Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 44

Which of the following pairs of substances would make a good buffer solution? (A) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) and \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) (B) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{LiOH}(a q)\) (C) \(\mathrm{HCl}(a q)\) and \(\mathrm{KCl}(a q)\) (D) \(\mathrm{HF}(a q)\) and \(\mathrm{NH}_{3}(a q)\)

Short Answer

Expert verified
The pair of substances that makes a good buffer solution is (A) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) and \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\).

Step by step solution

01

Identify the weak acids in each option

\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (Option A), \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (Option B), \(\mathrm{HCl}\) (Option C), and \(\mathrm{HF}\) (Option D) are all acids. However, only \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(\mathrm{HF}\) are weak acids.
02

Identify the conjugate bases for the weak acids

For an acid, its conjugate base is formed by removing a H+ ion from it. So the conjugate base for \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\) and for \(\mathrm{HF}\) is \(\mathrm{F}^-\).
03

Match the weak acids with its conjugate base

Observing the given options carefully, the only option that comprises of a weak acid and its conjugate base is option (A), which consists of \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (weak acid) and \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (conjugate base). \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is just \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^-\) with a sodium ion, which does not affect its ability to act as a base.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Even though it is a noble gas, xenon is known to form bonds with other elements. Which element from the options below would xenon most likely be able to bond with? (A) Lithium (B) Argon (C) Fluorine (D) Carbon

Which substance would have the highest boiling point? (A) Ethanol, because it is the most asymmetrical (B) Acetone, because of the double bond (C) Ethylene glycol, because it has the most hydrogen bonding (D) All three substances would have very similar boiling points because their molar masses are similar.

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.

A gas sample with a mass of 10 grams occupies 5.0 liters and exerts a pressure of 2.0 atm at a temperature of \(26^{\circ} \mathrm{C} .\) Which of the following expressions is equal to the molecular mass of the gas? The gas constant, \(R,\) is \(0.08(\mathrm{L} \times \mathrm{atm}) / \mathrm{mol} \times \mathrm{K}\) ). (A) \((0.08)(299) \mathrm{g} / \mathrm{mol}\) (B) \(\frac{(299)(0.50)}{(2.0)(0.08)} \mathrm{g} / \mathrm{mol}\) (C) \(\frac{299}{0.08} \mathrm{g} / \mathrm{mol}\) (D) \((2.0)(0.08) \mathrm{g} / \mathrm{mol}\)

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ When the cell is connected, which of the following reactions takes place at the anode? (A) \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) (B) \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}\) (C) \(2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2}(g)\) (D) \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}+2 e^{-}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Physical 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.

Sign-up for free