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

##### 1 Answer

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

When I use

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**

#"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

The **second**

#"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"# .what you should take.That's

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

**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

**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

This falls outside the range of the actual