Question

The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at 25.008C is given by the equation

${\mathbf{E}}_{\mathbf{meas}}\mathbf{=}{\mathbf{E}}_{\mathbf{ref}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{05916}\mathbf{pH}$

where ${\mathbf{E}}_{\mathbf{ref}}$%ref contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that localid="1663752320533" ${\mathbf{E}}_{\mathbf{ref}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{ }\mathbf{V}$and that localid="1663752329236" ${\mathbf{E}}_{\mathbf{meas}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{480}\mathbf{V}$.

a. What is the uncertainty in the values of pH and [H⁺] if the uncertainty in the measured potential is ±1 mV (±0.001 V)?

b. To what accuracy must the potential be measured for the uncertainty in pH to be ±0.02 pH unit?

Short answer

(a) Uncertainty in pH and H⁺ are$\pm 0.017\text{and}\pm \text{0.06}\times {\text{10}}^{-4}$ .

(b) Potential must be measured to the accuracy of $\pm 0.005$.

Step-by-step solution

Calculating the uncertainty in values of pH

Determining the value of pH

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{meas}}& =& {\mathrm{E}}_{\mathrm{ref}}+0.05916\mathrm{pH}\\ 0.480\mathrm{V}& =& 0.250\mathrm{V}+0.05916\mathrm{pH}\\ \mathrm{pH}& =& 3.888\\ & & \end{array}$

$\begin{array}{rcl}& =& \max +\min \mathrm{Ph}\\ & =& \pm (3.905-3.883)\\ & =& \pm 0.017\end{array}$

Hence, the uncertainty in values of pH will be ±0.017

Determining the uncertainty in values of H+

Determining the value of H⁺,

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]={10}^{-\mathrm{pH}} \\ \left[{\mathrm{H}}^{+}\right]=1.29\times {10}^{-4} \\ \mathrm{Max}+\min \left[{\mathrm{H}}^{+}\right], \\ {\left[{\mathrm{H}}^{+}\right]}_{\max }=1.24\times {10}^{-4}\mathrm{m} \\ \left[{\mathrm{H}}^{+}\right]\mathrm{uncertainty}=\pm 0.06\times {10}^{-4}\mathrm{p} \end{gathered}$$

Hence, the uncertainty in values of [H⁺] will be $\pm 0.06\times {10}^{-4}\mathrm{p}$

Finding out the extent of accuracy necessary for uncertainty in pH

$\begin{array}{rcl}0.480\mathrm{V}& =& 0.250\mathrm{V}+0.05916\mathrm{pH}\\ \mathrm{pH}& =& 3.89\mathrm{V}\\ \mathrm{V}& =& \pm \frac{0.02}{3.89}\\ \mathrm{V}& =& \pm 0.005\end{array}$

Hence potential must be measured to the accuracy of ±0.005