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V\m~X#GN~>=R5XbPR=?w2tt\b{^,K .yT `wT7k1AE2m5_ k*"A 3|so3iE6G0Z8Zuq7 eAazPTe2g1*CK*zZeZoF Di(w2~("e$"( *6#cG1zPm2,48u0uC(`XHF|TD3A XP@           %C 9  OB   CDEFGHIJ KL NOPQR SUXYZ[\]^_`abcd fghistu z_\R$xJ='Ƿ:$$$$$$$$$$$$$$2$qoRszˣ]"2$#_ϧ 2oקj2$Kd :J~r%2$Py;e2$ ɇܖfkcr2$| nfs-I!W2$2e %f"f2$yUٔ/h2$OȓYf" 2$DA@*-W|ZUF2$Ic*Dr-M-E2$8#]ROȧcw2$uK|w:T9[$2$#(?r'xVE~5$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$2$l,ZC?$F|$$$2$ 9p^հk / 2$'7w#:2$Ƞ ]J38]2$Sē[|Nl*e2$c,lS7j(4me2$ i$2$0HN3Pwz2$r;ءEא c AA1? 3wf̙@̙ Tg44d4dܽeO 0ppp@ g4+d+d$eO 0>p p<4BdBdv 04e?O ʚ;Ķ5ʚ;<4!d!d 0<4dddd 0""BRANCHTO08HOTSPOTTYPENextSlide>$DEFINEDINNAVIGATOR False0___PPT10 pp2___PPT9/ 0? +O =dEquilibrios heterogneos  Unidad 2. Segunda parte   OPk?Equilibrios heterogneos  ,Se habla de reaccin homognea cuando tanto reactivos como productos se encuentran en el mismo estado fsico. En cambio, si entre las sustancias que intervienen en la reaccin se distinguen varias fases o estados fsicos, hablaremos de reacciones heterogneas. Por ejemplo, la reaccin: CaCO3(s) CaO(s) + CO2(g) se trata de un equilibrio heterogneo. Aplicando la ley de accin de masas se cumplir que:P$  ]&   r m@Eq. heterogneos (cont).@     Sin embargo, las concentraciones (moles/V) de ambas sustancias slidas (CaCO3 y CaO) son constantes, al igual que las densidades de sustancias puras (masa/V) son tambin constantes. Por ello, agrupando las constantes en una sola a la que llamaremos KC se tiene: KC = [CO2] Anlogamente: KP = p(CO2) ATENCIN!: En la expresin de KC de la ley de accin de masas slo aparecen las concentraciones de gases y sustancias en disolucin, mientras que en la expresin de KP nicamente aparecen las presiones parciales de las sustancias gaseosas.Z"# I]         I@H     nAEjemplo: En un recipiente se introduce cierta cantidad de carbamato amnico, NH4CO2NH2 slido que se disocia en amoniaco y dixido de carbono cuando se evapora a 25C. Sabiendo que la constante KP para el equilibrio NH4CO2NH2(s) 2 NH3(g) + CO2(g) y a esa temperatura vale 2,310-4, calcula KC y las presiones parciales en el equilibrio. BTfG   m      #   . L Equilibrio: NH4CO2NH2(s) 2 NH3(g) + CO2(g) n(mol) equil. n  x 2x x Luego p(NH3) = 2 p(CO2) ya que la presin parcial es directamente proporcional al n de moles. KP = 2,3x10-4 = p(NH3)2 xp(CO2) = 4p(CO2)3 Despejando se obtiene que: p(CO2) = 0,039 atm : p(NH3) =0,078atm.Z+Z          H        333 33333 33         w 0      pBReacciones de precipitacin  Son reacciones de equilibrio heterogneo slido-lquido. La fase slida contiene una sustancia poco soluble (normalmente una sal) La fase lquida contiene los iones producidos en la disociacin de la sustancia slida. Normalmente el disolvente suele ser agua. Z  qC Solubilidad   iEs la mxima concentracin molar de soluto en un determinado disolvente, es decir, la molaridad de la disolucin saturada de dicho soluto. Depende de: La temperatura. Normalmente es mayor a mayor temperatura debido a la mayor energa del cristal para romper uniones entre iones. Energa reticular. Si la energa de solvatacin es mayor que la reticular U se favorece la disolucin. A mayor carcter covalente mayor U y por tanto menor solubilidad. La entropa. Al diluirse una sal se produce un sistema ms desordenado por lo que aunque energticamente no est favorecida la disolucin sta puede llegar a producirse.tq 2;   #  rD<Producto de solubilidad (KS o PS) en electrolitos de tipo AB@=$ $$ $$<  En un electrolito de tipo AB el equilibrio de solubilidad viene determinado por: AB(s) A+(ac) + B-(ac) Conc. inic. (mol/l): c 0 0 Conc. eq. (mol/l): c s s La concentracin del slido permanece constante. Y la constante de equilibrio tiene la expresin: Ejemplo: AgCl(s) Ag+(ac) + Cl -(ac) KS = [Ag+] x [Cl-] = s2  s es la solubilidad de la sal.Q> 6q      X        !    ,  Q1Determinacin de la formacin o no de precipitado22& 2 vEEjemplo: Deduce si se formar precipitado de cloruro de plata cuyo KS = 1,7 x 10-10 a 25C al aadir, a 250 mL de cloruro de sodio 0,02 M, 50 mL de nitrato de plata 0,5 M.f;  <  r P        AgCl(s) Ag+(ac) + Cl-(ac) KS = [Ag+] x [Cl-] = s2 Igualmente: Como entonces precipitar. 5 2  F& (      Z     D  |F3Producto de solubilidad en otro tipo de electrolito3  Tipo A2B: A2B (s) 2 A+(ac) + B2-(ac) Conc. inic. (mol/l): c 0 0 Conc. eq. (mol/l): c 2s s Y la constante de equilibrio tiene la expresin: Las misma expresin ser para electrolitos tipo AB2. Tipo AaBb: AaBb (s) a Ab+(ac) + b Ba-(ac) Conc. inic. (mol/l): c 0 0 Conc. eq. (mol/l): c as bs.}jM    Eb     E4  "       #  "     G%Factores que afectan a la solubilidad&&&%  0Adems de la temperatura, existen otros factores que influyen en la solubilidad por afectar a la concentracin de uno de los iones de un electrolito poco soluble. Estos son: Efecto ion comn. pH. Formacin de un cido dbil. Formacin de una base dbil. Formacin de complejos estables. Reacciones redox.zZZ:Z3Z  :32  m  HEfecto ion comn.  Si a una disolucin saturada de un electrolito poco soluble aadimos otra sustancia que aporta uno de los iones, la concentracin de ste aumentar. Lgicamente, la concentracin del otro ion deber disminuir para que el producto de las concentraciones de ambos permanezca constante. Como el equilibrio se desplaza a la izquierda, la solubilidad, que mide la mxima concentracin de soluto disuelto, disminuir.P  IEjemplo: Cul ser la solubilidad del cloruro de plata si aadimos nitrato de plata, sal soluble, hasta una concentracin final 0,002 M?2f  :AgCl(s) Ag+(ac) + Cl -(ac) Al aadir el AgNO3, la [Ag+] aumenta hasta 2 x10-3 M, pues se puede despreciar la concentracin que haba antes, ya que ks es muy baja. En consecuencia, el equilibrio se desplaza a la izquierda y la [Cl-], es decir, la nueva solubilidad, debe disminuir.JZ0Z(0Z(vZ    F P3       ^  N  2  JEjercicio: En equilibrio de disolucin de bromuro de plata cuya Ks=5,2 x 10-13 cul ser la nueva solubilidad si a litro de disolucin saturada de AgBr se aaden 0,2 ml de una disolucin 0,001 M de bromuro de potasio? f/   (.2@    Equilibrio: AgBr (s) Ag+(ac) + Br-(ac) Conc. eq. (mol/l): c s s n(Br-)0 = 0,5 L x7,2x10-7 mol/L = 3,6x10-7 mol n(Br-)aad = 0,0002 L x 0,001 mol/L = 2x10-7 mol Conc. inic. (mol/l): c 7,2x10-7 1,12x10-6 Conc. eq. (mol/l): c 7,2x10-7 -x 1,12x10-6 -x KS = 5,2 x 10-13 = (7,2x10-7 -x)(1,12x10-6 -x) De donde: x = 3,2 x 10-7 s = (7,2 x 10-7 -3,2 x 10-7) M =S (a 2              !                f                              !                                   K1Influencia del pH por formacin de un cido dbil21.1  0Equilibrio solubil.: AB(s) A- (ac) + B+ (ac) Equilibrio acidez: HA(ac) A- (ac) + H+ (ac) Si el anin A- en que se disocia un electrolito poco soluble forma un cido dbil HA, al aumen-tar la acidez o [H+] el equilibrio de disociacin del cido se desplazar hacia la izquierda. En consecuencia, disminuir [A-], con lo que se solubilizar ms electrolito AB. Ejemplo: al aadir un cido fuerte sobre el ZnCO3, se formar H2CO3, cido dbil, y al disminuir [CO32-], se disolver ms ZnCO3, pudindose llegar a disolver por completo.  c g2(   !   +t     )  Q  | L8Cambio en la solubilidad por formacin de una base dbil99(8  Suele producirse a partir de sales solubles que contienen el catin NH4+. NH4Cl(s) Cl- (ac) + NH4+ (ac) Equil base: NH4OH (ac) NH4+ (ac) + OH- (ac) Los NH4+ reaccionan con los OH- formndose NH4OH al desplazar el equilibrio de la base hacia la izquierda. Es el mtodo usual de disolver hidrxidos poco solubles tales como el Mg(OH)2. Equil. Solub.: Mg(OH)2(s) Mg2+(ac) + 2 OH-(ac) En consecuencia, disminuir [OH-], con lo que se solubilizar ms Mg(OH)2.&F                 ( K              N M Formacin de un complejo estable   Un ion complejo es un ion formado por ms de un tomo o grupo de tomos. Ejemplos: [Al(OH)4]-, [Zn(CN)4]2-, [AlF6]3- , [Ag(NH3)2]+. De esta manera, se pueden disolver precipitados. Por ejemplo, si se aade cianuro de sodio a electrolitos insolubles de cinc, como el Zn(OH)2, formndose el complejo [Zn(CN)4]2 -, que es muy estable. As, disminuir drsticamente la concentracin de Zn2+, con lo que se disolver ms Zn(OH)2. Igualmente, pueden disolverse precipitados de AgCl aadiendo amoniaco.I          H ItT         NOxidacin o reduccin de iones  Si alguno de los iones que intervienen en un equilibrio de solubilidad se oxida o se reduce como consecuencia de aadir un oxidante o reductor, la concentracin de este ion disminuir. En consecuencia, el equilibrio del electrolito insoluble se desplazar hacia al derecha, disolvindose en mayor cantidad. 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O +D=' & = @B +D  0 |R(  |x | c $)4`    | c $* 4C"?<$  0  " |0  f0e0eA r?3 ?4"<    H r$D 0 l  ` <  |` < ,$D  0 | <A w?3 ?` <  w | <A ~?3 ?C"?  ~H | 0޽h ? 33Պ3ff3s?k?___PPT10K?..   + {D>' = @B D^>' = @BA?%,( < +O%,( < +D' =%(D`' =%(D' =A@BBBB0B%(E5' =1B BPB`B,54*3>B ppt_c='`B@BPB<*|QD' =1:Bvisible*o3>+B#style.visibility<*|Q%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|QD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|QD' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|QpD' =1:Bvisible*o3>+B#style.visibility<*|Qp%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|QpD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|QpD' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|pD' =1:Bvisible*o3>+B#style.visibility<*|p%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|pD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|pD' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|D' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|D' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|D_' =%(D' =%(D' =A@BBBB0B%(E' =4 B`BPB`B?<*%()/%,( < +D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*|%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|D' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|6D' =1:Bvisible*o3>+B#style.visibility<*|6%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|6D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|6D' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|6SD' =1:Bvisible*o3>+B#style.visibility<*|6S%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|6SD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|6SD' =%(D`' =%(D' =A@BBBB0B%(E5' =1B B`BPB,54*3>B ppt_c='`B@BPB<*|SxD' =1:Bvisible*o3>+B#style.visibility<*|Sx%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*|SxD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*|Sx+p+0+|0 ++0+|0 +  0 QIx(  xr x S B4F   > x <DL ll ,$ 0 &Cmo saber si se formar precipitado?<' !&  M x <L8,$ 0 WMezclamos dos disoluciones que contienen dos iones que pueden formar una sal insoluble.X XW   x <P ,$ 0 Q = KPS Equilibrio : disolucin saturada Q > KPS Se desplaza hacia la izquierda : precipita Q < KPS No precipita : disolucin no saturada. x + 6 ,`   1  <  +  H x 0޽h ? fff  ___PPT10 .LH+PDD ' & = @B D ' = @BA?%,( < +O%,( < +D' =%(%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-o6Bdissolve*<3<*xD' =%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-o6Bdissolve*<3<*xD' =%(D' =%(D@' =A@BB BB0B%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D' =-o6Bdissolve*<3<*x++0+x0 ++0+x0 ++0+x0 +:  0  &(  x  c $Pm4F     c $Tn47Z<$ 0   0 Z0e0eA^ ? ?49- H $ 0   0A ? ? w 8 $D  0  c $A ??'8 $D 0   c $A] ??  8 $D 0   c $A_ ?? :u8 $D  0   c $A ??ET 8 $D  0H  0޽h ? 33Պ3ff344___PPT104..}`;a+,D)4' x= @B D3' = @BA?%,( < +O%,( < +D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*5%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*5D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*5DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*57%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*57D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*57D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*:F%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*:FD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*:FD' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*HK%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*HKD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*HKD' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* DA' =%(D' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*Kq%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*KqD' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*KqD' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* +p+0+0 ++0+0 +IJ  0 RJ 0 (  ~  s *4NEe     s *L 4C"?T<$  0  ( 0 # l0e0eA ?3 ?4"<  p H $D 0   <A ?3 ?G 8 $D  0l  9S 9 S,$D  0   BA ?3 ?z9S    BA ?3 ?C"? 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33Պ3ff3___PPT10i. +D=' & = @B +  0 0(  x  c $4F   x  c $d4  H  0޽h ? 33Պ3ff3___PPT10i. )+D=' & = @B +%  0   R(  ~  s *,4F%     s *,4<$ 0    <A` *?3 ?C"?3[8 *$D 0  <Ad /?3 ?C"?l Y8 /$D 0  c $Aa -??:8 -$@ 0`  c $A 0??(hX 0N   # lA >?̙1?"` Z> >$D  0d3H  0޽h ? 33Պ3ff3___PPT10w.. )+dUD' = @B D' = @BA?%,( < +O%,( < +D ' =%(D ' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<* D{' =%(D#' =%(D' =A@BBBB0B%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D' =+4 8?dCB0-#ppt_w/2BCB#ppt_xB*Y3>B ppt_x<*D' =+4 8?\CB#ppt_yBCB#ppt_yB*Y3>B ppt_y<*Dn' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<* %(D' =+4 8?\CB#ppt_xBCB#ppt_xB*Y3>B ppt_x<* D' =+4 8?dCB1+#ppt_h/2BCB#ppt_yB*Y3>B ppt_y<* +8+0+0 +6  0 0(  x  c $(4`     c $4 <$ 0    <A ?3 ?C"?Fo8 $D 0N  # lAb A?̙1?"`c A$D   0d3  c $Ae D??>q8 D$D 0H  0޽h ? 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