# Equilibrium problem about sulfurous acid?

##### 1 Answer

Well, the first thing I would do is to set up all the necessary components: reaction, ICE table, and the general point of the question. Here, all the ICE tables are in molarity.

In general, the question wants you to recognize/approximate that:

- sometimes you can ignore the second dissociation of a polyprotic acid, particularly if it's the last dissociation it can do.
- The
#"pH"# is thus primarily dictated by the first dissociation of a diprotic acid, especially if the first proton is of a not-strong acid (#K_(a1) < 1# ). - The predominant species will reflect the relative magnitudes of the
#"pH"# vs.#"pK"_(ai)# . If#"pH" < "pK"_(ai)# , then the acid the#"pK"_(ai)# represents will exist in its acidic form, and vice versa. You can verify this using the Henderson-Hasselbalch equation, but you should be able to do this conceptually on exams to check your work.

**DISCLAIMER:** *LONG ANSWER!*

I assume by now you know how to set one up and use it for a monoprotic acid, and this is just an extension into a diprotic acid. It's a matter of keeping your big picture straight, and knowing what assumptions to make.

#"H"_2"SO"_3(aq) rightleftharpoons "HSO"_3^(-)(aq) + "H"^(+)(aq)#

#"I"" ""0.01"" "" "" "" "0" "" "" "" "" "0#

#"C"" "-x" "" "" "+x" "" "" "" "+x#

#"E"" "0.01 - x" "" "x" "" "" "" "" "x#

#K_(a1) = (x^2)/(0.01 - x) = 1.5 xx 10^(-2)# By using the quadratic formula, which must be done because the

#K_(a1)# is not small enough at this low concentration, you should get

#x = "0.00686 M"#

So, the species in solution after **dissociation**

#"0.00314 M H"_2"SO"_3# #"0.00686 M HSO"_3^(-)# #"0.00686 M H"^(+)#

Now, the

Here, we know that the second dissociation will be very insignificant, but I'll do it anyway to show you how small it is and why we would just ignore it.

#"HSO"_3^(-)(aq) rightleftharpoons "SO"_3^(2-)(aq) + "H"^(+)(aq)#

#"I"" ""0.00686"" "" "" "" "0" "" "" "" "0.00686#

#"C"" "-x" "" "" "" "+x" "" "" "+x#

#"E"" "0.00686 - x" "" "x" "" "" "" "0.00686 + x#

#K_(a2) = 9.1 xx 10^(-8) = (x(0.00686 + x))/(0.00686 - x)#

Here, you CAN use the small

#K_(a2) ~~ (x(0.00686 + cancel(x)))/(0.00686 - cancel(x)) ~~ x#

#= 9.1 xx 10^(-8)# #"M"# (

#0.0013%# dissociation)

So, after considering **dissociation** **overall**:

#color(blue)("0.00314 M H"_2"SO"_3)# again, approximately, since the percent dissociation was so small.#(0.00686 - 9.1 xx 10^(-8)) "M HSO"_3^(-) ~~ color(blue)("0.00686 M HSO"_3^(-))# #color(blue)(9.1 xx 10^(-8) "M SO"_3^(2-))# , but practically we consider it zero...#"0.00686 M H"^(+) + 9.1 xx 10^(-8) "M H"^(+) ~~ color(blue)("0.00686 M H"^(+))#

After all that work to check the stages in variation of

#color(blue)("pH") = -log["H"^(+)] = color(blue)(2.16)#

At

- At
#"pH"# #0.5# , the solution is**more**acidic than the first#"pKa"# of#1.81# (which, if#"pH" = 1.81# , marks the half-equivalence point for the#"H"_2"SO"_3//"HSO"_3^(-)# equilibrium)the second*and*#"pKa"# of#7.04# (which, if#"pH" = 7.04# , marks the half-equivalence point for the#"HSO"_3^(-)//"SO"_3^(2-)# equilibrium).

This tells us that the solution will be dominated by the

mostacidic form of#color(blue)(bb("H"_2"SO"_3))# , i.e.itself.

- At
#"pH"# #5.5# , the solution is**more**basic than#"pKa" = 1.81# (belonging to#"H"_2"SO"_3# ) and**more**acidic than#"pKa" = 7.04# (belonging to#"HSO"_3^(-)# ).

So, the form of

#"H"_2"SO"_3# that will dominate ismore basic than#"H"_2"SO"_3# butmore acidic than#"SO"_3^(2-)# . Therefore,#color(blue)(bb("HSO"_3^(-)))# dominates here.

- I suppose you could figure this one out at this point; the
#"pH"# of#9# is*more basic than**both*#"pKa"# s, so the**most**basic species dominates, i.e.#color(blue)(bb("SO"_3^(2-)))# .