Well, the very first thing i would do is to set up all the essential components: reaction, ice table, and also the general point of the question. Here, every the ice cream tables room in molarity.

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

sometimes you have the right to ignore the 2nd dissociation the a polyprotic acid, specifically if it"s the critical dissociation it have the right to do.The #"pH"# is thus primarily dictated by the very first dissociation of a diprotic acid, particularly if the very first proton is that a not-strong mountain (#K_(a1) ).The predominant types will reflect the loved one magnitudes that the #"pH"# vs. #"pK"_(ai)#. If #"pH" , then the acid the #"pK"_(ai)# represents will certainly exist in its acidic form, and also vice versa. You deserve to verify this utilizing the Henderson-Hasselbalch equation, but you should have the ability to do this conceptually on exams to check your work.

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DISCLAIMER: LONG ANSWER!

I assume by now you know exactly how to set one up and also use it because that a monoprotic acid, and also this is just an expansion into a diprotic acid. It"s a matter of keeping your large picture straight, and also knowing what assumptions to make.

#a)#

#"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 utilizing the quadratic formula, which need to be done due to the fact that the #K_(a1)# is not tiny enough at this short concentration, you have to get

#x = "0.00686 M"#

So, the species in equipment after dissociation #bb1# are:

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

Now, the #"HSO"_3^(-)# should dissociate again, this time utilizing #K_(a2) = 9.1 xx 10^(-8)# (not skipping the #"H"^(+)# the was generated currently from the vault equilibrium step).

Here, we understand that the 2nd dissociation will certainly be really insignificant, yet I"ll carry out it quiet to display you how small it is and also why we would certainly just overlook 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 deserve to use the tiny #x# approximation, because #K_(a2)# is very little (#10^(-5)# is commonly a an excellent cutoff), and also you get:

#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 #bb2# in addition to dissociation #1#, we still have overall:

#color(blue)("0.00314 M H"_2"SO"_3)# again, approximately, since the percent dissociation to be 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 virtually we consider it zero...#"0.00686 M H"^(+) + 9.1 xx 10^(-8) "M H"^(+) ~~ color(blue)("0.00686 M H"^(+))#

#b)#

After all that work to inspect the step in sports of #<"H"^(+)>#, the #"pH"# should be easy.

#color(blue)("pH") = -log<"H"^(+)> = color(blue)(2.16)#

#c)#

At #"pH"# #0.5#, #5.5#, and #9#, let"s think about this conceptually without any type of calculations.

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

This tells us that the equipment will be overcame by the most acidic kind of #color(blue)(bb("H"_2"SO"_3))#, i.e. itself.

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

So, the type of #"H"_2"SO"_3# the will conquer is more simple than #"H"_2"SO"_3# however more acidic than #"SO"_3^(2-)#. Therefore, #color(blue)(bb("HSO"_3^(-)))# restrict here.

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I suppose you might figure this one the end at this point; the #"pH"# that #9# is more straightforward than both #"pKa"#s, so the most simple species dominates, i.e. #color(blue)(bb("SO"_3^(2-)))#.