Question

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?

Answer 1

`"pH" = 10.1`

Explanation:

! LONG ANSWER !!

The idea here is that aspartic acid, `"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 `"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 `pK_a` values, which are listed as

`{(pK_(a1) = 1.99), (pK_(a2) = 3.90), (pK_(a3) = 9.90) :}`

http://academics.keene.edu/rblatchly/Chem220/hand/npaa/aawpka.htm

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 `pK_(a1)`. Notice that the molecule carries an overall `(color(red)(1+))` charge from the protonated `-"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 `pK_(a1)` to just below `pK_(a2)`, a proton from one of the two carboxyl groups starts to comes off, leaving behind an `alpha`-carboxilic acid group, `"COO"^(-)`.

At this point, the molecule is neutral, since the `(color(red)(1+))` charge on the protonated amino group is balanced by the `(color(red)(1-))` charge on the `alpha`-carboxilic acid group.

The same thing happens as the pH increases from `pK_(a2)` to just below `pK_(a3)` - the proton from the second carboxyl group comes off, leaving behind a second `alpha`-carboxilic acid group.

At this point, the molecule carries a `(color(red)(1-))` charge, since you now get two `(color(red)(1-))` charges and only one `(color(red)(1+))` charge.

Finally, at pH values that exceed `pK_(a3)`, the last acidic proton comes off, leaving behind the amine group, `-"NH"_2`.

At this point, the molecule carries a `(color(red)(2-))` net charge.

So, you know that your solution contains the `"HA"^(-)` and `"A"^(2-)` forms of the molecule. More specifically, it contains `"8 mmoles"` of `"HA"^(-)` and `"10 mmoles"` of `"A"^(2-)`.

Since `"A"^(2-)` is the conjugate base of `"HA"^(-)`, you can say that you're dealing with a buffer solution that contains a weak acid, `"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 `"HA"^(-)` is your acidic molecule, you will need to use `pK_(a3)` in the above equation.

Right from the start, you can tell that the pH of the solution before adding the base is higher than `pK_(a3)`, 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, `"OH"^(-)`, in solution.

The hydroxide anions will then neutralize the `"HA"^(-)` molecules to produce water and `"A"^(2-)`, the conjugate base of the acid form.

`"HA"_text((aq])^(-) + "OH"_text((aq])^(-) -> "A"_text((aq])^(2-) + "H"_2"O"_text((l])`

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

This means that `"1 mmole"` of hydroxide anions will consume `"1 mmol"` of `"HA"^(-)` and produce `"1 mmol"` of `"A"^(2-)`.

After the reaction is complete, the solution will contain

`n_(OH^(-)) = "0 moles" ->` completely consumed by the reaction

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

`n_(A^(2-)) = "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)`