/*! 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 102 For solutions of a weak acid, a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 102

For solutions of a weak acid, a graph of \(\mathrm{pH}\) versus the logarithm of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

Short Answer

Expert verified
The magnitude of the slope of the line for a graph of pH versus the logarithm of the initial acid concentration of a weak acid is given by \( |m| = \frac{1}{2} \log K_a \) where K_a is the acid dissociation constant.

Step by step solution

01

Write the equilibrium expression for the weak acid dissociation

For a generic weak acid (HA), its equilibrium dissociation in water can be represented as follows: \( HA + H_2 O \rightleftharpoons H_3O^{+} + A^{-} \) At equilibrium, we can define the acid dissociation constant (Ka) as the ratio of the concentration of the products to the concentration of the reactants (excluding water): \( K_a = \frac{[H_3O^+ ][A^- ]}{[HA]} \)
02

Define the variables for a weak acid dissociation

Let the initial concentration of the weak acid be represented by "c" moles/L, and let the change in concentration due to dissociation be "x" moles/L. At equilibrium, the concentrations of different species are as follows: - HA: \( c-x \) moles/L - \( H_3O^+ \): \( x \) moles/L - \( A^- \): \( x \) moles/L Now, substitute these concentration values into the Ka expression: \( K_a = \frac{x^2}{c-x} \)
03

Use the definition of pH and relationship with initial concentration

The pH of a solution is defined as the negative logarithm of the H鈦�(or \( H_3O^+ \))concentration: \( pH = -\log [H_3O^+] \) Since \( [H_3O^+] = x \), we can rewrite the pH equation as: \( pH = -\log x \) The logarithm of the initial concentration is given by \( \log c \). The problem states that there is a linear relationship between pH and \( \log c \), which means that we can represent it as the following equation: \( pH = m\log c + b \) where m is the slope of the line and b is the y-intercept.
04

Rearrange equations to express dissociation in terms of initial concentration

First, solve the pH equation for "x": \( x = 10^{-pH} \) Now, substitute this expression for "x" in the Ka expression: \( K_a = \frac{(10^{-pH})^2}{c-10^{-pH}} \) Next, solve this last equation for pH: \( pH = -\log \sqrt{\frac{K_a(c-10^{-pH})}{c}} \)
05

Calculate the slope of the line

Now that we have the relationship between pH and initial concentration, we can identify the slope of the line (m) by comparing it to the equation mentioned in step 3. The slope of the line is given by the coefficient of \( \log c \), which in this case is: \(m = -\log \sqrt{\frac{K_a}{c}} = -\frac{1}{2} \log K_a\) Thus, the magnitude of the slope of the line is: \( |m| = \frac{1}{2} \log K_a \)

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

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right)\), a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). What is the \(\mathrm{pH}\) of a solution obtained by dissolving two extra-strength aspirin tablets,

What is the pH of a solution that is \(2.5 \times 10^{-9} \mathrm{M}\) in \(\mathrm{NaOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of \(9.95\). Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\). Calculate \(\left[\mathrm{H}^{+}\right]\)and \(\left[\mathrm{OH}^{-}\right]\)for a neutral solution at this temperature.

(a) Give the conjugate base of the following Bronsted-Lowry acids: (i) \(\mathrm{HIO}_{3}\), (ii) \(\mathrm{NH}_{4}^{+}\). (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-}\), (ii) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

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.