Problem 127 - General Chemistry: Principles and Modern Applications

Question

Write a plausible chemical equation to represent the reaction of \((\mathrm{a}) \mathrm{Cl}_{2}(\mathrm{g}) \quad\) with cold \(\quad \mathrm{NaOH}(\mathrm{aq})\) (b) \(\mathrm{NaI}(\mathrm{s})\) with hot \(\mathrm{H}_{2} \mathrm{SO}_{4}(\text { concd aq }) ;\) (c) \(\mathrm{Cl}_{2}(\mathrm{g})\) with \(\mathrm{KI}_{3}(\mathrm{aq}) ; \quad\) (d) \(\quad \mathrm{NaBr}(\mathrm{s}) \quad\) with hot \(\mathrm{H}_{3} \mathrm{PO}_{4}\) \((\text { concd aq })\) (e) \(\mathrm{NaHSO}_{3}(\mathrm{aq})\) with \(\mathrm{MnO}_{4}^{-1}(\mathrm{aq})\) in dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The equations representing the reactions are:\n1) \( \mathrm{Cl}_{2} + 2 \mathrm{NaOH} \to \mathrm{NaClO} + \mathrm{NaCl} + \mathrm{H}_{2} \mathrm{O} \); \n\n2) \( \mathrm{NaI} + \mathrm{H}_{2} \mathrm{SO}_{4} \to \mathrm{HI} + \mathrm{NaHSO}_{4} \); \n\n3) \( \mathrm{Cl}_{2} + 2 \mathrm{KI}_{3} \to 2 \mathrm{I}_{2} + 2 \mathrm{KCl} \); \n\n4) \( \mathrm{NaBr} + \mathrm{H}_{3} \mathrm{PO}_{4} \to \mathrm{HBr} + \mathrm{NaH}_{2} \mathrm{PO}_{4} \); \n\n5) \( 2 \mathrm{MnO}_{4}^{-1} + 5 \mathrm{NaHSO}_{3} + 3 \mathrm{H}_{2} \mathrm{SO}_{4} \to 2 \mathrm{MnSO}_4 + 5 \mathrm{NaHSO}_4 + 3 \mathrm{H}_2 \mathrm{O} \)

Step 1: Reaction of \( \mathrm{Cl}_{2}(\mathrm{g}) \) with \( \mathrm{NaOH}(\mathrm{aq}) \)

Chlorine reacts with cold aqueous sodium hydroxide to produce bleach or sodium hypochlorite and sodium chloride. The chemical equation is: \[ \mathrm{Cl}_{2} + 2 \mathrm{NaOH} \to \mathrm{NaClO} + \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O} \]

Step 2: Reaction of \( \mathrm{NaI}(\mathrm{s}) \) with hot \( \mathrm{H}_{2}\mathrm{SO}_{4}(\mathrm{aq}) \)

Sodium iodide reacts with hot concentrated sulfuric acid to produce hydrogen iodide and sodium hydrogen sulphate. The chemical equation is: \[ \mathrm{NaI} + \mathrm{H}_{2}\mathrm{SO}_{4} \to \mathrm{HI} + \mathrm{NaHSO}_{4} \]

Step 3: Reaction of \( \mathrm{Cl}_{2}(\mathrm{g}) \) with \( \mathrm{KI}_{3}(\mathrm{aq}) \)

Chlorine gas reacts with potassium iodide to produce iodine and potassium chloride. The chemical equation is: \[ \mathrm{Cl}_{2} + 2 \mathrm{KI}_{3} \to 2 \mathrm{I}_{2} + 2\mathrm{KCl} \]

Step 4: Reaction of \( \mathrm{NaBr}(\mathrm{s}) \) with hot \( \mathrm{H}_{3}\mathrm{PO}_{4}(\mathrm{aq}) \)

Sodium bromide reacts with hot concentrated phosphoric acid to produce hydrogen bromide and sodium hydrogen phosphate. The chemical equation is: \[ \mathrm{NaBr} + \mathrm{H}_{3}\mathrm{PO}_{4} \to \mathrm{HBr} + \mathrm{NaH}_{2}\mathrm{PO}_{4} \]

Step 5: Reaction of \( \mathrm{NaHSO}_{3}(\mathrm{aq}) \) with \( \mathrm{MnO}_{4}^{-1}(\mathrm{aq}) \) in dilute \( \mathrm{H}_{2}\mathrm{SO}_{4}(\mathrm{aq}) \)

Sodium hydrogen sulphite reacts with permanganate ion in dilute sulfuric acid to produce manganese sulphate, sodium sulphate and water. The chemical equation is: \[ 2 \mathrm{MnO}_{4}^{-1} + 5 \mathrm{NaHSO}_{3} + 3 \mathrm{H}_{2}\mathrm{SO}_{4} \to 2 \mathrm{MnSO}_{4} + 5 \mathrm{NaHSO}_{4} + 3 \mathrm{H}_{2} \mathrm{O} \]

Key Concepts

Chlorine and sodium hydroxide reaction

When chlorine gas (\( \mathrm{Cl}_{2} \)) meets cold, aqueous sodium hydroxide (\( \mathrm{NaOH} \)), a fascinating chemical reaction takes place. This reaction is often used in the production of common household bleach, and here's why. Chlorine acts as a disinfectant due to its ability to break down cell walls and proteins. When it reacts with sodium hydroxide, the products formed include sodium hypochlorite (\( \mathrm{NaClO} \)), sodium chloride (\( \mathrm{NaCl} \)), and water (\( \mathrm{H}_{2}\mathrm{O} \)).

The balanced chemical equation for this reaction is:

• \[ \mathrm{Cl}_{2} + 2 \mathrm{NaOH} \to \mathrm{NaClO} + \mathrm{NaCl} + \mathrm{H}_{2}\mathrm{O} \]

Understanding this reaction is crucial for industries that produce disinfectants and bleaches. Sodium hypochlorite is the active ingredient in bleach, making it invaluable in sanitation and cleaning processes. The formation of sodium chloride, or table salt, is also a noteworthy byproduct.

This reaction embodies the transformation of toxic chlorine gas into useful and safer products, showcasing the importance of chemical reactions in everyday life.

Sodium iodide and sulfuric acid reaction

Let's delve into the reaction between sodium iodide (\( \mathrm{NaI} \)) and concentrated sulfuric acid (\( \mathrm{H}_{2}\mathrm{SO}_{4} \)). This interaction is an excellent example of how powerful acids react with salts to form new compounds. Here, sodium iodide, when reacted with hot, concentrated sulfuric acid, yields hydrogen iodide (\( \mathrm{HI} \)) and sodium hydrogen sulfate (\( \mathrm{NaHSO}_{4} \)).

The chemical equation representing this process is:

• \[ \mathrm{NaI} + \mathrm{H}_{2}\mathrm{SO}_{4} \to \mathrm{HI} + \mathrm{NaHSO}_{4} \]

This reaction highlights an important property of sulfuric acid — its ability to act as a strong dehydrating agent. Upon heating, it not only produces a gaseous product (\( \mathrm{HI} \)), which can be collected, but also stays in solution as sodium hydrogen sulfate.

Hydrogen iodide, when dissolved in water, forms hydroiodic acid, which is used in both organic synthesis and as a reducing agent. The versatility of products formed from this reaction allows for numerous applications, demonstrating the transformation potential inherent in chemical reactions.

Potassium iodide and chlorine reaction

When potassium iodide (\( \mathrm{KI} \)) meets chlorine gas (\( \mathrm{Cl}_{2} \)), a vivid reaction occurs, characterized by the distinctive change in color. This reaction is intriguing because it epitomizes a simple displacement reaction where a more-reactive element displaces a less-reactive one.

The chemical equation for this reaction is:

• \[ \mathrm{Cl}_{2} + 2 \mathrm{KI} \to 2 \mathrm{I}_{2} + 2 \mathrm{KCl} \]

Chlorine, being more electronegative, displaces iodine to form potassium chloride (\( \mathrm{KCl} \)) and liberates iodine (\( \mathrm{I}_{2} \)) as a free element. The transition from a clear solution to one with a purple-brown hue, due to the liberated iodine, is visibly remarkable.

This reaction is significant for illustrating the oxidizing power of chlorine. It is also used in the laboratory to demonstrate the principles of redox chemistry, giving insights into how halogens interact in solution. Understanding such reactions is vital for fields that depend on halogen chemistry, such as pharmaceuticals, as these transformations form the basis for creating halogenated organic compounds.