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Chapter 20: Problem 86

You prepare \(1.00 \mathrm{L}\) of a buffer solution that is \(1.00 \mathrm{M}\) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) and \(1.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} .\) The solution is divided in half between the two compartments of an electrolytic cell. Both electrodes used are Pt. Assume that the only electrolysis is that of water. If 1.25 A of current is passed for 212 min, what will be the \(\mathrm{pH}\) in each cell compartment at the end of the electrolysis?

Short Answer

Expert verified
The pH in both the cathode and anode compartments will be 7.21 at the end of the electrolysis.

Step by step solution

01

Calculate the Total Charge

The total charge Q is obtained by the formula Q = I x t, where I = 1.25 A is current and t = 212 min x 60 s/min = 12720 s is time. Substituting those values into the formula gives Q = 1.25 A x 12720 s = 15900 C.
02

Calculate the Moles of Electrons

We know that each mole of electrons carries a charge of 1 Faraday which is approximately 96500 C/mol. Therefore, the number of moles of electrons transferred can be calculated by dividing the total charge by the charge of one mole of electrons. So, moles of electrons = 15900 C / 96500 C/mol = 0.1648 mol.
03

Calculate the Moles of Hydrogen and Hydroxide Ions

Under these conditions, each mole of electrons reduces one mole of water molecules, producing two moles of hydrogen ions \(H^{+}\) at the cathode and an equivalent amount of hydroxide ions \(OH^{-}\) at the anode. Therefore, we have produced 0.1648 mol of \(H^{+}\) and \(OH^{-}\) ions each.
04

Adjust the Buffer Concentrations

As the \(OH^{-}\) ions are produced, they react with the \(NaH_{2}PO_{4}\) to form \(Na_{2}HPO_{4}\) and water in the one compartment. Similarly, as \(H^{+}\) ions are produced, they react with the \(Na_{2}HPO_{4}\) to form \(NaH_{2}PO_{4}\) and water in the other compartment. Since we originally had 1 M concentration of each buffer in 0.5 L, we had 0.5 mol of each buffer. After electrolysis, the buffer concentrations adjust making \(Na_{2}HPO_{4}\) into 0.5 mol - 0.1648 mol = 0.3352 mol and \(NaH_{2}PO_{4}\) into 0.5 mol - 0.1648 mol = 0.3352 mol respectively, in the corresponding compartments.
05

Calculate the pH values

Given \(pKa_{2}\) for \(H_{2}PO_{4^{-}}/HPO_{4^{2-}}\) is 7.21, we can now calculate the pH in each compartment using the Henderson-Hasselbalch equation. For the first compartment: pH = \(pKa_{2}\) + log([\(HPO_{4^{2-}}\)] / [\(H_{2}PO_{4^{-}}\)]) = 7.21 + log(0.3352 / 0.3352) = 7.21. For the second compartment: pH = \(pKa_{2}\) + log([\(HPO_{4^{2-}}\)] / [\(H_{2}PO_{4^{-}}\)]) = 7.21 + log(0.3352 / 0.3352) = 7.21. So, the pH of both compartments remains the same.

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Most popular questions from this chapter

An aqueous solution of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is electrolyzed by means of Pt electrodes. (a) Which of the following gases should form at the anode: \(\mathrm{O}_{2}, \mathrm{H}_{2}, \mathrm{SO}_{2}, \mathrm{SO}_{3} ?\) Explain. (b) What product should form at the cathode? Explain. (c) What is the minimum voltage required? Why is the actual voltage needed likely to be higher than this value?

Only a tiny fraction of the diffusible ions move across a cell membrane in establishing a Nernst potential (see Focus On 20: Membrane Potentials), so there is no detectable concentration change. Consider a typical cell with a volume of \(10^{-8} \mathrm{cm}^{3},\) a surface area \((A)\) of \(10^{-6} \mathrm{cm}^{2},\) and a membrane thickness \((l)\) of \(10^{-6} \mathrm{cm}\) Suppose that \(\left[\mathrm{K}^{+}\right]=155 \mathrm{mM}\) inside the cell and \(\left[\mathrm{K}^{+}\right]=4 \mathrm{mM}\) outside the cell and that the observed Nernst potential across the cell wall is \(0.085 \mathrm{V}\). The membrane acts as a charge-storing device called a capacitor, with a capacitance, \(C,\) given by $$C=\frac{\varepsilon_{0} \varepsilon A}{l}$$ where \(\varepsilon_{0}\) is the dielectric constant of a vacuum and the product \(\varepsilon_{0} \varepsilon\) is the dielectric constant of the membrane, having a typical value of \(3 \times 8.854 \times 10^{-12}\) \(\mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\) for a biological membrane. The SI unit of capacitance is the firad, \(1 \mathrm{F}=1\) coulomb per volt \(=1 \mathrm{CV}^{-1}=1 \times \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-1}\) (a) Determine the capacitance of the membrane for the typical cell described. (b) What is the net charge required to maintain the observed membrane potential? (c) How many \(\mathrm{K}^{+}\) ions must flow through the cell membrane to produce the membrane potential? (d) How many \(\mathrm{K}^{+}\) ions are in the typical cell? (e) Show that the fraction of the intracellular \(K^{+}\) ions transferred through the cell membrane to produce the membrane potential is so small that it does not change \(\left[\mathrm{K}^{+}\right]\) within the cell.

\(\mathrm{Ni}^{2+}\) has a more positive reduction potential than \(\mathrm{Cd}^{2+}\) (a) Which ion is more easily reduced to the metal? (b) Which metal, Ni or Cd, is more easily oxidized?

In each of the following examples, sketch a voltaic cell that uses the given reaction. Label the anode and cathode; indicate the direction of electron flow; write a balanced equation for the cell reaction; and calculate \(E_{\mathrm{cell}}^{\circ}\). (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{Pb}^{2+}(\mathrm{aq})\) is displaced from solution by \(\mathrm{Al}(\mathrm{s})\) (c) \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\) \(\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g})\)

It is sometimes possible to separate two metal ions through electrolysis. One ion is reduced to the free metal at the cathode, and the other remains in solution. In which of these cases would you expect complete or nearly complete separation: (a) \(\mathrm{Cu}^{2+}\) and \(\mathrm{K}^{+} ;\) (b) \(\mathrm{Cu}^{2+}\) and \(\mathrm{Ag}^{+} ;\) (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Sn}^{2+} ?\) Explain.
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