# Oxalic acid is diprotic. What volume of 0.100M KOH is required to neutralize 25ml of 0.333M oxalic acid?

Mar 5, 2018

$\approx$ 34.2 ml of the $K O H$ solution

#### Explanation:

Oxalic acid is a weak acid that dissociates in two steps into Oxonium ions $\left[{H}_{3} {O}^{+}\right]$. To find how much $K O H$ is needed to neutralize the acid we must first determine the number of moles of Oxonium ions in the solution, as these will react in a 1:1 ratio with the Hydroxide ions to form water.

As it is a weak diprotic acid it has a ${K}_{a}$ value for both its acid form and anion form (hydrogen oxalate ion).

${K}_{a}$ (oxalic acid) =$5.4 \times {10}^{-} 2$
${K}_{a}$ (hydrogen oxalate ion)= $5.4 \times {10}^{-} 5$

Remember: ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[A n i o n\right]}{\left[A c i d\right]}$

Hence: $5.4 \times {10}^{-} 2 = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[A n i o n\right]}{\left[0.333\right]}$

$0.0179982 = \left(\left[{H}_{3} {O}^{+}\right] \times \left[A n i o n\right]\right)$

$\sqrt{0.01799282} = \left[{H}_{3} {O}^{+}\right] = \left[A n i o n\right]$
The concentration of the anion and the oxonium ion in the first dissociation are equal $\left(0.134157 m o l {\mathrm{dm}}^{-} 3\right)$

Then in the dissociation of the hydrogen oxalate ion we have:

$5.4 \times {10}^{-} 5 = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[A n i o n\right]}{\left[0.1341573703\right]}$

As the anion formed in the first step is acting as an acid in the second dissociation, we use its concentration of the anion found from the first dissociation.

$7.244497996 \times {10}^{-} 6 = \left(\left[{H}_{3} {O}^{+}\right] \times \left[A n i o n\right]\right)$
$\sqrt{7.244497996 \times {10}^{-} 6} = \left[{H}_{3} {O}^{+}\right] = \left[A n i o n\right]$
Which is $\left(0.0026915605 m o l {\mathrm{dm}}^{-} 3\right)$

All of the dissociations took place in a volume of 25 ml $\left(0.025 {\mathrm{dm}}^{3}\right)$ Hence we can find the number of moles of Oxonium ions formed from both dissociations.

concentration $\left(m o l {\mathrm{dm}}^{-} 3\right)$= (Moles)/ (volume) $\left({\mathrm{dm}}^{3}\right)$

$0.134157 \times 0.025 = 0.0033539343$
and
$0.0026915695 \times 0.025 = 6.72890125 \times {10}^{-} 5$
Add them up to get: $0.0034212233 m o l$ of $\left[{H}_{3} {O}^{+}\right]$
This is the same amount of moles we need of $O {H}^{-}$ to neutralize the acid.
Hence the volume of $K O H$ we need is:

concentration $\left(m o l {\mathrm{dm}}^{-} 3\right)$= (Moles)/ (volume) $\left({\mathrm{dm}}^{3}\right)$

$0.1 = \frac{0.0034212233}{v}$
$v = 0.034212233 {\mathrm{dm}}^{3}$ or multiply it by 1000 get it in ml.
$v \approx 34.2 m l$