# Question 20332

Apr 26, 2015

The pH of the solution is 2.63.

So, you're titrating sulfurous acid, ${H}_{2} S {O}_{3}$, a weak acid, with potassium hydroxide, $K O H$, a strong base.

The equations for the two dissociation equilibrium reactions that will be established in solution will be

${H}_{2} S {O}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H S {O}_{3 \left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$, ${K}_{a 1} = 5.9 \cdot {10}^{- 3}$

$H S {O}_{3 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s S {O}_{3 \left(a q\right)}^{2 -} + {H}_{3} {O}_{\left(a q\right)}^{+}$, ${K}_{a 2} = 6.0 \cdot {10}^{- 8}$

Now, because you have a significant difference in magnitude between ${K}_{a 1}$ and ${K}_{a 2}$, you won't get any $S {O}_{3}^{2 -}$ anions in your solution until after all of the sulfurous acid has been converted to hydrogen sulfite, $H S {O}_{3}^{-}$.

In order for that to happen, you have to add enough moles of strong base to completely neutralize all of the sulfurous acid present in the initial solution.

If you don't add enough moles of strong base, the solution will essentially become a buffer, because you'll have both a weak acid, ${H}_{2} S {O}_{3}$, and its conjugate base, $H S {O}_{3}^{-}$, present in solution.

Use the molarities of the two solutions to determine how many moles of each you add together

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{{H}_{2} S {O}_{3}} = \text{0.10 M" * 70 * 10^(-3)"L" = "0.0070 moles }$ ${H}_{2} S {O}_{3}$

${n}_{K O H} = \text{0.10 M" * 50 * 10^(-3)"L" = "0.0050 moles }$ $K O H$

When you start addin the base, this is what will happen

${H}_{2} S {O}_{3 \left(a q\right)} + K O {H}_{\left(a q\right)} \to K H S {O}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

The net ionic equation will be

${H}_{2} S {O}_{3 \left(a q\right)} + O {H}_{\left(a q\right)}^{-} \to H S {O}_{3 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)}$

You have less moles of $K O H$ than you have of ${H}_{2} S {O}_{3}$, which means that the strong base will be consumed by this reaction.

The number of moles of ${H}_{2} S {O}_{3}$ will decrease by the same number of moles of base added. At the same time, you'll produce
some conjugate base,$H S {O}_{3}^{-}$. So, you get

${n}_{K O H \text{remaining}} = 0$
n_(H_2SO_3"remaining") = 0.0070 - 0.0050 = "0.0020 moles"
n_(HSO_3^(-)"produced") = "0.0050 moles"

The final volume of the solution will be

${V}_{\text{final}} = {V}_{{H}_{2} S {O}_{3}} + {V}_{K O H}$
${V}_{\text{final" = 70 + 50 = "120 mL}}$

Your solution will now contain 0.0020 moles of weak acid and 0.0050 moles of conjugate base. Use the Henderson-Hasselbalch equation to determine the pH of the resulting solution

pH_"sol" = pK_(a1) + log((["conjugate base"])/(["weak acid"]))

Since you still have some sulfurous acid present in solution, the first equilibrium reaction will be established and you'll get

$p {K}_{a 1} = - \log \left({K}_{a 1}\right) = - \log \left(5.9 \cdot {10}^{- 3}\right) = 2.23$

The concentrations the weak acid and conjugate base will be

$\left[{H}_{2} S {O}_{3}\right] = \text{0.0020 moles"/(120 * 10^(-3)"L") = "0.01667 M}$

$\left[H S {O}_{3}^{-}\right] = \text{0.0050 moles"/(120 * 10^(-3)"L") = "0.04167 M}$

Now plug all your values into the H-H equation

pH_"sol" = 2.23 + log((0.04167cancel("M"))/(0.01667cancel("M"))) = color(green)(2.63)#