Question

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

Answer 1

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

When I use `"0.11 M NaOH"` and `"0.1 M"` triprotic acid, ideally, the first HALF equivalence point would be at `"5.8 mL"`, and the second HALF equivalence point would be at `"16.0 mL"`.

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

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

The first `K_a` of citric acid is `7.4 xx 10^(-3)`, so you should expect the first half-equivalence point to be near

`"pH" ~~ "pKa"_1 = -log(7.4 xx 10^(-3)) = 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 `"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 xx 10^(-5)`.

`"pH" ~~ "pKa"_2 = -log(1.7 xx 10^(-5)) = 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"`

`~~` `"18.3 mL"`

and the ACTUAL point nearest that is at `"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 `color(blue)("pKa"_1)` of about `color(blue)(2.2 - 2.4)`, as compared to a literature value of `color(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 `color(blue)("pKa"_2)` of about `color(blue)(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 `color(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 `color(blue)("pKa"_3)` around `color(blue)(5.8 - 6.0)`.

This falls outside the range of the actual `"pKa"_3`, which is `color(red)(6.40)`. But at least you have data you can talk about...