/*! 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} Q42P Calculate the pH of a solution p... [FREE SOLUTION] | 91影视

91影视

Find study content
Learning Materials

Discover learning materials by subject, university or textbook.

Explanations

Textbooks
All Subjects

Anthropology

Archaeology

Architecture

Art and Design

Bengali

Biology

Business Studies

Greek

Chemistry

Chinese

Combined Science

Computer Science

Economics

Engineering

English

English Literature

Environmental Science

French

Geography

German

History

Hospitality and Tourism

Human Geography

Italian

Japanese

Law

Macroeconomics

Marketing

Math

Media Studies

Medicine

Microeconomics

Music

Nursing

Nutrition and Food Science

Physics

Polish

Politics

Psychology

Religious Studies

Sociology

Spanish

Sports Sciences

Translation

All Subjects

Biology

Business Studies

Chemistry

Combined Science

Computer Science

Economics

English

Environmental Science

Geography

History

Math

Physics

Psychology

Sociology
Features
Features

Discover all of these amazing features with a free account.

Flashcards

91影视 AI

Notes

Study Plans

Study Sets
What鈥檚 new?

Flashcards
Study your flashcards with three learning modes.

Study Sets
All of your learning materials stored in one place.

Notes
Create and edit notes or documents.

Study Plans
Organise your studies and prepare for exams.
91影视
Discover

All the hacks around your studies and career - in one place.

Find a degree

Magazine
Featured

Magazine
Trusted advice for anyone who wants to ace their studies & career.

The largest student job board with the most exciting opportunities.

Verified student deals from top brands.

Discover our mobile app to take your studies anywhere.

Learning Materials

Features

Discover

Chapter 9: Q42P (page 209)

Calculate the pH of a solution prepared by mixing $0.0800 mol$ of chloroacetic acid plus $0.0400 mol$ of sodium chloroacetate in $1.00 L$ of water.
(a) First do the calculation by assuming that the concentrations of HA and $A^{-}$ equal their formal concentrations.
(b) Then do the calculation, using the real values of and in the solution.
(c) Using first your head, and then the Henderson-Hasselbalch equation, find the pH of a solution prepared by dissolving all the following compounds in one beaker containing a total volume of $1.00 L : 0.180 {molCICHCOO}_{2} H, 0.020 mol {CICHCHO}_{2} {Na}_{2} 0.080 {molHNO}_{3,}$ and $0.080 molCa {(OH)}_{2}$ . Assume that $Ca {(OH)}_{2}$ dissociates completely.

Short Answer

Expert verified

a) The pH is $2.56$ .

b) The pH is $2.61$ .

c) The pH is $2.86$ .

Step by step solution

01

Concept used.

The pH of the buffer determined by the concentration ratio of $[A^{-}]$ and $[HA]$ by the following Henderson-Hasselbalch equation: $pH = {pK}_{a} + \log ([A^{-}] / [HA])$

02

First do the calculation by assuming that the concentrations of HA and A- equal their formal concentrations.

a)

The pH,

$pH = {pK}_{a} + \log ([A^{-}] / [HA]) pH = {pK}_{a} + \log ([sodium chloroacetate] / [chloroacetic acod]) pH = 2.865 + \log (0.04 / 0.08) = 2.56$

03

The calculation using the real values of HA and A- in the solution.

b)

Calculate the pH by using the real values of $[HA]$ and $[A^{-}]$ :

$K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]} = \frac{[H^{+}] (0.04 - [H^{+}]}{0.08 - [H^{+}]} [H^{+}] = 2.48 脳 10^{3} M pH = - \log [H^{+}] = - \log (2.48 脳 10^{3}) = 2.61$

04

The pH of a solution prepared by dissolving all the following compounds in one beaker.

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!

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

Which of the following bases would be most suitable for preparing a buffer of $pH 9.00$ ?

(i) ${NH}_{3}$ (ammonia, $K_{b} = 1.76 脳 10^{- 5}$ );

(ii) role="math" localid="1654764490555" $C_{6} H_{5} {NH}_{2}$ (aniline,role="math" localid="1654764515634" $C_{6} H_{5} {NH}_{2}$ );

(iii) $H_{2} {NNH}_{2} (hydrazine, K_{b} = 1.05 脳 10^{- 6}))$ );

(iv) $C_{6} H_{s} {NH}_{2}$ (pyridine, $K_{b} = 1.58 脳 10^{- 9})$ ).

When is a weak acid weak and when is a weak acid strong? Show that the weak acid $99 %$ dissociatedrole="math" localid="1654854665468" $F = (0.0102) K_{a}$ will be 92%dissociated when dissolved in water if the formal concentration is one-tenth $(F = Ka / 10)$ of Show that the fraction of dissociation is 27%when $F = 10 Ka$ . At what formal concentration will the acid be 99%dissociatedCompare your answer with the left -hand curve in Figure 9-2

(a) Write the chemical reactions whose equilibrium constants are $^{K_{b}}$ and $^{K_{a}}$ for imidazole and imidazole hydrochloride respectively.

(b) Calculate the pH of a solution prepared by mixing $^{1.00 g}$ of imidazole with $^{1.00 g}$ of imidazole hydrochloride and diluting torole="math" localid="1655093511290" $^{100.0 mL}$ .

(c) Calculate the pH of the solution if $^{2.30 mL}$ of $^{1.07 {MHCIO}_{4}}$ are added.

(d) How many milliliters of $^{1.07 {MHCIO}_{4}}$ should be added to of imidazole to give a pH ofrole="math" localid="1655093364034" $^{6.993}$ ?

Write the chemical reaction whose equilibrium constant is

(a) $K_{a}$ for benzoic acid, $C_{6} H_{5} {CO}_{2} H$

(b) $K_{b}$ for benzoate ion, $C_{6} H_{5} {CO}_{2}^{2 -}$

(c) $K_{b}$ for aniline, $C_{6} H_{5} {NH}_{2}$

(d) $K_{a}$ for anilinium ion, $C_{6} H_{5} {NH}_{3}^{+}$

A solution contains 63 different conjugate acid-base pairs.

Among them is acrylic acid and acrylate ion, with the equilibrium ratio [acrylate]/[acrylic acid]5 0.75. What is the pH of the solution?

$H_{2} C = {CHCO}_{2} H {pK}_{a} = 4.25$

See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

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.