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## Q. 14.3

Using the Selectivity Coefficient

A fluoride ion-selective electrode has a selectivity coefficient $K_{F^{−} , OH^{−}}^{Pot}$ = 0.1. What will be the change in electrode potential when 1.0 × $10^{−4}$ M $F^{−}$ at pH 5.5 is raised to pH 10.5?

## Verified Solution

From Equation 14-10, we find the potential with negligible $OH^{−}$ at pH 5.5:

$E = constant \pm \frac{0.059 16}{z_{A}} \log[\mathcal{A_{A}} + \sum\limits_{x}{K_{A,X}^{Pot}} \mathcal{A_{X}}]$           (14-10)

$E = constant − 0.059 16 \log [1.0 × 10^{−4}] = constant + 236.6$ mV

At pH 10.50, $[OH^{−}] = 3.2 × 10^{−4}$ M, so the electrode potential is

$E = constant − 0.059 16 \log [1.0 × 10^{−4} + (0.1)(3.2 × 10^{−4})]$

= constant + 229.5 mV

The change is 229.5 − 236.6 = −7.1 mV, which is quite significant. If you didn’t know about the pH change, you would think that the concentration of $F^{−}$ had increased by 32%.

Test Yourself     Find the change in potential when 1.0 × $10^{−4}$ M $F^{−}$ at pH 5.5 is raised to pH 9.5. (Answer: −0.8 mV)