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The \(\mathrm{pH}\) of a solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(6.0 .\) Some chlorine gas is bubbled into this solution. Which of the following is correct? (a) The \(\mathrm{pH}\) of resultant solution becomes \(8.0\) (b) Hydrogen gas is liberated from resultant solution (c) The \(\mathrm{pH}\) of resultant solution becomes less than \(6.0\) oxygen gas is liberated (d) \(\mathrm{Cl}_{2} \mathrm{O}\) is formed in the resultant solution

Short Answer

Expert verified
(c) The pH of the resultant solution becomes less than 6.0, oxygen gas is liberated.

Step by step solution

01

Understanding the Problem

We have a solution of \(\mathrm{H}_2\mathrm{O}_2\) with a \(\mathrm{pH}\) of 6.0. Chlorine gas \((\mathrm{Cl}_2)\) will be bubbled into this solution. We need to predict what happens to the \(\mathrm{pH}\) and other chemical reactions after this addition from the given options.
02

Examining the Behavior of \(\mathrm{H}_2\mathrm{O}_2\) with Acidic and Basic Solutions

\(\mathrm{H}_2\mathrm{O}_2\) can function as both an oxidizing and reducing agent. The initial \(\mathrm{pH}\) of 6.0 indicates a near-neutral solution, hinting that \(\mathrm{H}_2\mathrm{O}_2\) can slightly ionize to form \(\mathrm{H}^+\). Chlorine gas addition alters this equilibrium.
03

Reaction of \(\mathrm{Cl}_2\) with \(\mathrm{H}_2\mathrm{O}_2\)

Chlorine gas is known to form hypochlorous acid and hydrochloric acid upon reacting with water: \[ \mathrm{Cl}_2 + \mathrm{H}_2O \rightarrow \mathrm{HCl} + \mathrm{HOCl} \]Both products (\(\mathrm{HCl}\) and \(\mathrm{HOCl}\)) are acidic.
04

Effects of the Reaction on \(\mathrm{pH}\)

The addition of \(\mathrm{HCl}\) and \(\mathrm{HOCl}\) will increase the concentration of \(\mathrm{H}^+\) ions, making the resultant solution more acidic. Therefore, the \(\mathrm{pH}\) will decrease from 6.0 to a value lower than 6.0.
05

Identification of Other Products

The oxidation of \(\mathrm{H}_2\mathrm{O}_2\) or other complex reactions in the presence of \(\mathrm{Cl}_2\) could lead to oxygen gas being released. This would occur as \(\mathrm{H}_2\mathrm{O}_2\) decomposes, typically forming \(\mathrm{O}_2\) gas.

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Key Concepts

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

Hydrogen Peroxide Behavior
Hydrogen peroxide, or \(\mathrm{H}_2\mathrm{O}_2\), is a fascinating chemical due to its dual ability to act as both an oxidizing and reducing agent. In more basic terms, this means it can either gain or lose electrons during reactions. This versatility plays a significant role in its interactions with other compounds.
In water, hydrogen peroxide slightly ionizes which results in the formation of \(\mathrm{H}^+\) ions. This behavior is crucial when discussing pH, as these ions suggest it can create a mildly acidic environment. The initial pH of \(6.0\) for the \(\mathrm{H}_2\mathrm{O}_2\) solution indicates a near-neutral terrain, which shows the delicate state of equilibrium in hydrogen peroxide solutions. Thus, any additions, like chlorine gas, can easily tip this balance.
When dealing with reactions, one must consider how the inherent properties of \(\mathrm{H}_2\mathrm{O}_2\) might affect the resulting solution. This sensitivity makes the compound a center of many significant chemical discussions.
Chlorine Gas Reactions
Chlorine gas, \(\mathrm{Cl}_2\), is well-known for its reactive nature, particularly in aqueous environments. When \(\mathrm{Cl}_2\) is introduced into water, it forms hypochlorous acid (\(\mathrm{HOCl}\)) and hydrochloric acid (\(\mathrm{HCl}\)). These reactions can be expressed with the simple equation:
  • \(\mathrm{Cl}_2 + \mathrm{H}_2O \rightarrow \mathrm{HCl} + \mathrm{HOCl}\)

This transformation is a result of the dissolution of \(\mathrm{Cl}_2\) where it undergoes hydrolysis, a process that reveals the acidic potential of chlorine gas. The products of this reaction, \(\mathrm{HCl}\) and \(\mathrm{HOCl}\), both contribute \(\mathrm{H}^+\) ions to the solution, emphasizing the shift toward a more acidic environment.
Given these details, the interaction between chlorine gas and water significantly impacts the pH of the solution involved. This mechanism is a vital aspect of understanding the ramifications of bubbling chlorine into an \(\mathrm{H}_2\mathrm{O}_2\) solution.
Reaction Products in Chemistry
The complexity of chemical reactions unveils numerous products, particularly when reactive species like \(\mathrm{H}_2\mathrm{O}_2\) and \(\mathrm{Cl}_2\) interact.
During this reaction, not only do acidic compounds emerge but the oxidation of hydrogen peroxide can lead to the liberation of oxygen gas. As \(\mathrm{H}_2\mathrm{O}_2\) often decomposes in various reactions, oxygen ( \(\mathrm{O}_2\)) is likely produced, which may escape as gas bubbles.
This liberation is a common feature in reactions involving peroxides, and it represents an essential aspect of the by-products of these many-step chemical processes. The understanding of how new products like \(\mathrm{HCl}\), \(\mathrm{HOCl}\), and \(\mathrm{O}_2\) come about is integral in estimating the behavior of the resultant solution.
Acid-Base Equilibrium
Acid-base equilibrium describes the balance between acidic and basic ions in a solution. It is a fundamental concept in chemistry, dictating the pH of the solution.
As chlorine gas reacts with water to produce \(\mathrm{HCl}\) and \(\mathrm{HOCl}\), both being acids, the equilibrium shifts to a more acidic stance. Consequently, the addition of these acids increases the concentration of \(\mathrm{H}^+\) ions. This increase lowers the pH of the solution, a direct illustration of deviation in equilibrium.
The initial hydrogen peroxide solution was hovering around pH 6.0, but with the production of more \(\mathrm{H}^+\) and the subsequent overwhelming presence of acidic compounds, a darker shade of acidity should be expected.
In chemical practice, tracking how acids or bases shift the pH of solutions provides insight into reaction outcomes and solution characteristics, which is fundamental for properly predicting and controlling chemical environments.

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Most popular questions from this chapter

Equilibrium constant for the reaction: \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) is \(81 .\) If the velocity constant of the forward reaction is \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \mathrm{sec}^{-1}\), what is the velocity constant (in \(\mathrm{L} \mathrm{mol}^{-1}\) sec. \(^{-1}\) ) for the backward reaction: (a) 13122 (b) 2 (c) 261 (d) 243

For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant: (a) Becomes one-fourth (b) Is doubled (c) Is halved (d) Remains same

The value of \(\mathrm{K}_{\mathrm{p}}\) in the reaction: \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{P}\left(\mathrm{CO}_{2}\right)\) (b) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{MgCO}_{3}\right)}{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}\)

An amount of solid \(\mathrm{NH}_{4}\) HS in placed in a flask already containing ammonia gas at a certain temperature and \(0.50\) atm pressure. Ammonium hydrogen sulphide decomposes to yield \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure is the flask rises to \(0.84 \mathrm{~atm}\), the equilibrium constant for \(\mathrm{NH}_{4} \mathrm{HS}\) decomposition at this temperature is: (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

The relation between \(K_{n}\) and \(K_{c}\) for the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (c) \(K_{p}^{p}=K_{c} /(R T)^{2}\) (d) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} / \mathrm{RT}\)

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