[Bert95]

Hello everyone,

I have an exercise here from D. C. Harris' Quantitative Chemical Analysis and I don't get the given solution (the exercise is No. 9 of chapter 12, or at least it is in the first German edition). The task is to

calculate pH - considering activity coefficients - after addition of 4.0 ml NaOH (0.100 M) to 10.0 ml of a solution of triethylammoniumbromide (0.100 M).

First thing I've done is to solve the problem without activities: after addition of 4 ml strong base 40 % of the initial amount of the acid is turned into corresponding base (triethylamin). pK_S of triethylammonium is 10.715 (according to the annex of the same book). Using the Henderson-Hasselbalch-equation I got:



In the chapter dealing with activities it is said, that using activity coefficients, the Henderson-Hasselbalch-equation has to be written as follows:



Furthermore it is said that a value of one can be assumed as the activity coefficient of uncharged species (so in this exercise that would be the case for the base ethylamin). There is also a table with activity coefficients at different ionic strenghts (µ); triethylammonium can be found in the last row:



Then I tried to calculate ionic strength () and got the first problem. How can this be done, if you don't know pH and with it [H⁺] and [OH^-]? Influence on ionic strength of these ions might be negligible (?), however, I used the result of the calculation without activities as an approximation. My calculation of the other ionic concentrations resulted in:

[Na⁺] = 0.1 M x 4/14 = 0.0286 M
[Br^-] = 0.1 M x 10/14 = 0.0714 M
[Et₃NH⁺] = 0.6 x 0.1 M x 10/14 = 0.0429 M

All in all I got µ=0.0716 M. I used this result with the table and interpolated between µ=0.05 M and µ=0.1 M, resulting in =0.81. And with all this I finally calculated



Unfortunately the result given in the book is pH=9.72 and I don't really see how to get there. I hope someone here can help me.

Greetings from Germany, Bert

PS: I hope my English is good enough to make the problem understandable.

[Borek]

Then I tried to calculate ionic strength () and got the first problem. How can this be done, if you don't know pH and with it [H⁺] and [OH^-]?

You already know that pH will be not much different from 10.5. If so, concentration of H+ is 3x10^-11 and concentration of OH- around 3x10^-4. You can use these values to calculate ionic strength. However, as they are orders of magnitude lower than concentrations of other ions, their effect is neglectable (especially taking into account fact that the table doesn't contain value for 0.07, but only for 0.05 and 0.1, so accuracy of ionic strength accuracy doesn't matter much).

Unfortunately the result given in the book is pH=9.72 and I don't really see how to get there.

You won't get there. As far as I can tell your result is a correct one (or at least very close to the correct one). 9.72 is wrong for sure, must be an error in the book.

I got 10.65, using slightly different methodology, so my result is not directly comparable.

[Bert95]

Thank you for your help. I thought through the problem again and I think it is indeed very likely to be an error in the book. Usually activity coefficients will be <1, so the logarithmic term increases. Therefore the result should be greater than the result of the calculation without activities.

[enahs]

Your answer differs so much from the book,  because you are using the wrong pKa value according to the book. In my 4th edition of the same book, English version, they list the pKa as 9.8.

However, in the book also used the Na+ effective activity instead of the concentration of the triethyl amine in the numerator. No sure why. I have not done anything with activity coefficients in forever!

[Bert95]

Your answer differs so much from the book,  because you are using the wrong pKa value according to the book. In my 4th edition of the same book, English version, they list the pKa as 9.8.

Well, I looked up the pKa in two other sources and also found 10.72 for triethylammonium. Interestingly, pKa for trimethylammonium is listet as 9.8 in my book. Maybe there is an error in the exercise text.

However, in the book also used the Na+ effective activity instead of the concentration of the triethyl amine in the numerator. No sure why.

I'm not shure about that, too.