HF + H20 ↔ F- +  H3O+          (concentr) Ca: 0,0010 mol/L (M)              Ka(HF)=10-3,25

equilibrioCa-x              x        x     Ka=[F-]·[H3O+]/[HF]   ;  10-3,25=(x · x)/(0,001 - x)=x²/10-3      x2=10-6,25-10-3,25·x

                                                  x2+10-3,25·x-10-6,25=0    x=[-10-3,25+√(10-3,25)2+4·10-6,85]/2= 5,2·10-4

α=(x)/Ca= (5,2·10-4mol/L)/10-3mol/L=5,2·10-1=0,52→52%        [H20+]=x=5,2·10-4mol/L→ pH=-log[H3O+]=3,27

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       CH3NH2 + H2O ↔ CH3NH3+ + HO-  (metanamina) (conc) Cb: 0,010 mol/L (M)  Kb(CH3NH2)=10-3,34 Kw=10-14

eq:    Cb-x                         x             x       Kb=[CH3NH3+]·[HO-]/[CH3NH2]   ;   10-3,34=(x · x)/(10-2-x)=x2/(10-2-x)

                                      x2=10-5,34-10-3,34·x     x2+10-3,34·x-10-5,34=0      x=[-10-3,34+√(10-3,34)2+4·10-5,34]/2=10-2,7

                                            [HO-]=x    pOH=-log[OH-]=2,71      pH + pOH=14    pH=14-2,71=11,28

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  HCN + H2O ↔ CN- + H3O+  }                                             Ca=10-3mol/L (M)       Ka(HCN)=10-9,31     Kw=10-14

           2H2O ↔ HO- + H3O+   }    [H3O+]=[CN-]+[HO-]

Ka(HCN)=[CN-]·[H3O+]/[HCN] }   [CN-]=Ka([HCN]/[H3O+]) }    [H3O+]=Ka ([HCN]/[H3O+] + Kw/[H3O+]

                 Kw=[H3O+]·[HO-]  }          [HO-]=Kw/[H3O+]    }    [H3O+]2=Ka[HCN] + Kw ;  [H3O+]2=10-9,31·10-3+10-14

                                                                                         [H3O+]2=10-12,31+10-14  ;   [H3O+]=√10-12,31+10-14

pH=-log[H3O+]= 6,15

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a) pH de 30mL de HCL d Concentr 0,02M       HCl + H2O ↔(α=1)↔  Cl- + H3O+     →[Cl-]=[H3O+]=Ca 

                                                                [H3O+]=0,02mol/L      pH=-log[H3O+]       pH=-log 0,02=1,7

b) pH de 70mL de NaOH d concentr 0,01M    NaOH ↔(100%)↔ Na+ + HO-         →[HO-]=Cb=0,01mol/L

                                                                        pOH=-log[HO-]=-log10-2=2    pH=14 - pOH=12

c) pH mezcla?  30mL d HCl 0,02M }HCl + NaOH ↔ NaCl + H2O

                   70mL d NaOH 0,01M }    H+ + HO- ↔ H2O                         -ΔnHCL=-ΔnNaOH

               nHCL=CHCl x VHCl=0,02mol/L x0,03L= 6·10-4mol HCl }       Reacciona todo el HCl y sobra NaOH

    nNaOH=CNaOH x VNaOH=0,01mol/L x 0,07L=7·10-4mol NaOH }         nf(NaOH)=7·10-4 - 6·10-4=1·10-4mol

el NaOH sobrant produc 1 concentrac d [HO-]= nNaOH/Vtotal=1·10-4mol/0,1L     [HO-]=10-3mol/L   pOH=3   pH=11

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a) 50 mL d HCl 0,01M  HCl + H2O ↔(100% α=1)↔ Cl- + H3O+  →[H3O+]=Ca=10-2M pH=-log[H3O+]=-log·10-2=2

b) 50 mL d Ba(OH)2 0,01M   Ba(OH)2    ↔     Ba2+ + 2OH-                 [OH-]=2Cb=0,02mol/L  

                                       Cb=0,01mol/L       0,01    2x0,01mol/L   pOH= -log[HO-]=1,7   pH=14-pOH=12,3

c) pH mezcla?  2HCl + Ba(OH)2 ↔ BaCl2 + 2H2O      -ΔnHCl/2=-ΔnBa(OH)2/1

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