/*! 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 50 For each of the following reacti... [FREE SOLUTION] | 91Ó°ÊÓ

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For each of the following reactions, identify the Bronsted-Lowry acids and bases and the conjugate acid-base pairs: (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons\) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)\) (b) \(\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HCO}_{3}^{-}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{SO}_{3}^{2-}(a q)\) (d) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{OH}(a q)\)

Short Answer

Expert verified
(a) Acids: CH3CO2H, NH4+; Bases: NH3, CH3CO2-. (b) Acids: H3O+, HCO3-; Bases: CO3^2-, H2O. (c) Acids: HSO3-, H3O+; Bases: H2O, SO3^2-. (d) Acids: H2O, H2SO3; Bases: HSO3-, OH-.

Step by step solution

01

Analyze Reaction (a)

In the reaction \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \), the acetic acid \( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \) donates a proton to become \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \), making it the Bronsted-Lowry acid. \( \mathrm{NH}_{3} \) accepts a proton to form \( \mathrm{NH}_{4}^{+} \), making it the Bronsted-Lowry base. The conjugate acid-base pairs are \((\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{CH}_{3} \mathrm{CO}_{2}^{-})\) and \((\mathrm{NH}_{3}, \mathrm{NH}_{4}^{+})\).
02

Analyze Reaction (b)

For the reaction \( \mathrm{CO}_{3}^{2-} + \mathrm{H}_{3} \mathrm{O}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{O} + \mathrm{HCO}_{3}^{-} \), \(\mathrm{H}_{3} \mathrm{O}^{+}\) donates a proton to become \( \mathrm{H}_{2} \mathrm{O} \), making it the Bronsted-Lowry acid, and \( \mathrm{CO}_{3}^{2-} \) accepts a proton to form \( \mathrm{HCO}_{3}^{-} \), making it the Bronsted-Lowry base. The conjugate acid-base pairs are \((\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{H}_{2} \mathrm{O})\) and \((\mathrm{CO}_{3}^{2-}, \mathrm{HCO}_{3}^{-})\).
03

Analyze Reaction (c)

In the reaction \( \mathrm{HSO}_{3}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+} + \mathrm{SO}_{3}^{2-} \), \( \mathrm{HSO}_{3}^{-} \) donates a proton to \( \mathrm{H}_{2} \mathrm{O} \), thus \( \mathrm{HSO}_{3}^{-} \) is the Bronsted-Lowry acid and \( \mathrm{H}_{2} \mathrm{O} \) acts as the base. The conjugate pairs are \((\mathrm{HSO}_{3}^{-}, \mathrm{SO}_{3}^{2-})\) and \((\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{3} \mathrm{O}^{+})\).
04

Analyze Reaction (d)

In the reaction \( \mathrm{HSO}_{3}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3} + \mathrm{OH}^{-} \), \( \mathrm{H}_{2} \mathrm{O} \) donates a proton to form \( \mathrm{OH}^{-} \), making it the Bronsted-Lowry acid, and \( \mathrm{HSO}_{3}^{-} \) accepts the proton to form \( \mathrm{H}_{2} \mathrm{SO}_{3} \), making it the base. The conjugate pairs are \((\mathrm{HSO}_{3}^{-}, \mathrm{H}_{2} \mathrm{SO}_{3})\) and \((\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-})\).

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Key Concepts

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

Acid-Base Reactions
Acid-base reactions are processes where acids and bases interact to form a chemical equilibrium. According to the Bronsted-Lowry theory, an acid is a substance that donates a proton ( H^+ ), and a base is a substance that accepts a proton. This concept of proton transfer is the crux of acid-base reactions.
In a reversible reaction, such as the ones in the exercises provided, one molecule will act as an acid by donating a proton, while the other will act as a base by accepting that proton. This exchange ultimately leads to the formation of a conjugate acid and a conjugate base, creating conjugate acid-base pairs. In this dynamic equilibrium, both forward and reverse reactions occur simultaneously, and the roles of acids and bases can switch based on the direction of the reaction. Understanding the delicate balance of shifting protons in acid-base reactions is essential for grasping chemical equilibrium.
Conjugate Acid-Base Pairs
Conjugate acid-base pairs are directly linked to the Bronsted-Lowry definition. When an acid donates a proton, it transforms into its conjugate base. Similarly, when a base accepts a proton, it becomes its conjugate acid. To illustrate this, consider an acid-base reaction:
  • In Reaction (a), the acetic acid (\( \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \)) donates a proton, forming its conjugate base \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \). \( \mathrm{NH}_{3} \) accepts a proton, becoming its conjugate acid \( \mathrm{NH}_{4}^{+} \).
  • In Reaction (b), \( \mathrm{H}_{3} \mathrm{O}^{+} \) loses a proton to become \( \mathrm{H}_{2} \mathrm{O} \), its conjugate base, while \( \mathrm{CO}_{3}^{2-} \) becomes \( \mathrm{HCO}_{3}^{-} \), its conjugate acid.
These relationships are always observed in pairs, and recognizing them can help you track the flow of protons throughout the reaction. Understanding conjugate acid-base pairs is essential for predicting the behavior of substances in different conditions.
Proton Transfer
Proton transfer is the key mechanism at play in Bronsted-Lowry acid-base reactions. It involves the movement of a proton ( H^+ ) from an acid to a base. The ability of a substance to donate or accept a proton is what classifies it as either an acid or a base. Then it's all about what happens after that crucial proton has moved.
In this process, we see that the acid becomes its conjugate base after losing the proton, and the base transforms into its conjugate acid after gaining the proton. This transfer can go backwards and forwards, especially in reversible reactions, and it's the heart of how these reactions work. The dynamic exchange of protons is what allows the chemical reaction to be reversible, and this process helps maintain chemical balance and stability in various systems.
Acids and Bases Identification
Identifying acids and bases in chemical reactions is crucial, and the Bronsted-Lowry theory offers a clear approach to this. You begin by looking for the substance that donates a proton; this is your acid. Conversely, the substance that receives the proton is your base. In the given exercises, identifying these components helps us understand the reactions:
  • In Reaction (c), \( \mathrm{HSO}_{3}^{-} \) is identified as the acid as it donates a proton to \( \mathrm{H}_{2} \mathrm{O} \).
  • In Reaction (d), \( \mathrm{H}_{2} \mathrm{O} \) acts as the acid, while \( \mathrm{HSO}_{3}^{-} \) serves as the base.
With practice, identifying acids and bases becomes intuitive, making it easier to predict reaction outcomes. This skill is fundamental for any chemistry student to develop, ensuring you can manage and predict chemical interactions effectively.

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Most popular questions from this chapter

Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic acid \(\mathrm{H}_{3} \mathrm{PO}_{4-}\)

You may have been told not to mix bleach and ammonia. The reason is that bleach (sodium hypochlorite) reacts with ammonia to produce toxic chloramines, such as \(\mathrm{NH}_{2} \mathrm{Cl}\). For example, in basic solution: \(\mathrm{OCl}^{-}(a q)+\mathrm{NH}_{3}(a q) \longrightarrow \mathrm{OH}^{-}(a q)+\mathrm{NH}_{2} \mathrm{Cl}(a q)\) (a) The following initial rate data for this reaction were obtained in basic solution at \(25^{\circ} \mathrm{C}:\) $$ \begin{array}{lccc} \hline \mathrm{pH} & \text { Initial }\left[\mathrm{OCl}^{-}\right] & \text {Initial }\left[\mathrm{NH}_{3}\right] & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 12 & 0.001 & 0.01 & 0.017 \\ 12 & 0.002 & 0.01 & 0.033 \\ 12 & 0.002 & 0.03 & 0.100 \\ 13 & 0.002 & 0.03 & 0.010 \\ \hline \end{array} $$ What is the rate law for the reaction? What is the numerical value of the rate constant \(k\), including the correct units? (b) The following mechanism has been proposed for this reaction in basic solution: \(\mathrm{H}_{2} \mathrm{O}+\mathrm{OCl} \rightleftharpoons \mathrm{HOCl}+\mathrm{OH}^{-} \quad\) Fast, equilibrium constant \(K_{1}\) \(\mathrm{HOCl}+\mathrm{NH}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NH}_{2} \mathrm{Cl}\) Slow, rate constant \(\mathrm{k}_{2}\) Assuming that the first step is in equilibrium and the second step is rate- determining, calculate the value of the rate constant \(k_{2}\) for the second step. \(K_{a}\) for \(\mathrm{HOCl}\) is \(3.5 \times 10^{-8} .\)

Natural rain has a \(\mathrm{pH}\) of \(5.6\) due to dissolved atmospheric carbon dioxide at a current level of 400 ppm. Various models predict that burning fossil fuels will increase the atmospheric \(\mathrm{CO}_{2}\) concentration to between 500 and 1000 ppm (by volume) by the year \(2100 .\) Calculate the \(\mathrm{pH}\) of rain in a scenario where the \(\mathrm{CO}_{2}\) concentration is \(750 \mathrm{ppm}\). (a) First use Henry's Law to calculate the concentration of dissolved \(\mathrm{CO}_{2} .\) Solubility \(=k \cdot P\) (Section \(\left.12.4\right)\) and the Henry's Law constant \((k)\) for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.2 \times 10^{-2} \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{atm})\) (b) Next calculate the \(\mathrm{pH}\) of the rain. \(\mathrm{CO}_{2}\) reacts with water to product carbonic acid according to the equation: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$ Assume all the dissolved \(\mathrm{CO}_{2}\) is converted to \(\mathrm{H}_{2} \mathrm{CO}_{3}\). Acid dissociation constants for \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are \(K_{\mathrm{a} 1}=4.3 \times 10^{-7} ; K_{\Delta 2}=5.6 \times 10^{-11}\). (Worked Example 15.11 is a model for this calculation.) (c) Will rising \(\mathrm{CO}_{2}\) levels affect the acidity of rainfall?

A \(1.00 \times 10^{-3} \mathrm{M}\) solution of quinine, a drug used in treating malaria, has a \(\mathrm{pH}\) of \(9.75 .\) What are the values of \(K_{\mathrm{b}}\) and \(\mathrm{p} K_{\mathrm{b}}\) ?

Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{F}^{-}, \mathrm{HF}, \mathrm{CI}^{-}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in a solution that contains \(0.10 \mathrm{M}\) \(\mathrm{HF}\left(K_{\mathrm{a}}=3.5 \times 10^{-4}\right)\) and \(0.10 \mathrm{M} \mathrm{HCl}\)

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