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Chapter 5: Problem 4

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?

Short Answer

Expert verified
The weak electrolytes \(NH_3\) and \(CH_3COOH\) react together in an acid-base reaction to form ions, \(NH_4^+\) and \(CH3COO^-\), which increase the ion concentration in the solution. This leads to the solution becoming a good conductor of electricity.

Step by step solution

01

Understanding Weak Electrolytes

Weak electrolytes are substances that only partially ionize in their aqueous solutions, producing only a small number of ions. \(NH_3(aq)\) and \(CH_3COOH(aq)\) are both examples of weak electrolytes. When these are separate solutions, they generate few ions and conduct electricity only weakly.
02

Acid-Base Reactions

When these two solutions are mixed, an acid-base reaction occurs. \(NH_3\) is a weak base and \(CH_3COOH\) is a weak acid. They react to form ionized entities, that is, \(NH_4^+\) ions and \(CH3COO^-\) ions, which contribute to increased ion concentration in the solution.
03

Increased Conductivity

The increased number of ions in the solution leads to an increased ability to conduct electricity. This is the reason why, when these two weak electrolytes are mixed, the resulting solution becomes a good conductor of electricity.

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Most popular questions from this chapter

Balance these equations for reactions in acidic solution. (a) $\mathrm{IBr}+\mathrm{BrO}_{3}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{IO}_{3}^{-}+\mathrm{Br}^{-}+\mathrm{H}_{2} \mathrm{O}$ (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{3}+\mathrm{Sn} \longrightarrow$ $\mathrm{NH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Sn}^{2+}$ (c) $\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{S}+\mathrm{NO}$ (d) $\mathrm{H}_{5} \mathrm{IO}_{6}+\mathrm{I}_{2} \longrightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}$ (e) $\mathrm{S}_{2} \mathrm{F}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{S}_{8}+\mathrm{H}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+\mathrm{HF}$

A \(\mathrm{KMnO}_{4}(\) aq) solution is to be standardized by titration against \(\mathrm{As}_{2} \mathrm{O}_{3}(\mathrm{s}) .\) A \(0.1078 \mathrm{g}\) sample of \(\mathrm{As}_{2} \mathrm{O}_{3}\) requires \(22.15 \mathrm{mL}\) of the \(\mathrm{KMnO}_{4}(\) aq) for its titration. What is the molarity of the \(\mathrm{KMnO}_{4}(\) aq)? \(5 \mathrm{As}_{2} \mathrm{O}_{3}+4 \mathrm{MnO}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O}+12 \mathrm{H}^{+} \longrightarrow\) \(10 \mathrm{H}_{3} \mathrm{AsO}_{4}+4 \mathrm{Mn}^{2+}\)

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)

A \(25.00 \mathrm{mL}\) sample of \(0.132 \mathrm{M}\) \(\mathrm{HNO}_{3}\) is mixed with \(10.00 \mathrm{mL}\) of \(0.318 \mathrm{M} \mathrm{KOH} .\) Is the resulting solution acidic, basic, or exactly neutralized?

Balance these equations for redox reactions in basic solution. (a) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{ClO}_{3}^{-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{Cl}^{-}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{OCl}^{-} \longrightarrow \mathrm{FeO}_{4}^{2-}+\mathrm{Cl}^{-}\) (c) \(\mathrm{ClO}_{2} \longrightarrow \mathrm{ClO}_{3}^{-}+\mathrm{Cl}\) (d) \(\mathrm{Ag}(\mathrm{s})+\mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\)
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