[_cheers]

For the case of a doubly-charged cation, as the primary species, in the presence of a singly-charged interferring cation, the referenced potential of the electrode (in Volts at 25°C) may be given by:
E1,2 = constant + [ β×0.05916/(n)] log(a1 + k1,2×a22)
For a solution which is 0.0136 M in CaCl2 plus 0.0136 M in NaCl, the potential is 241.6 mV.


For the determination of Ca2+-activity in the presence of Na+, calculate the value of the constant in the above expression.

Ok so I calc'd a1,a2 and am assuming E1,2 =.2416V
Thats where i am stuck. How do I find the constant without K1,2 ?

Calculate the selectivity coefficient.

This I need some hints on. I cant find any decent literature on the subject. Thanks in advance.

[Aeon]

Here are some good references.

http://www.iupac.org/publications/pac/pdf/1995/pdf/6703x0507.pdf
http://www.iupac.org/publications/pac/2000/7210/7210pdfs/7210umezawa_1852.pdf

Both IUPAC papers.

Wang's Analytical Electrochemistry has a good chapter on ion selective electrodes (ISEs).
http://books.google.com/books/about/Analytical_electrochemistry.html?id=ZgUa4WbLMKkC