Select the (a) best and (b) poorest electrical conductors from the following solutions, and explain the reason for your choices: \(0.10 \mathrm{M} \mathrm{NH}_{3} ; 0.10 \mathrm{M} \mathrm{NaCl} ; 0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid); \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol).

Short Answer

Expert verified
The best electrical conductor among the given solutions is 0.10 M \(NaCl\) and the poorest electrical conductor is 0.10 M \(CH_{3}CH_{2}OH\).

Step by step solution

01

Identify whether the chemicals in solution ionize

The first step is to figure out which of these chemicals ionize when they are put in solution. \(NH_{3}\) produces few ions, \(NaCl\) readily ionizes into \(Na^{+}\) and \(Cl^{-}\), \(CH_{3}COOH\) (acetic acid) partially ionizes to form \(CH_{3}COO^{-}\) and \(H^{+}\), and \(CH_{3}CH_{2}OH\) (Ethanol) does not ionize in solution.
02

Rank the solutions based on ion production

Once we've identified which chemicals ionize, and how readily they ionize, we can rank the solutions from most to least conductive. In order of how many ions they would produce, our ranking looks like this: \(NaCl\) > \(CH_{3}COOH\) > \(NH_{3}\) > \(CH_{3}CH_{2}OH\).
03

Identify the best and the poorest conductors

Based on our ranking in the previous step, we can now select the best and poorest conductors from the solutions: the best electrical conductor is 0.10 M \(NaCl\) and the poorest conductor is 0.10 M \(CH_{3}CH_{2}OH\). This is because \(NaCl\) completely ionizes in solution, producing the maximum number of ions possible for maximum conductivity, while ethanol does not ionize at all, producing a solution with virtually no ions to conduct electricity.

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Most popular questions from this chapter

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) $\mathrm{Ca}^{2+}+2 \mathrm{I}^{-}+2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \longrightarrow$ (b) $\mathrm{Ba}^{2+}+\mathrm{S}^{2-}+2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \longrightarrow$ (c) $2 \mathrm{K}^{+}+\mathrm{S}^{2-}+\mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow$

The following reactions do not occur in aqueous solutions. Balance their equations by the half-equation method, as suggested in Are You Wondering \(5-2\) (a) \(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) \longrightarrow\) \(\mathrm{CO}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{8}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g})+\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)\)

A \(110.520 \mathrm{g}\) sample of mineral water is analyzed for its magnesium content. The \(\mathrm{Mg}^{2+}\) in the sample is first precipitated as \(\mathrm{MgNH}_{4} \mathrm{PO}_{4},\) and this precipitate is then converted to \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7},\) which is found to weigh 0.0549 g. Express the quantity of magnesium in the sample in parts per million (that is, in grams of \(\mathrm{Mg}\) per million grams of \(\mathrm{H}_{2} \mathrm{O}\) ).

Assuming the volumes are additive, what is the \(\left[\mathrm{Cl}^{-}\right]\) in a solution obtained by mixing \(225 \mathrm{mL}\) of \(0.625 \mathrm{M}\) \(\mathrm{KCl}\) and \(615 \mathrm{mL}\) of \(0.385 \mathrm{M} \mathrm{MgCl}_{2} ?\)

Sodium hydroxide used to make standard \(\mathrm{NaOH}(\mathrm{aq})\) solutions for acid-base titrations is invariably contaminated with some sodium carbonate. (a) Explain why, except in the most precise work, the presence of this sodium carbonate generally does not seriously affect the results obtained, for example, when \(\mathrm{NaOH}(\mathrm{aq})\) is used to titrate HCl(aq). (b) Conversely, show that if \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) comprises more than \(1 \%\) to \(2 \%\) of the solute in NaOH(aq), the titration results are affected.

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