# Equilibrium problem about sulfurous acid?

Jul 12, 2017

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 $\text{pH}$ is thus primarily dictated by the first dissociation of a diprotic acid, especially if the first proton is of a not-strong acid (${K}_{a 1} < 1$).
• The predominant species will reflect the relative magnitudes of the $\text{pH}$ vs. ${\text{pK}}_{a i}$. If ${\text{pH" < "pK}}_{a i}$, then the acid the ${\text{pK}}_{a i}$ 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.

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.

a)

${\text{H"_2"SO"_3(aq) rightleftharpoons "HSO"_3^(-)(aq) + "H}}^{+} \left(a q\right)$

$\text{I"" ""0.01"" "" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-x" "" "" "+x" "" "" "" } + x$
$\text{E"" "0.01 - x" "" "x" "" "" "" "" } x$

${K}_{a 1} = \frac{{x}^{2}}{0.01 - x} = 1.5 \times {10}^{- 2}$

By using the quadratic formula, which must be done because the ${K}_{a 1}$ is not small enough at this low concentration, you should get

$x = \text{0.00686 M}$

So, the species in solution after dissociation $\boldsymbol{1}$ are:

• ${\text{0.00314 M H"_2"SO}}_{3}$
• ${\text{0.00686 M HSO}}_{3}^{-}$
• ${\text{0.00686 M H}}^{+}$

Now, the ${\text{HSO}}_{3}^{-}$ should dissociate again, this time using ${K}_{a 2} = 9.1 \times {10}^{- 8}$ (not ignoring the ${\text{H}}^{+}$ that was generated already from the previous equilibrium step).

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.

${\text{HSO"_3^(-)(aq) rightleftharpoons "SO"_3^(2-)(aq) + "H}}^{+} \left(a q\right)$

$\text{I"" ""0.00686"" "" "" "" "0" "" "" "" } 0.00686$
$\text{C"" "-x" "" "" "" "+x" "" "" } + x$
$\text{E"" "0.00686 - x" "" "x" "" "" "" } 0.00686 + x$

${K}_{a 2} = 9.1 \times {10}^{- 8} = \frac{x \left(0.00686 + x\right)}{0.00686 - x}$

Here, you CAN use the small $x$ approximation, since ${K}_{a 2}$ is very small (${10}^{- 5}$ is usually a good cutoff), and you get:

${K}_{a 2} \approx \frac{x \left(0.00686 + \cancel{x}\right)}{0.00686 - \cancel{x}} \approx x$

$= 9.1 \times {10}^{- 8}$ $\text{M}$

(0.0013% dissociation)

So, after considering dissociation $\boldsymbol{2}$ in addition to dissociation $1$, we still have overall:

• $\textcolor{b l u e}{{\text{0.00314 M H"_2"SO}}_{3}}$ 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^(-))
• $\textcolor{b l u e}{9.1 \times {10}^{- 8} {\text{M SO}}_{3}^{2 -}}$, but practically we consider it zero...
• "0.00686 M H"^(+) + 9.1 xx 10^(-8) "M H"^(+) ~~ color(blue)("0.00686 M H"^(+))

b)

After all that work to check the stages in variation of $\left[{\text{H}}^{+}\right]$, the $\text{pH}$ should be easy.

$\textcolor{b l u e}{{\text{pH") = -log["H}}^{+}} = \textcolor{b l u e}{2.16}$

c)

At $\text{pH}$ $0.5$, $5.5$, and $9$, let's consider this conceptually without any calculations.

• At $\text{pH}$ $0.5$, the solution is more acidic than the first $\text{pKa}$ of $1.81$ (which, if $\text{pH} = 1.81$, marks the half-equivalence point for the ${\text{H"_2"SO"_3//"HSO}}_{3}^{-}$ equilibrium) and the second $\text{pKa}$ of $7.04$ (which, if $\text{pH} = 7.04$, marks the half-equivalence point for the ${\text{HSO"_3^(-)//"SO}}_{3}^{2 -}$ equilibrium).

This tells us that the solution will be dominated by the most acidic form of $\textcolor{b l u e}{\boldsymbol{{\text{H"_2"SO}}_{3}}}$, i.e. itself.

• At $\text{pH}$ $5.5$, the solution is more basic than $\text{pKa} = 1.81$ (belonging to ${\text{H"_2"SO}}_{3}$) and more acidic than $\text{pKa} = 7.04$ (belonging to ${\text{HSO}}_{3}^{-}$).

So, the form of ${\text{H"_2"SO}}_{3}$ that will dominate is more basic than ${\text{H"_2"SO}}_{3}$ but more acidic than ${\text{SO}}_{3}^{2 -}$. Therefore, $\textcolor{b l u e}{\boldsymbol{{\text{HSO}}_{3}^{-}}}$ dominates here.

• I suppose you could figure this one out at this point; the $\text{pH}$ of $9$ is more basic than both $\text{pKa}$s, so the most basic species dominates, i.e. $\textcolor{b l u e}{\boldsymbol{{\text{SO}}_{3}^{2 -}}}$.