# What is the Ka value of this citric acid + NaOH titration?

Jul 3, 2017

Hm, that's odd... Here's what I was expecting to see:

When I use $\text{0.11 M NaOH}$ and $\text{0.1 M}$ triprotic acid, ideally, the first HALF equivalence point would be at $\text{5.8 mL}$, and the second HALF equivalence point would be at $\text{16.0 mL}$.

Here, the first and second equivalence points are hard to see. They look indistinguishable from each other.

And here's what you got (obviously, turn in your own data):

The first ${K}_{a}$ of citric acid is $7.4 \times {10}^{- 3}$, so you should expect the first half-equivalence point to be near

${\text{pH" ~~ "pKa}}_{1} = - \log \left(7.4 \times {10}^{- 3}\right) = 2.13$.

If we assume you did this absolutely perfectly, then the first equivalence point (of THREE!) should be near

2 xx overbrace("2.0 mL")^(V_("half-equiv")) = "4.0 mL".

And in your case it's around $\text{5.0 mL}$ (you can see that from how the points clump together). That's probably as good an approximation as you can make, as there is (barely) a semblance of a first equivalence point near there.

The second ${K}_{a}$ of citric acid is $1.7 \times {10}^{- 5}$.

${\text{pH" ~~ "pKa}}_{2} = - \log \left(1.7 \times {10}^{- 5}\right) = 4.77$.

There is in fact no easy way to find where this is... but there is a workaround. Here's my best estimate:

Knowing the first and third equivalence points at least somewhat, the second equivalence point is ideally right in the middle... each proton is just like any other.

V_(eq2) ~~ ("31.6 mL" - "5.0 mL")/2 + "5.0 mL"

$\approx$ $\text{18.3 mL}$

and the ACTUAL point nearest that is at $\text{17.55 mL}$. That's what you should take.

So now, you've got to work from there to find each of the half-equivalence points.

Take the DIFFERENCE in volumes between each equivalence point, halve THAT, and add it to the left-hand volume, and do not just take the volume you read off of the $x$ axis:

1st half-equiv pt.

("5.0 mL" - "0.0 mL")/2 + "0.0 mL" = "2.5 mL",

corresponding to a $\textcolor{b l u e}{{\text{pKa}}_{1}}$ of about $\textcolor{b l u e}{2.2 - 2.4}$, as compared to a literature value of $\textcolor{red}{2.13}$.

With your data, that's the best you can do. Trace it out to the $y$ axis and verify this.

2nd half-equiv pt.

("17.55 mL" - "5.0 mL")/2 + "5.0 mL" = "11.3 mL",

corresponding to a $\textcolor{b l u e}{{\text{pKa}}_{2}}$ of about $\textcolor{b l u e}{3.8 - 4.0}$. Trace it out to the $y$ axis and verify this.

Turns out this is not a good estimate to the literature value of $\textcolor{red}{4.77}$, but you have no other way to find this, because the first and second equivalence points seem to blend together too well.

3rd half-equiv pt.

This is also somewhat of a guesstimate.

("31.6 mL" " "- overbrace("17.55 mL")^"estimated from earlier")/2 + "17.55 mL" ~~ "24.6 mL",

giving you a $\textcolor{b l u e}{{\text{pKa}}_{3}}$ around $\textcolor{b l u e}{5.8 - 6.0}$.

This falls outside the range of the actual ${\text{pKa}}_{3}$, which is $\textcolor{red}{6.40}$. But at least you have data you can talk about...