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What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed (b) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) are formed (c) \(\mathrm{CrO}_{4}^{2-}\) is oxidized to \(+7\) state to \(\mathrm{Cr}\) (d) \(\mathrm{CrO}_{4}^{2-}\) is reduced to \(+3\) state of \(\mathrm{Cr}\)

Short Answer

Expert verified
Option (a) is correct: \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) and \( \mathrm{H}_2\mathrm{O} \) are formed.

Step by step solution

01

Understand Potassium Chromate Solution

Potassium chromate, \( \mathrm{K}_2\mathrm{CrO}_4 \), in an aqueous solution exists as chromate ions (\( \mathrm{CrO}_4^{2-} \)). In solutions, chromate ions can participate in various chemical reactions depending on the pH of the solution.
02

Reaction with Dilute Nitric Acid

Dilute nitric acid provides an acidic environment that affects the chromate ions. In an acidic solution, chromate ions \( \mathrm{CrO}_4^{2-} \) tend to convert into dichromate ions \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \). The balanced chemical reaction for this conversion is: \[2\mathrm{CrO}_4^{2-} + 2\mathrm{H}^+ \rightarrow \mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}_2\mathrm{O}\]
03

Determine Reaction Products

From the chemical reaction, it can be seen that the addition of an excess of dilute nitric acid to a solution of potassium chromate forms dichromate ions (\( \mathrm{Cr}_2\mathrm{O}_7^{2-} \)) and water (\( \mathrm{H}_2\mathrm{O} \)). There is no reduction to \( \mathrm{Cr}^{3+} \), nor is there oxidation to a +7 state.
04

Select the Correct Answer

Based on the chemical reaction \( 2\mathrm{CrO}_4^{2-} + 2\mathrm{H}^+ \rightarrow \mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}_2\mathrm{O} \), option (a) is the correct outcome. \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) and \( \mathrm{H}_2\mathrm{O} \) are formed when potassium chromate is treated with an excess of dilute nitric acid.

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Key Concepts

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

Chromate to Dichromate Conversion
When potassium chromate (\( \mathrm{K}_2\mathrm{CrO}_4 \)) is dissolved in water, it produces chromate ions (\( \mathrm{CrO}_4^{2-} \)). However, these ions don't always remain as such. In an acidic environment, which is created by adding acids like nitric acid, chromate ions can transform into dichromate ions (\( \mathrm{Cr}_2\mathrm{O}_7^{2-} \)). This is a fascinating chemical reaction because it results in a visible color change from yellow to orange. Such transformations depend heavily on the pH of the solution and help illustrate how chemical species can exist in different forms based on environmental conditions. Understanding these conversions is key to mastering solution chemistry.
Acid-Base Reactions
Acid-base reactions are a foundational concept in chemistry. They involve the transfer of hydrogen ions (\( \mathrm{H}^+ \)) from acids to bases. In the context of our chromate-dichromate conversion, nitric acid acts as the proton donor, providing \( \mathrm{H}^+ \) to the chromate ions, which leads to the formation of dichromate ions. This role of acids as hydrogen donors is just one part of a broader concept. Acid-base reactions can lead to the formation of new compounds and often produce water as a byproduct, as seen in the conversion reaction. Recognizing these reactions helps in predicting and balancing chemical equations effectively.
Solution Chemistry
Solution chemistry involves understanding how substances dissolve, interact, and react in a liquid, typically water. It's about knowing how solutes like potassium chromate can exist as ions in a solution and how they can react when conditions change. In solution chemistry, the presence of other ions or a change in the concentration of hydrogen ions (\( \mathrm{H}^+ \)) can lead to significant changes in the behavior and composition of a solution. When discussing chromate ions in solution, their transformation into dichromate under acidic conditions is a classic example of how solution conditions can dictate the chemical form and properties of dissolved species.
Oxidation States
Oxidation states are numbers that represent the total number of electrons an atom gains, loses, or shares when it forms chemical bonds. In reactions, tracking oxidation states helps chemists identify what's being oxidized or reduced. In the chromate to dichromate conversion, chromate (\( \mathrm{CrO}_4^{2-} \)) and dichromate (\( \mathrm{Cr}_2\mathrm{O}_7^{2-} \)) ions both involve chromium in the +6 oxidation state, meaning there's no change in the oxidation number during this conversion despite the transformation. Recognizing and understanding these states aids in comprehending redox reactions, where the electron transfer is crucial.
Balancing Chemical Equations
Balancing chemical equations is a crucial skill for any chemistry student. It ensures the same number of each type of atom is present on both sides of the reaction equation, reflecting the conservation of mass. For the chromate-dichromate reaction, balancing involves ensuring that all chromate ions (\( \mathrm{CrO}_4^{2-} \)) and hydrogen ions (\( \mathrm{H}^+ \)) transform into the correct amounts of dichromate ions (\( \mathrm{Cr}_2\mathrm{O}_7^{2-} \)) and water (\( \mathrm{H}_2\mathrm{O} \)). By practicing this skill, you ensure a thoughtful and precise approach to predicting and understanding chemical reactions, and maintain the principle that matter cannot be created or destroyed.

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Most popular questions from this chapter

In the reaction \(\mathrm{Zn}+\mathrm{NaOH} \longrightarrow \Delta \longrightarrow\) the product A is (a) \(\mathrm{Zn}(\mathrm{OH})_{2}\) (b) \(2 \mathrm{Na} \mathrm{ZnO}_{2}\) (c) \(\mathrm{Na}_{2} \mathrm{ZnO}_{2}\) (d) none of these

What is the current order of spin only magnetic moment (in B.M.) of \(\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}\) and \(\mathrm{V}^{2+}\) ? (a) \(\mathrm{Mn}^{2+}>\mathrm{V}^{2+}>\mathrm{Cr}^{2+}\) (b) \(\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}\) (c) \(\mathrm{Mn}^{2+}>\mathrm{Cr}^{2+}>\mathrm{V}^{2+}\) (d) \(\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Mn}^{2+}\)

Cerium \((\mathrm{Z}=58)\) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect? (a) the common oxidation states of cerium are \(+3\) and \(+4\) (b) the \(+3\) oxidation state of cerium is more stable than the \(+4\) oxidation state (c) the \(+4\) oxidation state of cerium is not known in solutions (d) cerium (IV) acts as an oxidizing agent

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The pair of the compounds in which both the metals are in the highest possible oxidation state is (a) \(\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{MnO}_{2}\) (b) \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}, \mathrm{MnO}_{4}^{-}\) (c) \(\mathrm{TiO}_{3}, \mathrm{MnO}_{2}\) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}\)

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