/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Write the chemical equation and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 49

Write the chemical equation and the \(K_{a}\) expression for the dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: \((\mathbf{a}) \mathrm{HNO}_{2},\) (b) \(\mathrm{ClH}_{2} \mathrm{CCOOH}\).

Short Answer

Expert verified
(a) For HNO鈧� dissociation with H鈦� as a product: \[\mathrm{HNO}_{2}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\) With hydronium ion as a product: \[\mathrm{HNO}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\) (b) For ClH鈧侰COOH dissociation with H鈦� as a product: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\) With hydronium ion as a product: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\] \(K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\)

Step by step solution

01

Writing the chemical equation for HNO鈧� dissociation with H鈦� as a product

To write the chemical dissociation reaction for HNO鈧� in aqueous solution, we can represent it in the following form: \[\mathrm{HNO}_{2}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\]
02

Writing the chemical equation for HNO鈧� dissociation with hydronium ion as a product

For the dissociation reaction of HNO鈧� with hydronium ion as a product, we can write the equation as: \[\mathrm{HNO}_{2}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{NO}_{2}^{-}(aq)\]
03

Writing the \(K_{a}\) expression for the dissociation of HNO鈧�

The \(K_{a}\) expression is the equilibrium constant for the dissociation reaction of an acid. When the dissociation occurs with \(\mathrm{H}^{+}(aq)\) as a product, the \(K_{a}\) expression for HNO鈧� is given by: \[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\] And with the hydronium ion as a product, the expression is: \[K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\]
04

Writing the chemical equation for ClH鈧侰COOH dissociation with H鈦� as a product

For the dissociation reaction of ClH鈧侰COOH in aqueous solution, we can write the equation as follows: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\]
05

Writing the chemical equation for ClH鈧侰COOH dissociation with hydronium ion as a product

To write the dissociation reaction of ClH鈧侰COOH with the hydronium ion as a product, we can represent it in the following form: \[\mathrm{ClH}_{2}\mathrm{CCOOH}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq) + \mathrm{ClH}_{2}\mathrm{CCOO}^{-}(aq)\]
06

Writing the \(K_{a}\) expression for the dissociation of ClH鈧侰COOH

The \(K_{a}\) expression is the equilibrium constant for the dissociation reaction of an acid. When the dissociation occurs with \(\mathrm{H}^{+}(aq)\) as a product, the \(K_{a}\) expression for ClH鈧侰COOH is given by: \[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\] And with the hydronium ion as a product, the expression is: \[K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{ClH}_{2}\mathrm{CCOO}^{-}]}{[\mathrm{ClH}_{2}\mathrm{CCOOH}]}\]

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 91影视!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a central concept in chemistry that describes the state in which the concentrations of all reactants and products remain constant over time. This occurs when the rate of the forward reaction, where reactants convert into products, equals the rate of the reverse reaction, where products convert back into reactants.
This dynamic balance means that, even though the reaction appears to have stopped, reactions are still occurring albeit at equal rates in both directions.
  • It's important to realize that equilibrium does not imply equal concentrations of reactants and products.
  • Instead, it reflects a steady state where no observable changes in concentration occur.
Understanding chemical equilibrium is crucial because it influences various chemical processes, such as reactions in a biological system or industrial applications. This concept applies directly to the dissociation of acids in aqueous solutions, where certain concentrations are maintained due to this balance.
Equilibrium Constant (Ka)
The equilibrium constant, specifically the acid dissociation constant represented as \(K_a\), is a special value that quantifies the extent of dissociation of an acid in water. It reflects how strongly an acid releases its protons or forms \[ ext{H}^+ \] ions in a solution. Higher \(K_a\) values indicate stronger acids that dissociate more completely. This is crucial for predicting the behavior of acids in various chemical systems.
To construct the \(K_a\) expression, you divide the product of the concentrations of the dissociated ions by the concentration of the undissociated acid:
  • For an acid \[ ext{HA}(aq) ightleftharpoons ext{H}^+(aq) + ext{A}^-(aq) \] the \(K_a\) expression is:
\[ K_a = \frac{[ ext{H}^+][ ext{A}^-]}{[ ext{HA}]} \]
  • If the dissociation involves the formation of \[ ext{H}_3 ext{O}^+ \] ions, as in \[ ext{HA}(aq) + ext{H}_2 ext{O}(l) ightleftharpoons ext{H}_3 ext{O}^+(aq) + ext{A}^-(aq) \], the expression remains conceptually similar:
\[ K_a = \frac{[ ext{H}_3 ext{O}^+][ ext{A}^-]}{[ ext{HA}]} \] This expression helps chemists understand how an acid will behave in different solution environments, aiding in predictions about pH and reaction direction.
Aqueous Solution Chemistry
Aqueous solution chemistry involves the study of substances dissolved in water, creating a solution where water is the solvent. This branch of chemistry is fundamental because many chemical reactions, biological processes, and industrial operations occur in water or involve water solutions.
In the context of acid dissociation, understanding how acids behave in aqueous solutions is key. Water is a polar solvent, which means it can effectively stabilize ions produced during acid dissociation. This property makes water an excellent medium for studying ionic dissociation and allows different acids to release protons (\( ext{H}^+ \)) or form hydronium ions (\( ext{H}_3 ext{O}^+ \)).
  • Aqueous chemistry explains phenomena like the conductivity of solutions, which increases as ions dissociate.
  • It also contributes to the understanding of pH, a measure of acidity, which is particularly vital in processes like digestion and brewing.
Thus, aqueous solution chemistry provides a framework for analyzing numerous essential biochemical and industrial processes involving aqueous media.

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

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in \(200 \mathrm{~mL}\) of water?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2},\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((\mathbf{c}) 1.00\) \(\mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},(\mathbf{d})\) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}\)

(a) Write a chemical equation that illustrates the autoionization of water. (b) Write the expression for the ionproduct constant for water, \(K_{w} .(\mathbf{c})\) If a solution is described as basic, which of the following is true: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\), (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: \((\mathbf{a}) \mathrm{F}^{-}(\mathbf{b}) \mathrm{Br}^{-}(\mathbf{c}) \mathrm{HS}^{-}(\mathbf{d}) \mathrm{ClO}_{4}^{-}(\mathbf{e}) \mathrm{HCOO}^{-}\)

Which member of each pair produces the more acidic aqueous solution: \((\mathbf{a}) \mathrm{Zn} \mathrm{Br}_{2}\) or \(\mathrm{CdCl}_{2},\) (b) \(\mathrm{CuCl}\) or \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{NiBr}_{2} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation
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.