¨HC₂O₄^-£©+c£¨C₂O₄^2-£©=0.1mol?L^-1
¢ÚÎåÖÖÈÜÒºÖÐc£¨H₂C₂O₄£©ÓÉ´óµ½Ð¡ÅÅÁеÄ˳ÐòÊÇ

 

£®

Answer 1

¿¼µã£ºÀë×ÓŨ¶È´óСµÄ±È½Ï,Èõµç½âÖÊÔÚË®ÈÜÒºÖеĵçÀëƽºâ,ÑÎÀàË®½âµÄÓ¦ÓÃ

רÌ⣺µçÀëƽºâÓëÈÜÒºµÄpHרÌâ,ÑÎÀàµÄË®½âרÌâ

·ÖÎö£º£¨1£©pHԽС£¬H₂C₂O₄µÄŨ¶ÈÔ½´ó£»
£¨2£©²ÝËáÄÆÖеIJÝËá¸ùÀë×ÓÊÇÈõËáÒõÀë×Ó£¬Ò×Ë®½â£¬C₂O₄^2-ÓëCa²⁺ÐγɳÁµí£»
£¨3£©¸ù¾ÝµçºÉÊغãÈ·¶¨ÈÜÒºÖÐÇâÀë×ÓºÍÇâÑõ¸ùÀë×ÓŨ¶ÈÏà¶Ô´óС£¬´Ó¶øÈ·¶¨ÈÜÒºËá¼îÐÔ£»²ÝËáÇâÄÆÈÜÒº³ÊËáÐÔ£¬ÒªÊ¹ÈÜÒº³ÊÖÐÐÔ£¬ÔòÇâÑõ»¯ÄÆÓ¦¸ÃÉÔ΢¹ýÁ¿£»
£¨4£©¢ÙA£®NaHC₂O₄ÈÜÒºÏÔËáÐÔ£¬ËµÃ÷²ÝËáÇâ¸ùÀë×ÓµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£»
B£®ÈÜÒºbÖУ¬´æÔÚµçºÉÊغãºÍÎïÁÏÊغ㣻
C£®ÈκÎÈÜÒºÖж¼´æÔÚµçºÉÊغ㣻
D£®ÈκÎÈÜÒºÖж¼´æÔÚÎïÁÏÊغ㣻
 ¢Ú²ÝËá¸ùÀë×ÓË®½â³Ì¶È´óÓÚ²ÝËáÇâ¸ùÀë×Ó£¬²ÝËáÊÇÈõµç½âÖÊ£¬µçÀë³Ì¶È½ÏС£®

½â´ð£º ½â£º£¨1£©¸ù¾ÝͼÏóÖª£¬ÈÜÒºµÄpHԽС£¬²ÝËáµÄŨ¶ÈÔ½´ó£¬ËùÒÔÇúÏß1Ϊ²ÝËᣬ¹Ê´ð°¸Îª£ºH₂C₂O₄£»
£¨2£©²ÝËáÄÆÖеIJÝËá¸ùÀë×ÓÊÇÈõËáÒõÀë×Ó£¬Ò×Ë®½â£¬ËùÒÔ²ÝËáÄÆÈÜÒºÖУ¬

c(Na+)

c(C2O42-)

£¾2£¬¸Ã²ÝËáÄÆÈÜÒºÖеÎÈëÂÈ»¯¸ÆÈÜÒººó£¬C₂O₄^2-ÓëCa²⁺ÐγɳÁµí£¬ËùÒÔC₂O₄^2-Ũ¶È½µµÍ£¬

c(Na+)

c(C2O42-)

Ôö¼Ó£¬Àë×Ó·½³ÌʽΪC₂O₄^2-+Ca²⁺=CaC₂O₄¡ý£¬¹Ê´ð°¸Îª£º£¾£»C₂O₄^2-+Ca²⁺=CaC₂O₄¡ý£»
£¨3£©¸ù¾ÝµçºÉÊغãµÃc£¨Na⁺£©+c£¨H⁺£©=2c£¨C₂O₄^2-£©+c£¨HC₂O₄^-£©+c£¨OH^-£©£¬c£¨Na⁺£©=2c£¨C₂O₄^2-£©+c£¨HC₂O₄^-£©£¬Ôòc£¨H⁺£©=c£¨OH^-£©£¬ËùÒÔÈÜÒº³ÊÖÐÐÔ£»
ËáÇâÄÆÈÜÒº³ÊËáÐÔ£¬ÒªÊ¹ÈÜÒº³ÊÖÐÐÔ£¬ÔòÇâÑõ»¯ÄÆÓ¦¸ÃÉÔ΢¹ýÁ¿£¬ËùÒÔV£¨NaOH£©£¾10mL£¬
¹Ê´ð°¸Îª£ºÖУ»£¾£»
£¨4£©¢ÙA£®NaHC₂O₄ÈÜÒºÏÔËáÐÔ£¬ËµÃ÷²ÝËáÇâ¸ùÀë×ÓµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬Ôòc£¨C₂O₄^2-£©£¾c£¨H₂C₂O₄£©£¬¹Ê´íÎó£»
B£®ÈÜÒºbÖУ¬´æÔÚµçºÉÊغãºÍÎïÁÏÊغ㣬¸ù¾ÝÎïÁÏÊغãµÃc£¨Na⁺£©=c£¨HC₂O₄^-£©+c£¨H₂C₂O₄£©+c£¨C₂O₄^2-£©£¬¸ù¾ÝµçºÉÊغãµÃc£¨H⁺£©+c£¨Na⁺£©=c£¨HC₂O₄^-£©+2c£¨C₂O₄^2-£©+c£¨OH^-£©£¬ËùÒÔµÃc£¨H⁺£©+c£¨H₂C₂O₄£©=2c£¨C₂O₄^2-£©+c£¨OH^-£©£¬¹Ê´íÎó£»
C£®ÈκÎÈÜÒºÖж¼´æÔÚµçºÉÊغãc£¨NH₄⁺£©+c£¨H⁺£©=c£¨HC₂O₄^-£©+2c£¨C₂O₄^2-£©+c£¨OH^-£©£¬¹ÊÕýÈ·£»
D£®ÈκÎÈÜÒºÖж¼´æÔÚÎïÁÏÊغ㣬¸ù¾ÝÎïÁÏÊغãµÃc£¨H₂C₂O₄£©+c£¨HC₂O₄^-£©+c£¨C₂O₄^2-£©=0.1mol?L^-1£¬¹ÊÕýÈ·£»
¹ÊÑ¡CD£»
¢Ú²ÝËá¸ùÀë×ÓË®½â³Ì¶È´óÓÚ²ÝËáÇâ¸ùÀë×Ó¡¢ï§¸ùÀë×Ó´Ù½ø²ÝËá¸ùÀë×Ó»ò²ÝËáÇâ¸ùÀë×ÓË®½â£¬²ÝËáµçÀë³Ì¶È½ÏС£¬ËùÒÔÈÜÒºÖÐc£¨H₂C₂O₄£©ÓÉ´óµ½Ð¡ÅÅÁеÄ˳ÐòÊÇc£¾e£¾b£¾d£¾a£¬¹Ê´ð°¸Îª£ºc£¾e£¾b£¾d£¾a£®

µãÆÀ£º±¾Ì⿼²éÁËÀë×ÓŨ¶È´óС±È½Ï£¬Ã÷È·ÈÜÒºÖеÄÈÜÖʼ°ÆäÐÔÖÊÊǽⱾÌâ¹Ø¼ü£¬¸ù¾ÝµçºÉÊغ㡢ÎïÁÏÊغãÀ´·ÖÎö½â´ð£¬ÌâÄ¿ÄѶÈÖеȣ®