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Chapter 17: Problem 147

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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Most popular questions from this chapter

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-} .\) You read the voltage to be 0.76 \(\mathrm{V}\) a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}\) $$\mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?$$

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{lr} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & 1.50 \\ \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg} & -2.37 \end{array}$$ a. What is the standard potential for this cell? b. A nonstandard cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Mg}^{2+}\right]=\) \(1.00 \times 10^{-5} M .\) The cell potential is observed to be 4.01 V. Calculate \(\left[\mathrm{Au}^{3+}\right]\) in this cell. \mathscr{E}^{\circ}

Consider the following half-reactions: $$\begin{array}{ll} \operatorname{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \operatorname{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 \(M\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

It took \(150 .\) s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+.\)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 \mathrm{M} .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\) $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$
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