Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

\(\mathrm{CrO}_{3}\) dissolves in aqueous \(\mathrm{NaOH}\) to give (a) \(\mathrm{Cr}(\mathrm{OH})_{3}\) (b) \(\mathrm{Cr}(\mathrm{OH})_{2}\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (d) \(\mathrm{CrO}_{4}^{2-}\)

Short Answer

Expert verified
The main product is \(\mathrm{CrO}_{4}^{2-}\) (option d).

Step by step solution

01

Understand the Reactants

The problem involves chromium trioxide, \(\mathrm{CrO}_{3}\), and aqueous sodium hydroxide, \(\mathrm{NaOH}\). \(\mathrm{CrO}_{3}\) is a chromium(VI) oxide, and \(\mathrm{NaOH}\) is a strong base. We need to understand how these reactants interact in water.
02

Write the Reaction

When \(\mathrm{CrO}_{3}\) is dissolved in aqueous \(\mathrm{NaOH}\), a reaction occurs to form a chromate ion. The overall balanced reaction is:\[\mathrm{CrO}_{3} + 2 \mathrm{OH}^{-} \rightarrow \mathrm{CrO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}\]In this reaction, \(\mathrm{CrO}_{3}\) reacts with hydroxide ions \(\mathrm{OH}^{-}\) from \(\mathrm{NaOH}\) to form \(\mathrm{CrO}_{4}^{2-}\), the chromate ion, and water.
03

Determine the Product

According to the reaction from Step 2, the main product when \(\mathrm{CrO}_{3}\) dissolves in aqueous \(\mathrm{NaOH}\) is \(\mathrm{CrO}_{4}^{2-}\), which is the chromate ion. This matches option (d) given in the exercise.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chromium trioxide
Chromium trioxide, denoted as \(\mathrm{CrO}_{3}\), is a powerful oxidizing agent. It appears as a bright red crystalline solid. It's often utilized in the preparation of other chromium compounds and in chromium plating. In organic chemistry, it's sometimes used as a reagent to oxidize primary alcohols to aldehydes or carboxylic acids. Being an oxide of chromium, \(\mathrm{CrO}_{3}\) features chromium in a +6 oxidation state, which is particularly toxic. Hence, it requires careful handling. When dissolved in water or a basic solution like sodium hydroxide, \(\mathrm{CrO}_{3}\) undergoes significant transformation to produce chromate ions. This chemical behavior is essential in understanding its broader applications and its role in various chemical reactions.
Sodium hydroxide
Sodium hydroxide (\(\mathrm{NaOH}\)), commonly known as lye or caustic soda, is a strong base used in many industrial and chemical processes. It appears as white, solid pellets or a concentrated aqueous solution and is highly soluble in water. As a strong base, \(\mathrm{NaOH}\) dissociates completely in water to give sodium (\(\mathrm{Na^+}\)) and hydroxide ions (\(\mathrm{OH}^-\)). This disassociation is crucial to the reaction involving chromium trioxide. The hydroxide ions serve as key reactants, facilitating various chemical transformations, including the formation of chromate ions from chromium trioxide.When working with \(\mathrm{NaOH}\), it's important to note that it is highly corrosive. It can cause severe burns upon contact with skin or damage to materials. Therefore, proper safety measures, such as wearing gloves and goggles, are essential when handling this chemical.
Chemical reactions
Chemical reactions involve the transformation of reactants into products. In this context, we look at how \(\mathrm{CrO}_{3}\) reacts with \(\mathrm{NaOH}\) to yield chromate ions. The fundamental reaction is:
  • \(\mathrm{CrO}_{3} + 2 \mathrm{OH}^- \rightarrow \mathrm{CrO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}\)
Here, chromium trioxide reacts with hydroxide ions from sodium hydroxide. This leads to the formation of chromate ions and water.Understanding the stoichiometry is essential. Each molecule of \(\mathrm{CrO}_{3}\) reacts with two hydroxide ions to yield a single chromate ion. This balanced equation helps predict the outcome of mixing these substances under controlled conditions. Such reactions underpin numerous industrial and laboratory processes, including those involved in metal treatment and synthesis of compounds.
Chromate ion
The chromate ion, \(\mathrm{CrO}_{4}^{2-}\), is formed when chromium trioxide is dissolved in a basic solution like sodium hydroxide. It is characterized by a tetrahedral geometry around the chromium atom, where four oxygen atoms are symmetrically positioned. Chromate ions are often used in industrial settings for pigment production, corrosion resistance, and wood preservation. They can also serve as a precursor to other chromium compounds.One of the interesting aspects of chromate ions is their ability to exist in equilibrium with dichromate ions (\(\mathrm{Cr}_2\mathrm{O}_7^{2-}\)), depending on the pH of the solution. At higher pH levels, chromate ions are favored, while at lower pH, dichromate ions are more stable. This property makes the chromate-dichromate system a subject of study in acid-base equilibria.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which is not true about Libermann nitroso test? (a) Here the blue colour is due to formation of sodium salt of indophenol (b) It is given by nitrophenols (c) It is also given by \(p\)-substituted phenols (d) Both (b) and (c)

Borax \(\left[\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} .10 \mathrm{H}_{2} \mathrm{O}\right]\) when heated on platinum loop it gives a dark transparent glass like bead. The hot bead is dipped in the salt till it reacts with transition metal oxide. It produces characteristic bead of meta borate. $$ \begin{array}{ll} \text { Colour of the bead } & \text { Ion } \\ \text { Blue green or light blue } & \mathrm{Cu}^{2+} \\ \text { Yellow } & \mathrm{Fe}^{2+} \text { or } \mathrm{Fe}^{3+} \\ \text { Green } & \mathrm{Cr}^{3+} \\ \text { Violet } & \mathrm{Mn}^{2+} \\ \text { Dark blue } & \mathrm{Co}^{2+} \\ \text { Brown } & \mathrm{Ni}^{2+} \end{array} $$ The hybridisation of \(\mathrm{B}\) in borax is (a) sp (b) \(\mathrm{sp}^{2}\) (c) \(\mathrm{sp}^{3}\) (d) Both (B) and (C)

Phenol and benzoic acid can be distinguished by which tests? (1) Libermann nitroso test (2) Reaction with \(\mathrm{NaHCO}_{3}\) (3) Reaction with \(\mathrm{NaOH}\) (4) Ferric chloride test (a) 1,2 and 3 (b) 1,2 and 4 (c) 1,3 and 4 (d) 2,3 and 4

Identify the correct order of solubility of \(\mathrm{Na}_{2} \mathrm{~S}, \mathrm{CuS}\) and \(\mathrm{ZnS}\) in aqueous medium. (a) \(\mathrm{Na}_{2} \mathrm{~S}>\mathrm{CuS}>\mathrm{ZnS}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}>\mathrm{ZnS}>\mathrm{CuS}\) (c) \(\mathrm{CuS}>\mathrm{ZnS}>\mathrm{Na}_{2} \mathrm{~S}\) (d) \(\mathrm{ZnS}>\mathrm{Na}_{2} \mathrm{~S}>\mathrm{CuS}\)

Identify the products \(\mathrm{A}\) and \(\mathrm{B}\) here. (1) \(2 \mathrm{Ag}^{+}\)(excess) \(+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) \(\stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{A}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (2) \(2 \mathrm{NO}_{2}^{-}+2 \mathrm{I}^{-} \stackrel{\text { acid medium }}{\longrightarrow} \mathrm{B}+\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (a) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=\mathrm{N}_{2}\) (b) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=\mathrm{N}_{2} \mathrm{O}\) (c) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{O}, \mathrm{B}=2 \mathrm{NO}_{2}\) (d) \(\mathrm{A}=\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{~B}=2 \mathrm{NO}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free