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

[Et

_{3}NH

^{+}] = 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.