Question

1£®ÉèN_AΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÊýÖµ£¬Èç¹ûagijÆøÌåÖк¬ÓеķÖ×ÓÊýΪb£¬ÔòcgijÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýÔ¼ÊÇ$\frac{22.4bc}{a{N}_{A}}$£¨Óú¬N_AµÄʽ×Ó±íʾ£©£®
£¨2£©°ÑVLº¬ÓÐMgSO₄ÓëK₂SO₄µÄ»ìºÏÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬Ò»·Ý¼ÓÈ뺬amolNaOHµÄÈÜÒº£¬Ç¡ºÃʹþÀë×ÓÍêÈ«³ÁµíΪMg£¨OH£©₂£»ÁíÒ»·Ý¼ÓÈ뺬bmolBaCl₂µÄÈÜÒº£¬Ç¡ºÃʹSO₄^2-ÍêÈ«³ÁµíΪBaSO₄£¬ÔòÔ­»ìºÏÈÜÒºÖмØÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈΪ$\frac{2£¨2b-a£©}{V}$£®
£¨3£©Ä³Î¶ÈÏ£¬Èô½«Na₂SO₄ÈÜÒºÕô·¢180gË®£¬¿É±ä³É±¥ºÍÈÜÒº£®ÈôÏò¸ÃÈÜÒºÖмÓÈë14.2gÎÞË®Na₂SO₄Ò²¿ÉÇ¡ºÃ±ä³É±¥ºÍÈÜÒº£®ÈôÔÚ¸ÃÈÜÒºÖмÓÈëNa₂SO₄•10H₂OÐγɱ¥ºÍÈÜҺʱËùÐèNa₂SO₄•10H₂OµÄÖÊÁ¿Îª35.8g£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ýn=$\frac{N}{{N}_{A}}$¼ÆËãagÆøÌåµÄÎïÖʵÄÁ¿£¬ÖÊÁ¿Ö®±ÈµÈÓÚÆäÎïÖʵÄÁ¿Ö®±È£¬¾Ý´Ë¼ÆËãcg¸ÃÆøÌåµÄÎïÖʵÄÁ¿£¬¸ù¾ÝV=n¡ÁVmÀ´¼ÆËãÌå»ý£»
£¨2£©Á½µÈ·ÝÈÜÒºÖÐÒ»·Ý¼ÓBaCl₂µÄÈÜÒº·¢ÉúBa²⁺+SO₄^2-¨TBaSO₄¡ý£¬ÓÉ·½³Ìʽ¿ÉÖªn£¨SO₄^2-£©=n£¨Ba²⁺£©£¬ÁíÒ»·Ý¼ÓÇâÑõ»¯ÄÆÈÜҺʱ·¢ÉúMg²⁺+2OH^-¨TMg£¨OH£©₂¡ý£¬ÓÉ·½³Ìʽ¿É֪ÿ·ÝÈÜÒºÖÐn£¨Mg²⁺£©=$\frac{1}{2}$n£¨OH^-£©£¬
ÔÙÀûÓõçºÉÊغã¿É֪ÿ·ÝÖÐ2n£¨Mg²⁺£©+n£¨K⁺£©=2n£¨SO₄^2-£©£¬¾Ý´Ë¼ÆËãÿ·ÝÖÐn£¨K⁺£©£¬¸ù¾Ýc=$\frac{n}{V}$¼ÆËã¼ØÀë×ÓŨ¶È£»
£¨3£©¸ù¾ÝÌâÒâ·ÖÎö¿ÉÖª£¬½«14.2gÎÞË®Na₂SO₄ÈÜÓÚ180gË®¼´¿ÉÐγɱ¥ºÍÈÜÒº£®
Éèµ±Ïò¸ÃÈÜÒºÖмÓÈëxgNa₂SO₄•10H₂OÇ¡ºÃÐγɱ¥ºÍÈÜÒº£¬Õâ¼ÓÈëµÄxgNa₂SO₄•10H₂OÖк¬ÓеÄÈÜÖÊNa₂SO₄µÄÖÊÁ¿Îªx¡Á$\frac{142}{322}$g£¬ÒýÈëµÄË®µÄÖÊÁ¿Îªx¡Á$\frac{180}{322}$g£¬¹ÊÈܼÁË®µÄÖÊÁ¿Îª£¨180+x¡Á$\frac{180}{322}$£©g£¬¸ù¾Ý±ÈÀý¼ÆËã¼´¿É£®

½â´ð ½â£º£¨1£©a¿ËijÆøÌåÖк¬ÓеķÖ×ÓÊýΪb£¬Ôòc¿ËÆøÌ庬ÓеķÖ×ÓÊýΪ$\frac{cb}{a}$£¬c¿Ë¸ÃÆøÌåµÄÎïÖʵÄÁ¿Îª$\frac{\frac{bc}{a}}{N{\;}_{A}mol{\;}^{-1}}$=
$\frac{bc}{a{N}_{A}}$mol£¬ÔÚ±ê×¼×´¿öÏÂVm=22.4L/mol£¬Ôòcg¸ÃÆøÌåµÄÌå»ýΪ$\frac{bc}{a{N}_{A}}$mol¡Á22.4L/mol=$\frac{22.4bc}{a{N}_{A}}$L£»
¹Ê´ð°¸Îª£º$\frac{22.4bc}{a{N}_{A}}$£»
£¨2£©»ìºÏÈÜÒº·Ö³ÉÁ½µÈ·Ý£¬Ã¿·ÝÈÜҺŨ¶ÈÓëÔ­ÈÜҺŨ¶ÈÏàͬ£¬
Ò»·Ý¼ÓBaCl₂µÄÈÜÒº·¢ÉúBa²⁺+SO₄^2-¨TBaSO₄¡ý£¬ÓÉ·½³Ìʽ¿ÉÖªn£¨SO₄^2-£©=n£¨Ba²⁺£©=n£¨BaCl₂£©=bmol£¬
ÁíÒ»·Ý¼ÓÇâÑõ»¯ÄÆÈÜҺʱ·¢ÉúMg²⁺+2OH^-¨TMg£¨OH£©₂¡ý£¬ÓÉ·½³Ìʽ¿É֪ÿ·ÝÈÜÒºÖÐn£¨Mg²⁺£©=$\frac{1}{2}$n£¨OH^-£©=$\frac{1}{2}$amol£¬
ÓɵçºÉÊغã¿É֪ÿ·ÝÖÐ2n£¨Mg²⁺£©+n£¨K⁺£©=2n£¨SO₄^2-£©£¬¹Êÿ·ÝÖÐÈÜÒºn£¨K⁺£©=2bmol-2¡Á$\frac{1}{2}$amol=£¨2b-a£©mol£¬¹ÊÔ­ÈÜÒºÖмØÀë×ÓŨ¶È=$\frac{£¨2b-a£©mol}{\frac{V}{2}L}$=$\frac{2£¨2b-a£©}{V}$mol/L£»
¹Ê´ð°¸Îª£º$\frac{2£¨2b-a£©}{V}$£»
£¨3£©¸ù¾ÝÌâÒâ·ÖÎö¿ÉÖª£¬½«14.2gÎÞË®Na₂SO₄ÈÜÓÚ180gË®¼´¿ÉÐγɱ¥ºÍÈÜÒº£®
Éèµ±Ïò¸ÃÈÜÒºÖмÓÈëxgNa₂SO₄•10H₂OÇ¡ºÃÐγɱ¥ºÍÈÜÒº£¬ÓÉÓÚ¼ÓÈëµÄxgNa₂SO₄•10H₂OÖк¬ÓеÄÈÜÖÊNa₂SO₄µÄÖÊÁ¿Îªx¡Á$\frac{142}{322}$g=$\frac{142x}{322}$g£¬½á¾§Ë®ÈÜÓÚË®ºóÒ²±äΪÈܼÁ£¬¼´ÒýÈëµÄË®µÄÖÊÁ¿Îªx¡Á$\frac{180}{322}$g=$\frac{180x}{322}$g£¬¹ÊÈܼÁË®µÄÖÊÁ¿Îª£¨180+$\frac{180x}{322}$£©g£¬ÔòÓУº14.2g¡«180g
                                             $\frac{142x}{322}$g¡«£¨180+$\frac{180x}{322}$£©g
¹ÊÓУº$\frac{\frac{14.2g}{142xg}}{322}$=$\frac{180g}{£¨\frac{180x}{322}+180£©g}$£¬½âµÃ£º½âµÃx=35.8g£¬
¹Ê´ð°¸Îª£º35.8g£®

µãÆÀ ±¾Ì⿼²éÁËÎïÖʵÄÁ¿ÓйؼÆË㣬Ã÷È·ÒÔÎïÖʵÄÁ¿ÎªºËÐļÆË㹫ʽ£¬ÊìϤÎïÖʵÄÁ¿Å¨¶È¸ÅÄîÊǽâÌâ¹Ø¼ü£¬½øÐб¥ºÍÈÜÒºµÄÓйؼÆË㣬ӦעÒâµ½µ±¼ÓÈë´ø½á¾§Ë®µÄÎïÖʵÄʱºò£¬½á¾§Ë®ÈÜÓÚË®ºóÒ²±äΪÈܼÁ£¬ÄѶÈÖеȣ®