Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the following will decrease with dilution at a given temperature? (a) \(\mathrm{pH}\) of \(10^{-3} \mathrm{M}\) acetic acid solution (b) \(\mathrm{pH}\) of \(10^{-3} \mathrm{M}\) aniline solution (c) degree of dissociation of \(10^{-3} \mathrm{M}\) acetic acid (d) degree of dissociation of \(10^{-3} \mathrm{M}\) aniline solution

Short Answer

Expert verified
Dilution will decrease the pH of a 10^{-3} M acetic acid solution (a) and increase the degree of dissociation for both acetic acid (c) and aniline (d), but pH will increase for aniline solution (b).

Step by step solution

01

Understanding Acetic Acid Solution

Acetic acid is a weak acid. The degree of dissociation for weak acids increases with dilution, leading to an increase in hydrogen ion concentration. Thus, pH decreases. For option (a), diluting the solution will increase the degree of dissociation and consequently decrease the pH.
02

Understanding Aniline Solution

Aniline is a weak base. Similar to weak acids, the degree of dissociation for a weak base increases with dilution, leading to an increase in hydroxide ion concentration. This results in an increase in pH. Therefore, for option (b), diluting the aniline solution would decrease the hydrogen ion concentration, which means an increase in pH, not a decrease.
03

Determining the Degree of Dissociation of Acetic Acid

Dilution increases the degree of dissociation for weak acids since the equilibrium shifts to compensate for the increase in the volume of the solution. Hence, option (c) the degree of dissociation of the acetic acid will increase, not decrease.
04

Determining the Degree of Dissociation of Aniline

Similarly, as with acetic acid, dilution of weak bases including aniline leads to an increased degree of dissociation. Therefore, for option (d), the degree of dissociation of aniline also increases with dilution.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acetic Acid Dissociation
Acetic acid (\(CH_3COOH\)) is known as a weak acid, which means it does not fully dissociate in water. The dissociation of acetic acid in water can be represented by the chemical equation: \( CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \). The weak dissociation is quantified by the acid dissociation constant (\(K_a\)), which is a reflection of the acid's strength.

Upon dilution, the equilibrium of the dissociation reaction shifts to the right according to Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed, the system adjusts to restore the equilibrium. This shift results in an increased degree of dissociation as the reaction seeks to replace ions that have been 'spread out' by the additional water, resulting in more hydrogen ions (\(H^+\) ions) being produced. Consequently, as the concentration of hydrogen ions in the solution increases, the pH of the solution decreases, becoming more acidic.
Aniline Solution Behavior
Aniline (\(C_6H_5NH_2\) is categorized as a weak base, meaning it doesn't completely ionize in solution. Its behavior in water is noted by the reaction: \( C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^- \). Upon dilution, the system will shift to produce more hydroxide ions (\(OH^-\) ions) to restore equilibrium, thus increasing the degree of dissociation.

Because the concentration of hydroxide ions in the solution rises, the solution becomes less acidic and more basic, which is reflected in an increase in pH value. The pH is a measure of the hydrogen ion concentration; for bases, the relevant factor is the presence of hydroxide ions, which reduce hydrogen ion concentration indirectly by their reaction with hydrogen ions to form water (\(H^+ + OH^- \rightarrow H_2O\) ), leading to an increase in pH upon dilution due to the decrease in hydrogen ion concentration.
Weak Acid and Base Dilution
When diluting solutions of weak acids or bases, the key concept to understand is that the degree of dissociation, which is the fraction of the solute molecules that dissociate into ions, changes. For both weak acids and bases, dilution leads to an increase in their degree of dissociation. This phenomenon occurs due to the system's response to the change in concentration—the equilibrium shifts to produce more ions to counteract the increased volume.

Diluting a weak acid increases the ratio of its dissociated ions, thus increasing its acidity, while diluting a weak base increases the ratio of its dissociated ions, thus increasing its basicity. It’s critical to distinguish between the effects of dilution on strong acids/bases, which are completely dissociated at all concentrations, and weak acids/bases, for which dilution has a notable effect on dissociation and consequently on the pH of the solution.
pH Calculation
pH is a measure of the acidity or basicity of an aqueous solution and is calculated as the negative logarithm of the hydrogen ion concentration (\( pH = -\text{log}[H^+] \)). The pH scale typically ranges from 0 to 14 in aqueous solutions, with 7 being neutral. A pH less than 7 indicates an acidic solution, while a pH greater than 7 indicates a basic solution.

The calculation of pH for weak acids or bases takes into account their dissociation constant (\(K_a\) for acids, and \(K_b\) for bases) and the initial concentration of the solute. Since weak acids and bases do not completely dissociate, these constants are necessary to determine the equilibrium concentration of ions. For diluted weak acids or bases, one must first calculate the new degree of dissociation according to the dilution before finding the pH. This process often requires the use of the equilibrium constant expression and ICE (Initial, Change, Equilibrium) tables to find the concentrations of the ions at equilibrium required to calculate pH.

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

Select the best indicator from the given table for titration of \(20 \mathrm{~mL}\) of \(0.02 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) with \(0.02 \mathrm{M} \mathrm{NaOH}\). Given \(\mathrm{pK}_{a}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.74\) \begin{tabular}{r|l|c|} \hline & Indicator & pH range \\ (I) & Bromothymol blue & \(6.0-7.6\) \\ (II) & Thymolphthalein & \(9.3-10.5\) \\ (III) & Malachite green & \(11.4-13\) \\ (IV) & M-Cresol purple & \(7.4-9.0\) \\ \hline \end{tabular} (a) I (b) II (c) III (d) IV

When \(100 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaCN}\) solution is titrated with \(0.1 \mathrm{M} \mathrm{HCl}\) solution the variation of \(\mathrm{pH}\) of solution with volume of \(\mathrm{HCl}\) added will be : (a) (b) (c) (d)

\(\mathrm{pK}_{a}\) of \(\mathrm{NH}_{4}^{+}\) is \(9.26 .\) Hence, effective range for \(\mathrm{NH}_{4} \mathrm{OH}-\mathrm{NH}_{4} \mathrm{Cl}\) buffer is abouts (a) \(8.26\) to \(10.26\) (b) \(4.74\) to \(5.74\) (c) \(3.74\) to \(5.74\) (d) \(8.26\) to \(9.26\)

Given \(K_{a}\) values of \(5.76 \times 10^{-10}\) and \(4.8 \times 10^{-10}\) for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{HCN}\) respectively. What is the equilibrium constant for the following reaction? $$ \mathrm{NH}_{4}^{+}(a q .)+\mathrm{CN}^{-}(a q .) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCN}(a q .) $$ (a) \(0.83\) (b) \(1.2\) (c) \(8.0 \times 10^{-11}\) (d) \(27.6 \times 10^{-10}\)

What is \(\left[\mathrm{NH}_{4}^{+}\right]\) in a solution that contain \(0.02 \mathrm{M} \mathrm{NH}_{3}\left(K_{b}=1.8 \times 10^{-5}\right)\) and \(0.01 \mathrm{M}\) ' \(\mathrm{KOH}\) ? (a) \(9 \times 10^{-6}\) (b) \(1.8 \times 10^{-5}\) (c) \(3.6 \times 10^{-5}\) (d) None of these

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.

Sign-up for free