## Chapter 14

## Q. 14.4

**Response of an Ion-Selective Electrode**

When a fluoride electrode was immersed in standard solutions (maintained at a constant ionic strength of 0.1 M with NaNO_{3}), the following potentials (versus S.C.E.) were observed:

[F^{−}] (M) | E(mV) |

1.00 × 10^{−5} | 100.0 |

1.00 × 10^{−4} | 41.5 |

1.00 × 10^{−3} | −17.0 |

Because the ionic strength is constant, the response should depend on the logarithm of the F^{−} concentration. Find [F^{−}] in an unknown that gave a potential of 0.0 mV.

## Step-by-Step

## Verified Solution

We fit the calibration data with Equation 14-12:

E = constant − β(0.059 16) \log \mathcal{A_{F^{−}}}(outside) **(14-12)**

\underset{y}{E} = m \underbrace{\log[F^{- }]}_{x} + b

Plotting E versus \log [F^{−}] gives a straight line with a slope m = −58.5 mV and a y-intercept b = −192.5 mV. Setting E = 0.0 mV, we solve for [F^{−}]:

0.0 mV = (−58.5 mV) \log[F^{−}] − 192.5 mV ⇒ [F^{−}] = 5.1 × 10^{−4} M

* Test Yourself* Find [F^{−}] if E = 81.2 mV. Is the calibration curve valid for E = 110.7 mV?

(

*2.1 × 10^{−4} M; no, because calibration points do not go above 100 mV)*

**Answer:**