The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{C}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text {ref }}\) 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 \(\mathscr{E}_{\text {ref }}=0.250 \mathrm{~V}\) and that \(\mathscr{C}_{\text {tme\pi }}=0.480 \mathrm{~V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}(\pm 0.001 \mathrm{~V})\) ? b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

Step by step solution

01

Write the given equation

The Nernst equation relating the measured potential with pH is given by: \[\mathscr{C}_{\text{meas}} = \mathscr{E}_{\text{ref}} + 0.05916 \cdot \text{pH}\]

02

Rearrange the equation to find pH

Let's rearrange the equation to find the expression for pH: \[\text{pH} = \frac{\mathscr{C}_{\text{meas}} - \mathscr{E}_{\text{ref}}}{0.05916}\]

03

Insert the given values and find pH

Assume that \(\mathscr{E}_{\text {ref}}=0.250 \text{ V}\) and \(\mathscr{C}_{\text {meas }}=0.480 \text{ V}\). Plug in these values to find the pH value: \[\text{pH} = \frac{0.480 - 0.250}{0.05916} \approx 3.89\]

04

Calculate the uncertainty in pH

The uncertainty in the measured potential is given, \(\pm 0.001 \text{ V}\). Use the same rearranged equation to find the uncertainty in pH: \[\Delta \text{pH} = \frac{\Delta \mathscr{C}_{\text{meas}}}{0.05916} = \frac{0.001}{0.05916} \approx \pm0.017\]

05

Calculate the uncertainty in H+ concentration

The relationship between pH and H+ concentration is: \[\text{pH} = -\log_{10}[\text{H}^{+}]\] Now, to find the uncertainty in H+ concentration, first find the H+ concentration without uncertainty: \[[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-3.89} \approx 1.29 \times 10^{-4}\text{ M}\] Then, find the upper and lower limits of H+ concentration using uncertainty in pH: \[[\text{H}^{+}]_{\text{min}} = 10^{-(3.89+0.017)} \approx 1.11 \times 10^{-4}\text{ M}\] \[[\text{H}^{+}]_{\text{max}} = 10^{-(3.89-0.017)} \approx 1.49 \times 10^{-4}\text{ M}\] So, the uncertainty in H+ concentration is between \(1.11 \times 10^{-4}\text{ M}\) and \(1.49 \times 10^{-4}\text{ M}\). #b. Precision required in potential for the uncertainty in pH to be ±0.02 pH unit#

06

Use the relationship between potential uncertainty and pH uncertainty

To find the required precision in the potential measurement to achieve the desired uncertainty in pH, use the relationship obtained in step 4: \[\Delta \text{pH} = \frac{\Delta \mathscr{C}_{\text{meas}}}{0.05916}\] Let the required uncertainty in the potential be \(\Delta \mathscr{C}_{\text{req}}\).

07

Solve for the required potential uncertainty

Now, we need to find the potential uncertainty to maintain the pH uncertainty as \(\pm0.02\). \[\Delta \text{pH} = 0.02 = \frac{\Delta \mathscr{C}_{\text{req}}}{0.05916}\] Now, solve for \(\Delta \mathscr{C}_{\text{req}}\): \[\Delta \mathscr{C}_{\text{req}} = 0.05916 \times 0.02 \approx 0.00118 \text{ V}\] So, the potential must be measured with a precision of approximately \(\pm 0.00118\) V for the uncertainty in pH to be maintained at \(\pm 0.02\) pH unit.

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