- #1

- 343

- 1

_{2}or SO

_{3}

^{2-}or HSO

_{3}

^{-}in solution. I get the following equilibria:

SO

_{2}(aq) + H2O (l) ⇔ HSO

_{3}

^{-}(aq) + H

_{3}O

^{+}(aq) ... K

_{1}

HSO

_{3}

^{-}(aq) + H2O (l) ⇔ SO

_{3}

^{2-}(aq) + H

_{3}O

^{+}(aq) ... K

_{2}

And this equilibrium always going on:

2H

_{2}O (l) ⇔ H

_{3}O

^{+}(aq) + OH

^{-}(aq) ... K

_{w}

Other than that, to solve the system we would write two more equations:

Mass Balance (S): Initial Concentration of S = [SO

_{2}] + [SO

_{3}

^{2-}] + [HSO

_{3}

^{-}]

Charge Balance: [H

_{3}O

^{+}] = [OH

^{-}] + [HSO

_{3}

^{-}] + 2[SO

_{3}

^{2-}]

So it looks like the only real initial variable is the Initial Concentration of S that we put in our system. Then, does this mean that if we put [itex]c[/itex] M of SO

_{2}to begin with in the solution, the equilibrium concentrations of every species would be the same as if we had put [itex]c[/itex] M of Na

_{2}SO

_{3}or [itex]c[/itex] M of NaHSO

_{3}? And also the same as if we put [itex](1/3)c[/itex] M of Al

_{2}(SO

_{3})

_{3}? (as the initial concentration of S placed in the solution is the same one way or the other)