# Question #010a5

Apr 12, 2015

This is a titration of sulfurous acid, ${H}_{2} S {O}_{3}$, a weak diprotic acid, with sodium hydroxide, $N a O H$, a strong base.

Since sulfurous acid is a diprotic acid, you'll have two equilibria reactions ,each with its own acid dissociation constant and $p {K}_{a}$

${H}_{2} S {O}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + H S {O}_{3 \left(a q\right)}^{-}$, $p {K}_{a 1} = 1.85$

$H S {O}_{3 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + S {O}_{3 \left(a q\right)}^{2 -}$, $p {K}_{a 2} = 7.19$ So, starting from the lower-left corner. The first point is where pH is equal to $p {K}_{a 1}$, no base has been added, so just the ${H}_{2} S {O}_{3}$ form of the acid will be in solution.

Now you start adding the base. This will increase the pH enough to trigger the buffer region at which $\left[{H}_{2} S {O}_{3}\right] = \left[H S {O}_{3}^{-}\right]$.

The next step is the first equivalence point at which the first ${H}^{+}$ is completely dissociated from ${H}_{2} S {O}_{3}$ $\to$ now only $H S {O}_{3}^{-}$ remains in solution.

Adding more base will trigger another buffer region in which $\left[H S {O}_{3}^{-}\right] = \left[S {O}_{3}^{2 -}\right]$. Notice that the pH of the solution is equal to $p {K}_{a 2}$ (think Henderson-Hasselbalch equation).

Finally, adding more base will result in a second equivalence point at which all the ${H}^{+}$s have been dissociated from $H S {O}_{3}^{-}$, leaving only $S {O}_{3}^{2 -}$ in solution.