/*! 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 20 (a) Write an equation for the re... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 20

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (c) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q) ?\) What is its conjugate base?

Short Answer

Expert verified
a) The equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l)\) is: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq) + \mathrm{OH^{-}}(aq) \] b) The equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l)\) is: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq) + \mathrm{H}_{3}\mathrm{O^{+}}(aq) \] c) The conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is \(\mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq)\), and its conjugate base is \(\mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq)\).

Step by step solution

01

Write the equation of the reaction.

To write the equation where \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l)\), consider the behavior of a base according to Bronsted-Lowry definition: a base accepts a proton \((\mathrm{H}^{+})\). In the reaction with water, the base will accept a proton from \(\mathrm{H}_{2}\mathrm{O}(l)\). The equation will look like this: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq) + \mathrm{OH^{-}}(aq) \] b) H2C6H7O5- acting as an acid in H2O(l)
02

Write the equation of the reaction.

To write the equation where \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l)\), consider the behavior of an acid according to Bronsted-Lowry definition: an acid donates a proton \((\mathrm{H}^{+})\). In the reaction with water, the acid will donate a proton to \(\mathrm{H}_{2}\mathrm{O}(l)\). The equation will look like this: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq) + \mathrm{H}_{3}\mathrm{O^{+}}(aq) \] c) Conjugate acid and conjugate base
03

Identify the conjugate acid.

The conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is the species formed after accepting a proton. From the equation in (a), the conjugate acid is: \[ \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq) \]
04

Identify the conjugate base.

The conjugate base of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is the species formed after donating a proton. From the equation in (b), the conjugate base is: \[ \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq) \]

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Key Concepts

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

Understanding Conjugate Acids
In the Bronsted-Lowry acid-base theory, a conjugate acid forms when a base accepts a proton (H^{+}). Let's explore this idea using our example with H_{2}C_{6}H_{7}O_{5}^{-}. In reaction with water, it acts as a base and accepts a proton, transforming into its conjugate acid:
  • The water molecule donates a proton to H_{2}C_{6}H_{7}O_{5}^{-}.
  • This results in the formation of H_{3}C_{6}H_{7}O_{5}, the conjugate acid.
This reaction highlights the transferable nature of hydrogen ions and their role in forming conjugate acid-base pairs, illustrated by the equation:\[H_{2}C_{6}H_{7}O_{5}^{-} + H_{2}O \rightleftharpoons H_{3}C_{6}H_{7}O_{5} + OH^{-}\]The concept of conjugate acids helps us see the dynamic balance in chemical reactions.
Identifying Conjugate Bases
A conjugate base emerges when an acid donates a proton. In our case, H_{2}C_{6}H_{7}O_{5}^{-} acts as an acid and donates a proton to water, producing its conjugate base:
  • H_{2}C_{6}H_{7}O_{5}^{-} gives up an H^{+} to H_{2}O.
  • This forms HC_{6}H_{6}O_{5}^{2-}, the conjugate base.
This reaction can be displayed as:\[H_{2}C_{6}H_{7}O_{5}^{-} + H_{2}O \rightleftharpoons HC_{6}H_{6}O_{5}^{2-} + H_{3}O^{+}\]Understanding conjugate bases is crucial, as it shows how substances can shift roles in chemical reactions, altering their identity in the process. This flexibility is key in studying equilibria in acid-base chemistry.
Exploring Acid-Base Reactions
Acid-base reactions are foundational in chemistry, particularly through the lens of the Bronsted-Lowry theory. This theory describes how protons are exchanged between substances:
  • An acid is defined as a proton donor.
  • A base is a proton acceptor.
In the context of H_{2}C_{6}H_{7}O_{5}^{-} , these roles illustrate how molecules can behave differently depending on their environment:
  • In an acidic role, it donates a proton, generating hydronium ion H_{3}O^{+} and a conjugate base.
  • In a basic role, it accepts a proton, producing OH^{-} and a conjugate acid.
These reactions highlight the reversible nature of many chemical processes, demonstrating the interconversion between acids and bases.

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Most popular questions from this chapter

A particular sample of vinegar has a pH of \(2.90 .\) If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\) calculate the concentration of acetic acid in the vinegar.

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: \((\mathbf{a}) 0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{g}\) of \(\mathrm{KOH}\) in 500.0 mL of solution, ( c ) 10.0 \(\mathrm{mL}\) of 0.0105 \(\mathrm{MCa}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{mL},(\mathbf{d})\) a solution formed by mixing 20.0 \(\mathrm{mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with 40.0 \(\mathrm{mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}.\)

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 LiOH in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 1.00\) mL of 0.175 M NaOH diluted to \(2.00 \mathrm{L},\) (d) a solution formed by adding 5.00 \(\mathrm{mL}\) of 0.105 \(\mathrm{M} \mathrm{KOH}\) to 15.0 \(\mathrm{mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}.\)

Addition of the indicator methyl orange to an unknown solution leads to a yellow color. The addition of bromthymol blue to the same solution also leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) What is the range (in whole numbers) of possible pH values for the solution? (c) Is there another indicator you could use to narrow the range of possible pH values for the solution?

Phenylacetic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\right)\) is one of the substances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A 0.085\(M\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{COOH}\) has a pH of \(2.68 .\) Calculate the \(K_{a}\) value for this acid.
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