The measurement of \(\mathrm{pH}\) 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 $$\mathscr{E}_{\text { meas }}=\mathscr{E}_{\text { ref }}+0.05916 \mathrm{pH}$$ 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 \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that $\mathscr{E}_{\text { meas }}=0.480 \mathrm{V}$ a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the nncertainty in the measured potential is \(+1 \mathrm{mV}\) \(( \pm 0.001 \mathrm{V}) ?\) 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

Find pH value using the given equation

First, we need to find the value of pH using given values of \(\mathscr{E}_{meas}\) and \(\mathscr{E}_{ref}\). Using the given equation, we can find the pH value as follows: \[ \mathrm{pH} = \frac{\mathscr{E}_{meas} - \mathscr{E}_{ref}}{0.05916} \]

02

Calculate the pH using given values

Now, let's plug the given values of \(\mathscr{E}_{meas}\) and \(\mathscr{E}_{ref}\) into the equation: \[ \mathrm{pH} = \frac{0.480 V - 0.250 V}{0.05916} \approx 3.89 \]

03

Calculate the uncertainty in pH

The uncertainty in pH can be calculated by taking the derivative of the equation for pH with respect to the measured potential and multiplying it by the uncertainty in the measured potential: \[ \Delta \mathrm{pH} = \frac{d(\mathrm{pH})}{d(\mathscr{E}_{meas})} \times \Delta \mathscr{E}_{meas} \] Since \(\frac{d(\mathrm{pH})}{d(\mathscr{E}_{meas})} = \frac{1}{0.05916}\), we can plug in the given uncertainty of the measured potential: \[ \Delta \mathrm{pH} = \frac{1}{0.05916} \times 0.001 V \approx 0.02 \]

04

Calculate the uncertainty in H+ concentration

To find the uncertainty in H+ concentration, first, use the relationship between pH and H+ concentration: \[ [\mathrm{H}^{+}] = 10^{-\mathrm{pH}} \] Now, we can find the uncertainty in H+ concentration by taking the derivative of this equation with respect to pH and multiplying it by the uncertainty in pH: \[ \Delta [\mathrm{H}^{+}] = \frac{d([\mathrm{H}^{+}])}{d(\mathrm{pH})} \times \Delta \mathrm{pH} \] Since \(\frac{d([\mathrm{H}^{+}])}{d(\mathrm{pH})} = -10^{-\mathrm{pH}} \mathrm{ln}(10)\), we can plug in the calculated pH and uncertainty in pH: \[ \Delta [\mathrm{H}^{+}] = -10^{-3.89} \times 2.3026 \times 0.02 \approx 1.50 \times 10^{-5} \mathrm{M} \]

05

Find precision required for given uncertainty in pH (part b)

Now in part b, we need to determine the required precision for an uncertainty of \(±0.02\) pH unit. Using the relationship we found in step 3: \[ \Delta \mathrm{pH} = \frac{1}{0.05916} \times \Delta \mathscr{E}_{meas} \] Isolate the uncertainty in the measured potential and plug in the given uncertainty in pH: \[ \Delta \mathscr{E}_{meas} = 0.05916 \times 0.02 \approx 0.00118 V \] #Conclusion#: For part a), the uncertainty in pH is approximately \(±0.02\) and the uncertainty in H+ concentration is approximately \(±1.50 \times 10^{-5} \mathrm{M}\). For part b), the potential must be measured with a precision of approximately \(±0.00118 V\) for the uncertainty in pH to be \(±0.02\) pH unit.

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