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 solutions is a buffer? (a) 0.10\(M\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(0.10 \mathrm{MCH}_{3} \mathrm{CONa},(\mathbf{b}) 0.10 \mathrm{MCH}_{3} \mathrm{COOH}\) (c) 0.10 \(\mathrm{M} \mathrm{HCl}\) and \(0.10 \mathrm{M} \mathrm{NaCl},(\mathbf{d})\) both a and \(\mathrm{c},(\mathbf{e})\) all of a, \(\mathrm{b},\) and \(\mathrm{c} .\)

Short Answer

Expert verified
The correct answer is (a) 0.10 M CH3COOH and 0.10 M CH3COONa, as it consists of a weak acid (acetic acid) and its conjugate base (acetate).

Step by step solution

01

Identify the components of each solution

In the given exercise, we are given the following solutions: (a) 0.10 M CH3COOH and 0.10 M CH3COONa (b) 0.10 M CH3COOH (c) 0.10 M HCl and 0.10 M NaCl (d) Both a and c (e) All of a, b, and c We now need to identify the components of each solution and determine if they can form a buffer.
02

Analyze the solutions

(a) 0.10 M CH3COOH (acetic acid) and 0.10 M CH3COONa (sodium acetate): In this solution, CH3COOH is a weak acid and CH3COO- (from the dissociation of CH3COONa) is its conjugate base. Thus, this solution can act as a buffer. (b) 0.10 M CH3COOH (acetic acid): In this solution, there is only a weak acid present, and no conjugate base is available. Therefore, this solution cannot act as a buffer. (c) 0.10 M HCl (hydrochloric acid) and 0.10 M NaCl (sodium chloride): In this solution, HCl is a strong acid and completely dissociates into H+ and Cl-. Since HCl is not a weak acid, this solution cannot be a buffer.
03

Choose the correct answer

Based on our analysis in step 2, we can conclude that: (a) is a buffer (b) is not a buffer (c) is not a buffer Thus, the correct answer is (a) 0.10 M CH3COOH and 0.10 M CH3COONa.

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.

Acid-Base Equilibrium
Acid-base equilibrium is a fundamental concept in chemistry that deals with the balance between acids and bases in a solution. This balance is crucial for buffer solutions, which maintain a stable pH despite the addition of small amounts of acid or base.
To understand acid-base equilibrium, it is essential to know that acids donate protons (\( H^+ \)) while bases accept them. For a buffer, this equilibrium is achieved by having a weak acid and its conjugate base present in solution. The weak acid can donate \( H^+ \), while the conjugate base can accept \( H^+ \), thus resisting changes in pH.
A classic example of acid-base equilibrium is seen in the acetic acid (\( ext{CH}_3 ext{COOH} \)) and sodium acetate (\( ext{CH}_3 ext{COONa} \)) buffer system. Here, acetic acid partially dissociates to produce its conjugate base acetate (\( ext{CH}_3 ext{COO}^- \)), which maintains equilibrium by reacting with added acids or bases.
  • If a strong acid is added, the acetate ions react with the \( H^+ \) ions, neutralizing them.
  • If a strong base is added, the acetic acid donates \( H^+ \) to neutralize the hydroxide ions (\( ext{OH}^- \)).
This balance ensures that the pH of the solution remains relatively constant, which is vital for many chemical processes and biological functions.
Conjugate Acid-Base Pairs
Conjugate acid-base pairs are pairs of molecules or ions that are connected through the gain or loss of a proton (\( H^+ \)). In any acid-base reaction, an acid is transformed into its conjugate base, while a base is transformed into its conjugate acid.
For example, in the acetic acid-sodium acetate buffer system, acetic acid (\( ext{CH}_3 ext{COOH} \)) acts as the acid. When it donates a proton, it forms its conjugate base, acetate (\( ext{CH}_3 ext{COO}^- \)). Conversely, acetate can act as a base by accepting a proton to reform acetic acid.
Understanding conjugate acid-base pairs is crucial in recognizing how buffers stabilize pH. When a buffer solution is disturbed by the addition of a strong acid or base:
  • The conjugate base absorbs excess \( H^+ \) ions from strong acids, preventing significant pH changes.
  • The conjugate acid provides \( H^+ \) ions if a strong base is added, again stabilizing the pH.
This dynamic operation of conjugate pairs ensures that buffers effectively moderate pH changes, making them invaluable in both laboratory and biological contexts.
Weak Acids and Strong Acids
Weak and strong acids differ in their ability to dissociate in water. This distinction is key in understanding their roles in buffers. Weak acids, such as acetic acid, do not fully dissociate in water, which allows them to play a crucial role in buffer systems.
Weak Acids:
  • Partially dissociate in solution.
  • Exist in equilibrium with their conjugate base.
  • Suitable for buffers as their dissociation is reversible, allowing equilibrium to be maintained.
Strong acids, such as hydrochloric acid (\( ext{HCl} \)), behave differently. They fully dissociate in water, releasing all their protons.
  • This complete dissociation means they can't form effective buffers because there is no reverse reaction to restore equilibrium.
The inefficacy of strong acids in buffer solutions is evident in the fact that they do not provide the equilibrium stability that weak acids and their conjugate bases do. Thus, understanding the role of weak and strong acids is fundamental in selecting components for creating an effective buffer system.

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 solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2},\) is the same, \(4 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) . (a) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?\) (c) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2},\) what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

How many milliliters of 0.0850\(M \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) 40.0 \(\mathrm{mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3},\) (\mathbf{b} ) 35.0 \(\mathrm{mL}\) of \(0.0850 M \mathrm{CH}_{3} \mathrm{COOH},(\mathbf{c}) 50.0 \mathrm{mL}\) of a solution that contains 1.85 \(\mathrm{g}\) of \(\mathrm{HCl}\) per liter?

A 1.00 -L. solution saturated at \(25^{\circ} \mathrm{C}\) with lead(lI) iodide contains 0.54 \(\mathrm{g}\) of \(\mathrm{Pbl}_{2}\) . Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\) .

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) \((\mathbf{b} ) \)The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

A 20.0 -mL sample of 0.150 \(\mathrm{M} \mathrm{KOH}\) is titrated with 0.125 \(\mathrm{M}\) \(\mathrm{HClO}_{4}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of acid have been added: (a) \(20.0 \mathrm{mL},\) (b) 23.0 \(\mathrm{mL}\) \((\mathbf{c}) 24.0 \mathrm{mL},(\mathbf{d}) 25.0 \mathrm{mL},(\mathbf{e}) 30.0 \mathrm{mL}\)

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