Question

15£®ÏÖÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖпɵçÀë²úÉúÏÂÁÐÀë×Ó£¨¸÷ÖÖÀë×Ó²»Öظ´£©

ÑôÀë×Ó	H+  Na+  Al3+   Ag+  Ba2+
ÒõÀë×Ó	OH-  Cl-  CO32-  NO3-  SO42-

ÒÑÖª£º
¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£»C¡¢D¡¢EÈÜÒº³ÊËáÐÔ£®
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¨³Áµí°üÀ¨Î¢ÈÜÎÏÂͬ£©£®
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí£®
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©·Ö±ðд³öA¡¢EµÄ»¯Ñ§Ê½£ºANa₂CO₃¡¢EAl₂£¨SO₄£©₃ 
£¨2£©Ð´³öAÓëE·´Ó¦µÄÀë×Ó·½³Ìʽ£º2Al³⁺+3CO₃^2-+3H₂O¨T3CO₂¡ü+2Al£¨OH£©₃¡ý£»
£¨3£©ÒÑÖª£ºNaOH£¨aq£©+HNO₃£¨aq£©=NaNO₃£¨aq£©+H₂O£¨1£©£»¡÷H=-Q kJ•mol^-1£®Ð´³öBÓëCÏ¡ÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽOH^-£¨aq£©+H⁺£¨aq£©=H₂O£¨1£©¡÷H=-QkJ/mol»ò$\frac{1}{2}$Ba£¨OH£©₂£¨aq£©+HC1£¨aq£©=$\frac{1}{2}$BaC1₂£¨aq£©+H₂O£¨1£©¡÷H=-QkJ/mol
»òBa£¨OH£©₂£¨aq£©+2HC1£¨aq£©=BaC1₂£¨aq£©+2H₂O£¨1£©¡÷H=-2QkJ/mol£®
£¨4£©ÔÚ100mL 0.1mol•L^-1 EÈÜÒºÖУ¬ÖðµÎ¼ÓÈë40mL 1.6mol•L^-1 NaOHÈÜÒº£¬×îÖյõ½³ÁµíÎïÖʵÄÁ¿Îª0.016molmol£®

Answer 1

·ÖÎö ¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£¬½áºÏÀë×ӵĹ²´æ¿ÉÖª£¬Ó¦ÎªBa£¨OH£©₂¡¢Na₂CO₃ÖеÄÒ»ÖÖ£¬C¡¢D¡¢EÈÜÒº³ÊËáÐÔ£¬Ó¦ÎªAgNO₃¡¢ÁòËáÂÁ¡¢HClÖеÄÒ»ÖÖ£»
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¬ÔòAΪNa₂CO₃£¬BΪBa£¨OH£©₂£¬EΪAl₂£¨SO₄£©₃£¬CΪHCl£»
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí£¬ÔòDΪAgNO₃£¬È»ºó½áºÏÎïÖʵÄÐÔÖʼ°»¯Ñ§ÓÃÓïÀ´½â´ð£®

½â´ð ½â£º¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£¬¸ù¾ÝÀë×Ó¹²´æ¿ÉÖª£¬A¡¢BӦΪBa£¨OH£©₂¡¢Na₂CO₃ÖеÄÒ»ÖÖ£¬C¡¢D¡¢EÈÜÒº³ÊËáÐÔ£¬Ó¦ÎªAgNO₃¡¢Al₂£¨SO₄£©₃¡¢HClÖеÄÒ»ÖÖ£»
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£¬AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¬ÔòAΪNa₂CO₃£¬BΪBa£¨OH£©₂£¬EΪAl₂£¨SO₄£©₃£¬CΪHCl£»
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£¬CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí£¬ÔòDΪAgNO₃£¬
£¨1£©ÓÉÒÔÉÏ·ÖÎö¿ÉÖªAΪNa₂CO₃£¬BΪBa£¨OH£©₂£¬CΪHCl£¬DΪAgNO₃£¬EΪAl₂£¨SO₄£©₃£¬
¹Ê´ð°¸Îª£ºNa₂CO₃£»Al₂£¨SO₄£©₃£»
£¨2£©A¡¢E·¢ÉúÏ໥´Ù½øË®½â·´Ó¦£¬Àë×Ó·´Ó¦Îª2Al³⁺+3CO₃^2-+3H₂O¨T3CO₂¡ü+2Al£¨OH£©₃¡ý£¬
¹Ê´ð°¸Îª£º2Al³⁺+3CO₃^2-+3H₂O¨T3CO₂¡ü+2Al£¨OH£©₃¡ý£»
£¨3£©BÓëCµÄÏ¡ÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪOH^-£¨aq£©+H⁺£¨aq£©=H₂O£¨1£©¡÷H=-QkJ/mol»ò$\frac{1}{2}$Ba£¨OH£©₂£¨aq£©+HC1£¨aq£©=$\frac{1}{2}$BaC1₂£¨aq£©+H₂O£¨1£©¡÷H=-Q kJ/mol
»òBa£¨OH£©₂£¨aq£©+2HC1£¨aq£©=BaC1₂£¨aq£©+2H₂O£¨1£©¡÷H=-2QkJ/mol£¬
¹Ê´ð°¸Îª£ºOH^-£¨aq£©+H⁺£¨aq£©=H₂O£¨1£©¡÷H=-QkJ/mol»ò$\frac{1}{2}$Ba£¨OH£©₂£¨aq£©+HC1£¨aq£©=$\frac{1}{2}$BaC1₂£¨aq£©+H₂O£¨1£©¡÷H=-QkJ/mol
»òBa£¨OH£©₂£¨aq£©+2HC1£¨aq£©=BaC1₂£¨aq£©+2H₂O£¨1£©¡÷H=-2QkJ/mol£»
£¨4£©ÁòËáÂÁºÍÇâÑõ»¯ÄƵķ´Ó¦ÊµÖÊΪ£ºAl³⁺+3OH^-=Al£¨OH£©₃¡ý£¬Al£¨OH£©₃+OH^-=[Al£¨OH£©₄]^-£¬
100mL 0.1mol/LµÄÁòËáÂÁÈÜÒºÖУ¬º¬ÓÐÂÁÀë×ÓµÄÎïÖʵÄÁ¿Îª0.02mol£¬
40mL 1.6mol/L NaOHÈÜÒºÖк¬ÓÐÇâÑõ»¯ÄÆÎïÖʵÄÁ¿Îª0.064mol£¬
     Al³⁺+3OH^-=Al£¨OH£©₃¡ý£¬
0.02mol  0.06mol 0.02mol
0.02molÂÁÀë×ÓÏûºÄ0.06molµÄÇâÑõ»¯ÄÆÉú³É0.02molµÄÇâÑõ»¯ÂÁ£¬
Al£¨OH£©₃+OH^-=[Al£¨OH£©₄]^-£¬
0.004mol   0.004mol
È»ºó¶àÓà0.004molµÄÇâÑõ»¯ÄÆÓÖ½«0.004molµÄÇâÑõ»¯ÂÁÈܽ⣬
ËùÒÔÊ£ÓàÇâÑõ»¯ÂÁµÄÎïÖʵÄÁ¿ÊÇ0.02mol-0.004mol=0.016mol£¬
¹Ê´ð°¸Îª£º0.016mol£®

µãÆÀ ±¾Ì⿼²é֪ʶµã½Ï¶à£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬¸ù¾ÝÀë×ӵĹ²´æ¡¢Àë×ÓµÄÐÔÖÊÀ´ÍƶÏÎïÖÊÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬×ÛºÏÐÔÇ¿£¬ÄѶȽϴó£¬Ñ§ÉúÐèÊìϤˮ½â¡¢ÈÈ»¯Ñ§·´Ó¦·½³Ìʽ¡¢Àë×Ó·´Ó¦µÈ֪ʶÀ´½â´ð£®