/*! 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 35 Give the conjugate base for each... [FREE SOLUTION] | 91影视

91影视

Give the conjugate base for each of the following acids: (a) \(\mathrm{HClO}(a q)\) (b) \(\mathrm{NH}_{4}^{*}(a q)\) (c) \(\mathrm{HN}_{3}(a q)\) (d) \(\mathrm{HS}^{-}(a q)\)

Short Answer

Expert verified
(a) ClO鈦, (b) NH鈧, (c) N鈧冣伝, (d) S虏鈦.

Step by step solution

01

Understanding Conjugate Bases

A conjugate base is formed when an acid donates a proton (H鈦). The remaining species after this process is the conjugate base.
02

Identify Conjugate Base of HClO

The acid \(\mathrm{HClO}\) donates a proton (H鈦) to form its conjugate base. Removing a proton from \(\mathrm{HClO}\) results in the conjugate base, \(\mathrm{ClO}^{-}\).
03

Identify Conjugate Base of NH鈧勨伜

The acid \(\mathrm{NH}_4^{+}\) donates a proton (H鈦) to form its conjugate base. Removing a proton from \(\mathrm{NH}_4^{+}\) results in the conjugate base, \(\mathrm{NH}_3\).
04

Identify Conjugate Base of HN鈧

The acid \(\mathrm{HN}_3\) donates a proton (H鈦) to form its conjugate base. Removing a proton from \(\mathrm{HN}_3\) results in the conjugate base, \(\mathrm{N}_3^{-}\).
05

Identify Conjugate Base of HS鈦

The acid \(\mathrm{HS}^-\) donates a proton (H鈦) to form its conjugate base. Removing a proton from \(\mathrm{HS}^-\) results in the conjugate base, \(\mathrm{S}^{2-}\).

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.

Acid-base reactions
In chemistry, acid-base reactions are fundamental to understanding how different substances interact. These reactions involve the transfer of protons (\(\mathrm{H}^+\)) between molecules. An acid is a substance that can donate a proton, while a base is one that can accept a proton. The interaction between acids and bases can result in the formation of water and a salt when strong acids and bases react, although this is not always the case.
When you study acid-base reactions, you often deal with the concept of conjugate acid-base pairs. A conjugate pair consists of two substances that transform into one another by the gain or loss of a proton. For example, when an acid loses a proton, it becomes its conjugate base. This relationship is crucial for predicting the outcome of reactions and understanding the concept of pH balance in solutions.
  • Acids have conjugate bases formed by losing a proton.
  • Bases have conjugate acids formed by gaining a proton.
Understanding acid-base reactions makes it easier to grasp other chemical processes and predict the behavior of substances in different conditions.
Proton donation
Proton donation is a key concept in acid-base chemistry. It refers to the process by which an acid gives away a proton (\(\mathrm{H}^+\)) to another molecule. This transaction is the essence of what defines a substance as an acid under the Br酶nsted-Lowry theory.
During proton donation, an acid transforms into its conjugate base. For example, when hydrochloric acid (\(\mathrm{HCl}\)) donates a proton, it becomes chloride (\(\mathrm{Cl}^-\)), which is its conjugate base. Similarly, ammonium ion (\(\mathrm{NH}_4^+\)) will donate a proton to become ammonia (\(\mathrm{NH}_3\)). The ability of an acid to donate a proton is what drives the overall reaction process with a base. This donation can alter the chemical environment, dictating reactions that could affect solubility, reactivity, and other chemical properties.
  • Proton donation transforms acids into their conjugate bases.
  • The strength of an acid is linked to its propensity to donate protons.
Proton donation is a useful concept because it simplifies the study of chemical reactions and helps in determining the behavior of acids and their strength.
Chemical equations
Chemical equations provide a symbolic representation of chemical reactions. Within the context of acid-base reactions, they show the reactants and products, including the transformation from acids to conjugate bases. Equations allow chemists to visualize what happens when substances react, and they must balance to reflect the conservation of mass and charge.
For example, let's consider the proton donation of hydronium ion (\(\mathrm{HS}^-\)):\[\mathrm{HS}^- \longrightarrow \mathrm{S}^{2-} + \mathrm{H}^+\]Here, the hydronium ion donates a proton, forming a sulfide ion (\(\mathrm{S}^{2-}\)) as its conjugate base. The reaction illustrates how a chemical equation captures proton transfer.
Balanced chemical equations help predict the amount of reactants required and the expected products formed.
  • Equations represent the transformation and conservation of matter in reactions.
  • Balancing equations ensures no atoms are lost or gained, aligning with the law of conservation of mass.
By understanding chemical equations, you can discern the interactions and changes substances undergo in reactions and their importance in explaining changes in chemical compositions.

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

Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(3.00\) -M acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}(a q)\), solution. Calculate the \(\mathrm{pH}\) of the resulting solution when \(2.00\) milliliters of the \(3.00-\mathrm{M}\) acetic acid is diluted to make a \(250.0\) -milliliter solution. The value of \(K_{\mathrm{a}}\) for acetic acid at \(25^{\circ} \mathrm{C}\) is \(1.8 \times 10^{-5} \mathrm{M}\)

Nitrites, such as \(\mathrm{NaNO}_{2}\), are added to processed meats and hamburger both as a preservative and to give the meat a redder color by binding to hemoglobin in the red blood cells. When the nitrite ion is ingested, it reacts with stomach acid to form nitrous acid, \(\mathrm{HNO}_{2}(a q)\), in the stomach. Given that \(K_{\mathrm{a}}=5.6 \times 10^{-4} \mathrm{M}\) for nitrous acid at \(25^{\circ} \mathrm{C}\), compute the value of the ratio \(\left[\mathrm{HNO}_{2}\right] /\left[\mathrm{NO}_{2}^{-}\right]\) in the stomach following ingestion of \(\mathrm{NaNO}_{2}(a q)\) when \(\left[\mathrm{H}_{3} \mathrm{O}^{4}\right]\) is \(0.10 \mathrm{M}\). Assume a temperature of \(25^{\circ} \mathrm{C}\).

A solution of \(\mathrm{NaOH}(a q)\) contains \(6.3\) grams of \(\mathrm{NaOH}(s)\) per \(100.0\) milliliters of solution. Calculate the \(\mathrm{pH}\) and the \(\mathrm{pOH}\) of the solution at \(25^{\circ} \mathrm{C}\).

The \(\mathrm{pH}\) of a \(0.10\) -M aqueous solution of formic acid, \(\mathrm{HCOOH}\), is \(2.38\) at \(25^{\circ} \mathrm{C}\). Calculate the value of \(K_{\mathrm{a}}\) for formic acid.

The \(\mathrm{pH}\) of the contents of the human stomach can be as low as \(1.0 .\) Calculate the value of \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in the stomach when the \(\mathrm{pH}=1.0\).

See all solutions

Recommended explanations on Chemistry Textbooks

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.