/*! 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 53 Complete the following table: ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 53

Complete the following table: $$ \begin{array}{|l|l|} \hline \text { Acid } & \text { Conjugate Base } \\ \hline \mathrm{H}_{2} \mathrm{O} & \\ \hline & \mathrm{CN}^{-} \\ \hline \mathrm{HNO}_{2} & \\ \hline & \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \\ \hline \end{array} $$

Short Answer

Expert verified
H2O - OH-, HCN - CN-, HNO2 - NO2-, H3PO4 - H2PO4-

Step by step solution

01

Identify the Conjugate Base of H2O

The conjugate base of water (H2O) is obtained by removing one hydrogen ion (H+). This results in hydroxide ion (OH-). Hence, the conjugate base of H2O is OH-.
02

Identify the Acid for CN-

The conjugate acid of CN- is obtained by adding one hydrogen ion (H+) to it. This results in hydrogen cyanide (HCN). Hence, the acid that forms CN- as a conjugate base is HCN.
03

Identify the Conjugate Base of HNO2

The conjugate base of nitrous acid (HNO2) is obtained by removing one hydrogen ion (H+). This results in nitrite ion (NO2-). Hence, the conjugate base of HNO2 is NO2-.
04

Identify the Acid for H2PO4-

The conjugate acid of H2PO4- is obtained by adding one hydrogen ion (H+) to it. This results in phosphoric acid (H3PO4). Hence, the acid that forms H2PO4- as a conjugate base is H3PO4.
05

Complete the Table

Fill in the missing entries using the information obtained in the previous steps. The completed table is: | Acid | Conjugate Base ||--------|-----------------|| H2O | OH- || HCN | CN- || HNO2 | NO2- || H3PO4 | H2PO4- |

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.

Conjugate Acid-Base Pairs
In acid-base chemistry, acids and bases exist in pairs known as conjugate acid-base pairs. These pairs are related by the loss or gain of a hydrogen ion (H+). When an acid donates a hydrogen ion, it becomes its conjugate base. Conversely, when a base accepts a hydrogen ion, it becomes its conjugate acid.

For example:
  • Water (H鈧侽) acts as an acid when it donates a hydrogen ion to form its conjugate base, hydroxide ion (OH鈦�).
  • Cyanide ion (CN鈦�) acts as a base when it accepts a hydrogen ion to form its conjugate acid, hydrogen cyanide (HCN).
Understanding these pairs is essential for predicting the outcome of acid-base reactions.
Hydrogen Ion (H+)
The hydrogen ion, often represented as H鈦�, plays a crucial role in acid-base reactions. It is essentially a proton, and its presence or absence determines whether a molecule behaves as an acid or a base.

When an acid dissociates in water, it releases hydrogen ions into the solution. These H鈦� ions can then be accepted by a base, which explains why acids are described as proton donors and bases as proton acceptors.

For instance:
  • H鈧侽 loses an H鈦� to become OH鈦�, showing its role as an acid.
  • CN鈦� gains an H鈦� to become HCN, showing its role as a base.
This simple transfer of H鈦� is the basis for defining the properties of acids and bases.
Acid Dissociation
Acid dissociation refers to the process by which an acid releases a hydrogen ion (H鈦�) when it dissolves in water. This is often represented by the equation:

HA <=> H鈦� + A鈦�

Here, HA represents the acid and A鈦� is its conjugate base.

For example:
  • When HNO鈧� (nitrous acid) dissociates, it releases an H鈦�, forming its conjugate base, NO鈧傗伝 (nitrite ion).
  • Similarly, water (H鈧侽) can dissociate to release an H鈦�, forming OH鈦� (hydroxide ion).
    • Acid dissociation is a fundamental concept for understanding acid strength and the behavior of acids in different environments.

    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

    Write a balanced equation for the neutralization of each of the following: a. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{NaOH}(a q)\) b. \(\mathrm{HCl}(a q)\) and \(\mathrm{Fe}(\mathrm{OH})_{3}(s)\) c. \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) and \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\)

    State whether each of the following solutions is acidic, basic, or neutral: a. soda, \(\mathrm{pH} 3.22\) b. shampoo, \(\mathrm{pH} 5.7\) c. laundry detergent, \(\mathrm{pH} 9.4\) d. rain, \(\mathrm{pH} 5.83\) e. honey, \(\mathrm{pH} 3.9\) f. cheese, \(\mathrm{pH} 5.2\)

    Calculate the \(\mathrm{pH}\) of each solution given the following \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) or \(\left[\mathrm{OH}^{-}\right]\) values: a. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1 \times 10^{-8} \mathrm{M}\) b. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-6} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=4 \times 10^{-2} \mathrm{M}\) d. \(\left[\mathrm{OH}^{-}\right]=8 \times 10^{-3} \mathrm{M}\) e. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=4.7 \times 10^{-2} \mathrm{M}\) f. \(\left[\mathrm{OH}^{-}\right]=3.9 \times 10^{-6} \mathrm{M}\)

    Identify the Br酶nsted-Lowry acid-base pairs in each of the following equations: a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)\) b. \(\mathrm{CO}_{3}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{OH}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NH}_{3}(a q) \rightleftarrows \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)\)

    Calculate the \(\left[\mathrm{OH}^{-}\right]\) of each aqueous solution with the following \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]:\) a. coffee, \(1.0 \times 10^{-5} \mathrm{M}\) b. soap, \(1.0 \times 10^{-8} \mathrm{M}\) c. cleanser, \(5.0 \times 10^{-10} \mathrm{M}\) d. lemon juice, \(2.5 \times 10^{-2} \mathrm{M}\)
    See all solutions

    Recommended explanations on Chemistry Textbooks

    Nuclear Chemistry

    Read Explanation

    Inorganic Chemistry

    Read Explanation

    Organic Chemistry

    Read Explanation

    The Earths Atmosphere

    Read Explanation

    Kinetics

    Read Explanation

    Chemical Reactions

    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.