# If 1 mmole of NaOH is added into a solution in which there is 8 mmoles of aspartic acid with a net -1 charge and 10 moles of aspartic acid with a ned -2 charge, what would the pH of the solution be?

Jan 11, 2016

$\text{pH} = 10.1$

#### Explanation:

The idea here is that aspartic acid, ${\text{C"_4"H"_7"NO}}_{4}$, is a triprotic acid, which means that it contains three acidic protons that it can release in solution.

For simplicity, I'll use ${\text{H"_3"A}}^{+}$ as the notation for aspartic acid. Now, these acidic protons will come off at different pH values.

These pH values will depend on the acid's $p {K}_{a}$ values, which are listed as

$\left\{\begin{matrix}p {K}_{a 1} = 1.99 \\ p {K}_{a 2} = 3.90 \\ p {K}_{a 3} = 9.90\end{matrix}\right.$

Here's how a molecule of aspartic acid looks like when fully protonated

This is the form you get at pH values that are below $p {K}_{a 1}$. Notice that the molecule carries an overall $\left(\textcolor{red}{1 +}\right)$ charge from the protonated $- {\text{NH}}_{3}^{+}$ group.

Here's how a titration would look like for the titration of aspartic acid with a strong base

As the pH of the solution increases from $p {K}_{a 1}$ to just below $p {K}_{a 2}$, a proton from one of the two carboxyl groups starts to comes off, leaving behind an $\alpha$-carboxilic acid group, ${\text{COO}}^{-}$.

At this point, the molecule is neutral, since the $\left(\textcolor{red}{1 +}\right)$ charge on the protonated amino group is balanced by the $\left(\textcolor{red}{1 -}\right)$ charge on the $\alpha$-carboxilic acid group.

The same thing happens as the pH increases from $p {K}_{a 2}$ to just below $p {K}_{a 3}$ - the proton from the second carboxyl group comes off, leaving behind a second $\alpha$-carboxilic acid group.

At this point, the molecule carries a $\left(\textcolor{red}{1 -}\right)$ charge, since you now get two $\left(\textcolor{red}{1 -}\right)$ charges and only one $\left(\textcolor{red}{1 +}\right)$ charge.

Finally, at pH values that exceed $p {K}_{a 3}$, the last acidic proton comes off, leaving behind the amine group, $- {\text{NH}}_{2}$.

At this point, the molecule carries a $\left(\textcolor{red}{2 -}\right)$ net charge.

So, you know that your solution contains the ${\text{HA}}^{-}$ and ${\text{A}}^{2 -}$ forms of the molecule. More specifically, it contains $\text{8 mmoles}$ of ${\text{HA}}^{-}$ and $\text{10 mmoles}$ of ${\text{A}}^{2 -}$.

Since ${\text{A}}^{2 -}$ is the conjugate base of ${\text{HA}}^{-}$, you can say that you're dealing with a buffer solution that contains a weak acid, ${\text{HA}}^{-}$, and its conjugate base.

This means that you can use the Henderson - Hasselbalch equation to help you figure out the pH of the solution

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

Since ${\text{HA}}^{-}$ is your acidic molecule, you will need to use $p {K}_{a 3}$ in the above equation.

Right from the start, you can tell that the pH of the solution before adding the base is higher than $p {K}_{a 3}$, since you have more moles of conjguate base than you have of weak acid.

Now, sodium hydroxide is a strong base that dissociates in a $1 : 1$ mole ratio to produce hydroxide anions, ${\text{OH}}^{-}$, in solution.

The hydroxide anions will then neutralize the ${\text{HA}}^{-}$ molecules to produce water and ${\text{A}}^{2 -}$, the conjugate base of the acid form.

${\text{HA"_text((aq])^(-) + "OH"_text((aq])^(-) -> "A"_text((aq])^(2-) + "H"_2"O}}_{\textrm{\left(l\right]}}$

You have $1 : 1$ mole ratios between all the species that are taking part in the reaction.

This means that $\text{1 mmole}$ of hydroxide anions will consume $\text{1 mmol}$ of ${\text{HA}}^{-}$ and produce $\text{1 mmol}$ of ${\text{A}}^{2 -}$.

After the reaction is complete, the solution will contain

${n}_{O {H}^{-}} = \text{0 moles} \to$ completely consumed by the reaction

n_("HA"^(-)) = "8 mmoles" - "1 mmole" = "7 mmoles"

${n}_{{A}^{2 -}} = \text{10 mmoles" + "1 mmole" = "11 mmoles}$

This means that you have

"pH" = pK_(a3) + log( (["A"^(2-)])/(["HA"^(-)]))

Since the volume of the buffer is the same for all species, you can use the number of moles in the H - H equation.

Therefore,

"pH" = 9.90 + log( (11 color(red)(cancel(color(black)("mmoles"))))/(7color(red)(cancel(color(black)("mmoles"))))) = color(green)(10.1)