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Chapter 19: Problem 4

Classify solutions of the following compounds as strong or weak electrolytes: (a) \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{HNO}_{3} ;(\mathrm{c}) \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\) (d) \(\mathrm{K}\left[\mathrm{CH}_{3} \mathrm{CO}_{2}\right]\) (e) \(\mathrm{NH}_{4} \mathrm{Cl}\)

Short Answer

Expert verified
(a) Strong, (b) Strong, (c) Weak, (d) Strong, (e) Strong.

Step by step solution

01

Analyze the Nature of the Compound

Identify the nature of each compound to determine its ability to dissociate in water. Strong electrolytes are typically strong acids, strong bases, and soluble salts that dissociate completely in solution.
02

Evaluate Compound (a) AgNO_{3}

Silver nitrate (AgNO_{3}) is a soluble salt that dissociates completely into Ag^{+} and NO_{3}^{-} ions in solution, classifying it as a strong electrolyte.
03

Evaluate Compound (b) HNO_{3}

Nitric acid (HNO_{3}) is a strong acid that completely ionizes in solution. Therefore, it is classified as a strong electrolyte.
04

Evaluate Compound (c) CH_{3}CH_{2}CO_{2}H

Propionic acid (CH_{3}CH_{2}CO_{2}H) is a weak acid and only partially ionizes in solution, making it a weak electrolyte.
05

Evaluate Compound (d) K[CH_{3}CO_{2}]

Potassium acetate (K[CH_{3}CO_{2}]) is a salt that dissociates completely into potassium (K^{+}) and acetate ions (CH_{3}CO_{2}^{-}) in solution, classifying it as a strong electrolyte.
06

Evaluate Compound (e) NH_{4}Cl

Ammonium chloride (NH_{4}Cl) dissociates completely into ammonium (NH_{4}^{+}) and chloride (Cl^{-}) ions in solution, classifying it as a strong electrolyte.

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Key Concepts

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

Strong Electrolytes
Strong electrolytes are substances that dissolve in water to produce a solution that conducts electricity very well. This is because they dissociate completely into ions in solution. When a compound dissociates completely, it means that all of its molecules break apart into ions. These ions are charged particles, which are responsible for conducting electricity in the solution.

For example, in the case of
  • Silver nitrate (\[\mathrm{AgNO}_{3}\]): It dissolves in water, producing \[\mathrm{Ag}^{+}\] and \[\mathrm{NO}_{3}^{-}\] ions.
  • Nitric acid (\[\mathrm{HNO}_{3}\]): Being a strong acid, it ionizes completely, releasing hydrogen ions and nitrate ions.
  • Potassium acetate (\[\mathrm{K[CH}_{3}\mathrm{CO}_{2}]\]): Dissolves entirely into potassium and acetate ions.
  • Ammonium chloride (\[\mathrm{NH}_{4}\mathrm{Cl}\]): Forms ammonium and chloride ions upon dissolving.
These substances are efficient at conducting electricity in solutions due to their full ionization.
Weak Electrolytes
Weak electrolytes are substances that do not fully ionize in solution. When dissolved, only a small fraction of their molecules turn into ions. This incomplete ionization results in solutions that conduct electricity, but not strongly.

Let's consider the example of
  • Propionic acid (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CO}_{2}\mathrm{H}\]): When propionic acid is dissolved in water, only a few of its molecules donate hydrogen ions to form ions.
This partial dissociation makes propionic acid a weak electrolyte. In such a solution, ions coexist with non-ionized molecules, hence the ability to conduct electricity is weaker than in strong electrolytes.
Acid-Base Chemistry
Acid-base chemistry revolves around the interactions of acids and bases with water and with each other. Acids are substances that can donate protons (\[\mathrm{H}^{+}\] ions), while bases are substances that accept protons. The strength of an acid or base is determined by its ability to ionize or dissociate in water.

Strong acids, like
  • Nitric acid (\[\mathrm{HNO}_{3}\]): Completely ionizes in water, leading to a lot of protons available in the solution.
In contrast, weak acids, such as
  • Propionic acid (\[\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CO}_{2}\mathrm{H}\]): Only partially ionize, releasing fewer protons.
This distinction is crucial as acids and bases participate in various reactions, significantly affecting chemical reaction mechanisms and equilibrium states.
Ionic Dissociation
Ionic dissociation is the process through which ionic compounds separate into their respective ions when dissolved in a solvent such as water. This process is fundamental for the formation of electrolytic solutions that are capable of conducting electricity.

For any salt, the basic principle involves
  • Example with potassium acetate (\[\mathrm{K[CH}_{3}\mathrm{CO}_{2}]\]): It dissociates completely, resulting in potassium ions (\[\mathrm{K}^{+}\]) and acetate ions (\[\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}\]).
  • Example with ammonium chloride (\[\mathrm{NH}_{4}\mathrm{Cl}\]): Forms ammonium (\[\mathrm{NH}_{4}^{+}\]) and chloride (\[\mathrm{Cl}^{-}\]) ions in solution.
Through ionic dissociation, ionic compounds effectively split into ions that are solvated by water molecules, facilitating the flow of electric current.

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Most popular questions from this chapter

Write down expressions that define (a) resistivity, (b) conductivity and (c) conductance.

The ionic mobilities of \(\mathrm{K}^{+}\) and \(\left[\mathrm{MnO}_{4}\right]^{-}\) at \(298 \mathrm{K}\) are \(7.62 \times 10^{-8}\) and \(6.35 \times 10^{-8} \mathrm{m}^{2} \mathrm{s}^{-1} \mathrm{V}^{-1}\) Determine \(A_{\mathrm{m}}^{\infty}(298 \mathrm{K})\) for \(\mathrm{KMnO}_{4}\)

Explain why the molar conductivity at infinite dilution of a weak acid such as \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is high, but then decreases dramatically as the concentration is raised even slightly.

Which of the following compounds are sparingly soluble in water (a) \(\mathrm{NH}_{4} \mathrm{Br}\) (b) \(\mathrm{AgBr}\) (c) \(\mathrm{KOH}\) (d) \(\mathrm{BaSO}_{4}\) ? How does this affect conductivity measurements?

The ionic mobilities of \(\mathrm{Sr}^{2+}\) and \(\mathrm{Cl}^{-}\) at \(298 \mathrm{K}\) are \(6.16 \times 10^{-8}\) and \(7.91 \times 10^{-8} \mathrm{m}^{2} \mathrm{s}^{-1} \mathrm{V}^{-1}.\) Determine \(\Lambda_{\mathrm{m}}^{\infty}(298 \mathrm{K})\) for \(\mathrm{SrCl}_{2}.\)
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