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## 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.

## 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?
(Answer: 2.1 × $10^{−4}$ M; no, because calibration points do not go above 100 mV)