Question

$\displaystyle Se \ considera \ o \ solutie \ de \ H2SO3 \\ (k_{a_1}=1,5\cdot 10^{-2} mol/L \ , \ pk_{a_1}=1,81 \ , \ k_{a_2}=9,1\cdot 10^{-8}mol/L \ , \ pk_{a_2}=7,04) \\ de \ concentratie \ 10^{-2}M \ , \ cu \ temperatura \ de \ 25\textdegree C. \\ Determinati: \\ a)concentratia \ la \ echilibru \ a \ tuturor \ speciilor \ chimice \ din \ solutie \\ b)pH-ul \ solutiei \\ c)specia \ care \ exista \ predominant \ in \ solutie \ la \ pH =0,5 \ , \ la \ pH=5,5 \ si \ la \ pH=9.$

Answer 1

                          H2SO3        +       H2O       ⇄      H3O⁺      +       SO3H⁻
initial   :                    c
au reactionat :        cx
la echilibru :         c(1-x)                                         cx                        cx
    Ka₁ =  [H3O⁺]ₓ[SO3H⁻] /[H2SO3] = [H3O⁺]² /H2SO3] = c²x²/[c(1-x)] = cx²/(1-x)
1,5ₓ10⁻² = 10⁻²ₓx²/(1-x)     1,5 - 1,5x = x²     x² + 1,5x - !,5 = 0  Δ = 2,25+ 6 = 8,25
x = (- 1,5 + 2,872)/2 = 0,686     [H3O⁺]₁ = 0,686ₓ10⁻² (mol/l)
                        HSO3⁻ + H2O  ⇄   H3O⁺          +         SO3²⁻
initial  :            0,686·10⁻²
au reactionat :  0,686ₓ10⁻²ₓy
la echilibru :  0,686(1-y)ₓ10⁻²        0,686ₓyₓ10⁻²            0,686ₓyₓ10⁻²
[H3O⁺]₂ = √(Ka₂ₓc) = √(9,1ₓ10⁻⁸ₓ0,686ₓ10⁻²) = ≈2,5·10⁻⁵ mol/l
[H3O⁺] = 686ₓ10 ⁻⁵ + 2,5 ₓ10⁻⁵ = 688,5ₓ10⁻⁵   pH = 5 - lg688,5 = 2,162   
[H2SO3] = (1-0,686)ₓ10⁻² = 0,314ₓ10⁻² mol/l
[HSO3⁻] = 0,686ₓ10⁻² mol/l  ;  [SO3²⁻] = 2,5ₓ10⁻⁵ mol/l
c) pH = 0,5    [H₃O⁺] = 10⁻⁰⁾⁵ = 1/3,16 = 0,316 mol/l > 0,686·10⁻² ⇒ echilibru ←  ⇒ predomina H2SO3 
pH = 5,5    [H3O⁺] = 10⁻⁵⁾⁵ = 10⁻⁵/√10 = 0,316·10⁻⁵ predomina SO3H⁻
pH = 9    [H3O⁺] = 10⁻⁹   predomina SO3²⁻