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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 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?

Short Answer

Expert verified
The uncertainty in pH is calculated to be \(\Delta pH = 0.015\), and the uncertainty in the hydrogen ion concentration is \(\Delta [H+] = 5.4\times10^{-6} \,\mathrm{mol/L}\). To maintain an uncertainty of \(\pm 0.02\) pH unit, the required precision for the measured potential is approximately \(\pm 1.2\) mV.

Step by step solution

01

Identify the given values

We know the following values: 1. Reference potential, \(E_{ref} = 0.250\) V 2. Measured potential, \(E_{meas} = 0.480\) V 3. Uncertainty in the measured potential, \(\pm0.001\) V 4. The pH meter equation at 25°C: \(E_{meas} = E_{ref} + 0.05916 \cdot \mathrm{pH}\)
02

Calculate pH value

We can rearrange the pH meter equation to find the value of pH: \(pH = \frac{E_{meas} - E_{ref}}{0.05916}\) Substituting the given values: \(pH = \frac{0.480 - 0.250}{0.05916} = 3.887\)
03

Calculate uncertainty in pH

Uncertainty in \(E_{meas}\) is given as \(\pm0.001\) V. To find the uncertainty in pH the value, we utilize the same equation and calculate the pH for the minimum and maximum values of \(E_{meas}\), and then find the difference. Minimum \(E_{meas} = 0.480 - 0.001 = 0.479\) V Maximum \(E_{meas} = 0.480 + 0.001 = 0.481\) V Minimum pH value: \(pH_{min} = \frac{0.479 - 0.250}{0.05916} = 3.872\) Maximum pH value: \(pH_{max} = \frac{0.481 - 0.250}{0.05916} = 3.902\) Uncertainty in pH: \(\Delta pH = \frac{pH_{max} - pH_{min}}{2} = \frac{3.902 - 3.872}{2} = 0.015\)
04

Calculate uncertainty in [H+]

With the uncertainty in pH determined, we can calculate the uncertainty in the hydrogen ion concentration, [H+]. The relationship between pH and [H+] can be written as: \(pH = -\log{[H+]}\) So, we will find the concentration values for both minimum and maximum pH values: Minimum [H+]: \([H+]_{min} = 10^{-pH_{min}} = 10^{-3.872} = 1.345\times10^{-4} \,\mathrm{mol/L}\) Maximum [H+]: \([H+]_{max} = 10^{-pH_{max}} = 10^{-3.902} = 1.237\times10^{-4} \,\mathrm{mol/L}\) Uncertainty in [H+]: \(\Delta [H+] = \frac{[H+]_{max} - [H+]_{min}}{2} = \frac{1.345\times10^{-4} - 1.237\times10^{-4}}{2} = 5.4\times10^{-6} \,\mathrm{mol/L}\)
05

Find required precision for given pH uncertainty

We have to find the precision in potential measurement for an uncertainty of \(\pm 0.02\) in pH value. Using the same pH meter equation, let's denote the required potential difference as \(\Delta E_{calc}\). \(\Delta E_{calc} = 0.02\cdot 0.05916 = 0.0011832\) V The required precision for the measured potential that will result in a \(\pm 0.02\) pH unit uncertainty is \(\pm 0.0011832\) V (approximately \(\pm 1.2\) mV).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst equation
The Nernst equation is crucial in understanding how pH meters work. It links the measurable electrical potential to the hydrogen ion concentration in a solution. The equation is used to calculate the cell potential under non-standard conditions. The general form of the Nernst equation is given by: \[ E = E^0 - \left( \frac{RT}{nF} \right) \ln{Q} \] Where:
  • \( E \) is the cell potential at temperature \( T \)
  • \( E^0 \) is the standard cell potential
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
  • \( n \) is the number of moles of electrons exchanged
  • \( F \) is the Faraday constant
  • \( Q \) is the reaction quotient
At 25°C and focusing on pH measurements, the Nernst equation simplifies to the linear equation used in pH meters, showing a direct relationship between potential and pH: \[ E_{meas} = E_{ref} + 0.05916 \cdot \text{pH} \] This equation demonstrates how a change in the hydrogen ion concentration directly affects the electrical signal read by a pH meter.
glass electrode
The glass electrode is a pivotal component in pH measurement. It is selectively responsive to hydrogen ions, making it ideal for pH meters. The glass bulb at the tip of the electrode allows hydrogen ions to interact with the surface, causing an electrical potential difference. This difference is measured relative to a stable reference electrode. The glass electrode features several layers:
  • The outer hydrated gel layer where ion exchange occurs
  • The inner glass structure providing mechanical support
  • Filled with an electrolyte solution to complete the circuit
The interaction between hydrogen ions and the glass creates a potential that follows the Nernst equation, translating the ion concentration directly into a measurable electric signal. A well-functioning glass electrode is crucial for accurate pH measurement.
hydrogen ion concentration
Hydrogen ion concentration is fundamental to understanding solutions' acidity or basicity. The pH scale is logarithmic, relating to the concentration of hydrogen ions \([H^+]\) in a solution. pH is defined as: \[ \text{pH} = -\log_{10}[H^+] \] This means a lower pH corresponds to a higher hydrogen ion concentration, indicating an acidic solution. Conversely, a higher pH indicates lower \([H^+]\), suggesting a basic solution. Converting pH back to concentration can be done with: \[ [H^+] = 10^{-\text{pH}} \] In experiments, understanding the relationship between pH and \([H^+]\) is crucial, especially when certain pH readings imply significant changes in \([H^+]\). Accurate measurement of \([H^+]\) helps in many chemical, environmental, and biological studies.
measurement uncertainty
Measurement uncertainty is an inherent part of scientific experiments, including those involving pH measurements. It signifies the range within which the true value likely falls. For pH measurements, uncertainties can arise due to variations in potential measurement or electrode performance. When measuring pH, it’s important to consider factors that influence uncertainty:
  • Sensitivity of the pH meter
  • Variations in glass electrode response
  • Reference electrode stability
  • Temperature fluctuations
With a potential uncertainty of \(\pm 0.001\) V, the corresponding pH uncertainty can be estimated using the derivative of the pH equation with respect to the potential. For precision, instruments need to minimize potential measurement variance, ensuring better estimation of pH values.

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Most popular questions from this chapter

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12} .\) Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$

Which of the following statement(s) is/are true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(A l^{3+}+3 e^{-} \longrightarrow A l\) and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg},\) aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

It takes \(15 \mathrm{kWh}\) (kilowatt-hours) of electrical energy to produce \(1.0 \mathrm{kg}\) aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt \(1.0 \mathrm{kg}\) aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is \(10.7 \mathrm{kJ} / \mathrm{mol} \text { ( } 1 \text { watt }=1 \mathrm{J} / \mathrm{s} \text { ). }]\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)

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