## Chapter 14

## 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?

## Step-by-Step

## 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. (

*−0.8 mV)*

**Answer:**