/*! 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 6 The conjugate base of \(\mathrm{... [FREE SOLUTION] | 91影视

91影视

The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is: (a) \(\mathrm{HPO}_{4}^{2-}\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (c) \(\mathrm{PO}_{4}^{3-}\) (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\)

Short Answer

Expert verified
The conjugate base of \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) is \( \mathrm{HPO}_{4}^{2-} \) (option a).

Step by step solution

01

Understand the Concept of a Conjugate Base

A conjugate base is formed when an acid donates a proton (H鈦). In this context, you need to identify which species remains after a proton is removed from \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \).
02

Identify the Acid and its Components

The species \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) is the acid, and it contains two hydrogen atoms. When one hydrogen atom is removed as a proton, you are left with \( \mathrm{HPO}_{4}^{2-} \).
03

Apply the Concept to Find the Conjugate Base

When \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) loses one proton (H鈦), it becomes \( \mathrm{HPO}_{4}^{2-} \). This is the conjugate base formed from the acid \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \).
04

Verify by Checking Other Options

Verify the remaining options:- \( \mathrm{H}_{3} \mathrm{PO}_{4} \) is a form of the acid gained by adding a proton, not the conjugate base.- \( \mathrm{PO}_{4}^{3-} \) results from losing two protons, not just one.- \( \mathrm{P}_{2} \mathrm{O}_{5} \) is a completely unrelated chemical species.

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 processes where an acid donates a proton (H鈦) to a base. This interaction results in the formation of conjugate acid-base pairs. To better understand, imagine acids as donors that give away parts of their molecular structure, while bases are the recipients that accept these parts. This transfer of protons is the essence of acid-base chemistry.
  • Acid: A substance that can donate a proton to another substance.
  • Base: A substance that can accept a proton.
A typical acid-base reaction can be represented as: \[\text{Acid} + \text{Base} \rightleftharpoons \text{Conjugate Base} + \text{Conjugate Acid}\] This bidirectional arrow shows that these reactions can proceed in both directions and reach a state of equilibrium. Identifying the conjugate base formed in these reactions is crucial in many chemical applications.
Proton Donation
The concept of proton donation is central to understanding acid-base reactions. When we talk about acids donating protons, we're referring to the acids losing an H鈦 ion. This process transforms the acid into its conjugate base. For instance, in our original exercise, the species \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) is the acid.
  • As \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) donates a proton, it loses an H鈦 ion.
  • The remaining species is \( \mathrm{HPO}_{4}^{2-} \), which is its conjugate base.
The reaction involved can be written mathematically: \[\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \text{H}^+\] Proton donation is a dynamic and reversible process where the acid can potentially regain its lost proton by interacting with a base. Identifying and understanding which species arises from proton donation enables chemists to predict reaction behaviors.
Chemical Species Identification
Identifying chemical species is a critical skill in chemistry, as it helps in predicting and explaining reaction outcomes. When looking at chemical equations, it is essential to precisely identify each species involved to understand the overall process. In determining the conjugate base of \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \), recognizing the correct transformation helps in accurate identification.
  • \( \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \): The acid initially present.
  • \( \mathrm{HPO}_{4}^{2-} \): Product when the acid donates a proton, becoming the conjugate base.
Other options in the exercise illustrate how identifying species precisely prevents mistakes:
  • \( \mathrm{H}_{3} \mathrm{PO}_{4} \) is merely an addition of a proton, showing an acid form rather than a base.
  • \( \mathrm{PO}_{4}^{3-} \) involves losing two protons, thus a different scenario.
  • \( \mathrm{P}_{2} \mathrm{O}_{5} \) is unrelated, showcasing the importance of recognizing correct pairs.
By accurately identifying chemical species, students and chemists are ensured to understand reactions correctly, leading to more effective learning and application.

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

For the \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), the initial mole ratio of \(\mathrm{N}_{2}: \mathrm{H}_{2}\) is \(1: 3 .\) If at equilibrium only \(50 \%\) has reacted and equilibrium pressure is \(\mathrm{P}\). Find the value of \(\mathrm{P}_{\mathrm{NH}_{3}}\) at equilibrium. (a) \(\frac{\mathrm{P}}{3}\) (b) \(\frac{\mathrm{P}}{5}\) (c) \(\frac{\mathrm{P}}{9}\) (d) \(\frac{\mathrm{P}}{6}\)

The ratio of \(\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{c}}\) for the reaction: \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) is (a) 1 (b) RT (c) \((\mathrm{RT})^{1 / 2}\) (d) \((\mathrm{RT})^{-1 / 2}\)

For the hypothetical reactions, the equilibrium constant \((\mathrm{K})\) values are given: \(\mathrm{A} \rightleftharpoons \mathrm{B} \mathrm{K}_{1}=2\) \(\mathrm{B} \rightleftharpoons \mathrm{C} \mathrm{K}_{2}^{1}=4\) \(\mathrm{C} \rightleftharpoons \mathrm{D} \mathrm{K}_{3}^{2}=3\) The equilibrium constant \((\mathrm{K})\) for the reaction: \(\mathrm{A} \rightleftharpoons \mathrm{D}\) is (a) 3 (b) 6 (c) 12 (d) 24

For the reversible reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) At \(500^{\circ} \mathrm{C}\), the value of \(\mathrm{K}_{\mathrm{p}}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(\mathrm{K}_{\mathrm{c}}\), with concentration in mole \(\mathrm{L}^{-1}\), is: (a) \(1.44 \times 10^{-5} /(0.082 \times 500)^{-2}\) (b) \(1.44 \times 10^{-5} /(8.314 \times 773)^{-2}\) (c) \(1.44 \times 10^{-5}(0.082 \times 773)^{2}\) (d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\)

Le Chatelier's Principle is applicable to: (a) Heterogenous reaction (b) Homogenous reaction (c) Irreversible reaction (d) System in equilibrium

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.