[o0CY0o]

Fluoride ion-selective electrode measure F- lei the interference is OH- ion
Electrode selectivity coefficient K=0.1(i.e., repones to F- is 10 times as OH-)
How to calculate the maxium pH that can tolerate in analysis of 1x10^-5 M F-- solution if error is 1%?


In my mind, i think that E=Eo + 0.0592log( [F-] + 0.1[OH-] )
and since the error tolerated is 0.01
 0.0592log( [F-] + 0.1[OH-] ) < 0.01
However, when i subsitute  [F-] by 1x10^-5 the answer of pOH is 13.77.
Is there anything wrong with my calculation?

[Borek]

0.0592log( [F-] + 0.1[OH-] ) < 0.01

Error in potential, or error in concentration?