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Is it possible for a molecular substance to be a strong electrolyte? Explain.

Short Answer

Expert verified
Yes, if it's a strong acid, it can be a strong electrolyte by ionizing fully in solution.

Step by step solution

01

Understanding Electrolytes

Electrolytes are substances that conduct electricity when dissolved in water. They dissociate into ions, which are charged particles. Strong electrolytes completely ionize in solution, allowing the solution to conduct electricity effectively.
02

Types of Molecular Substances

Molecular substances are composed of molecules held together by covalent bonds. Unlike ionic compounds, they do not typically form ions when dissolved in solution.
03

Ionization of Acids

While most molecular substances do not conduct electricity, certain molecular acids can ionize in water. For example, strong acids such as hydrochloric acid (HCl) dissociate completely into hydrogen ions (H⁺) and chloride ions (Cl⁻), making them strong electrolytes.
04

Evaluating Possibility

Thus, while molecular substances are not typically considered strong electrolytes, a molecular substance like a strong acid can indeed ionize completely and conduct electricity 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 compounds that completely ionize or dissociate in water. When they dissolve, they separate into ions. These ions are electrically charged, allowing the solution to conduct electricity efficiently. This complete ionization distinguishes strong electrolytes from weak electrolytes, which only partially dissociate in water.
  • Examples of strong electrolytes include strong acids, strong bases, and most salts.
  • Hydrochloric acid (HCl), sodium hydroxide (NaOH), and potassium chloride (KCl) are common strong electrolytes.
Strong electrolytes are important in many chemical processes, as their ability to conduct electricity is crucial in reactions like electrolysis. They play a vital role in maintaining the electrical balance in biological systems, such as in nerves and muscles by affecting the movement of ions.
Ionization of Acids
The ionization of acids refers to the process in which an acid releases hydrogen ions ( \( H^+ \) ) when dissolved in water. This process is critical in determining whether an acid acts as a strong or weak electrolyte.When a strong acid, such as hydrochloric acid (HCl), ionizes, it dissociates completely:\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
  • This complete ionization is what makes HCl a strong electrolyte, as the presence of a large number of ions in the solution enhances conductivity.
  • Strong acids conduct electricity well because they produce a high concentration of ionized particles.
Other examples of strong acids that ionize completely include sulfuric acid ( \( H_2SO_4 \) ) and nitric acid ( \( HNO_3 \) ). In contrast, weak acids, like acetic acid ( \( CH_3COOH \) ), only partially ionize, resulting in fewer ions and limited conductivity.
Molecular Substances
Molecular substances consist of molecules formed by atoms sharing electrons through covalent bonds. These substances do not typically dissociate into ions when dissolved in solution, which affects their ability to conduct electricity.
  • Molecular substances include sugar ( \( C_{12}H_{22}O_{11} \) ) and water ( \( H_2O \) ).
  • Because they do not form ions, most molecular substances are non-electrolytes, meaning they do not conduct electricity in solution.
However, there are exceptions to this rule. Some molecular substances, such as strong acids, can ionize in water to become strong electrolytes. This unique characteristic allows certain molecular substances to break the typical mold.
Electricity Conduction in Solutions
Electricity conduction in solutions is primarily driven by the presence of ions, which are charged entities that facilitate the flow of electrical current. When a substance dissolves in water, its ability to conduct electricity is determined by how well it forms ions in the solution.
  • Strong electrolytes, which ionize completely, create a high concentration of ions and thus conduct electricity well.
  • Weak electrolytes produce fewer ions, resulting in poorer conductivity compared to strong electrolytes.
In the case of non-electrolytes, such as the majority of molecular substances, there is little to no ion formation, leading to minimal or no electricity conduction. Understanding this behavior is crucial for applications in chemistry and biology, as it influences reactions and processes in both areas.

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Most popular questions from this chapter

Which of the following solutions will not form a precipitate when added to \(0.10 \mathrm{M} \mathrm{BaCl}_{2} ?\) (a) \(0.10 \mathrm{MLiNO}_{3}\) (b) \(0.10 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\) (c) \(0.10 \mathrm{M} \mathrm{AgNO}_{3}\)

Individual solutions of \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) both conduct electricity, but the conductivity disappears when equal molar amounts of the solutions are mixed. Explain.

Brass is an approximately \(4: 1\) alloy of copper and zinc, along with small amounts of tin, lead, and iron. The mass percents of copper and zinc can be determined by a procedure that begins with dissolving the brass in hot nitric acid. The resulting solution of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Zn}^{2+}\) ions is then treated with aqueous ammonia to lower its acidity, followed by addition of sodium thiocyanate (NaSCN) and sulfurous acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)\) to precipitate copper(I) thiocyanate (CuSCN). The solid CuSCN is collected, dissolved in aqueous acid, and treated with potassium iodate \(\left(\mathrm{KIO}_{3}\right)\) to give iodine, which is then titrated with aqueous sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right)\). The filtrate remaining after CuSCN has been removed is neutralized by addition of aqueous ammonia, and a solution of diammonium hydrogen phosphate \(\left(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\right)\) is added to yield a precipitate of zinc ammonium phosphate \(\left(\mathrm{ZnNH}_{4} \mathrm{PO}_{4}\right)\). Heating the precipitate to \(900^{\circ} \mathrm{C}\) converts it to zinc pyrophosphate \(\left(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\right)\), which is weighed. The equations are (1) \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) (in acid) (2) \(\mathrm{Cu}^{2+}(a q)+\mathrm{SCN}^{-}(a q)+\mathrm{HSO}_{3}^{-}(a q) \longrightarrow\) \(\mathrm{CuSCN}(s)+\mathrm{HSO}_{4}^{-}(a q)\) (in acid) (3) \(\mathrm{Cu}^{+}(a q)+\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{I}_{2}(a q)\) (in acid) (4) \(\mathrm{I}_{2}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}(a q)\) (in acid) (5) \(\mathrm{ZnNH}_{4} \mathrm{PO}_{4}(s) \longrightarrow \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{NH}_{3}(g)\) (a) Balance all equations. (b) When a brass sample with a mass of \(0.544 \mathrm{~g}\) was subjected to the preceding analysis, \(10.82 \mathrm{~mL}\) of \(0.1220 \mathrm{M}\) sodium thiosulfate was required for the reaction with iodine. What is the mass percent copper in the brass? (c) The brass sample in part (b) yielded \(0.246 \mathrm{~g}\) of \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\). What is the mass percent zinc in the brass?

Assign oxidation numbers to each element in the following compounds: (a) \(\mathrm{NO}_{2}\) (b) \(\mathrm{SO}_{3}\) (c) \(\mathrm{COCl}_{2}\) (d) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (e) \(\mathrm{KClO}_{3}\) (f) \(\mathrm{HNO}_{3}\)

Classify each of the following reactions as a precipitation, acidbase neutralization, or oxidation-reduction: (a) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{HgI}_{2}(s)\) (b) \(2 \mathrm{HgO}(s) \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{KOH}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)\)

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