Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 4

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. Which of the following would create a good buffer when dissolved in formic acid? (A) \(\mathrm{NaCO}_{2} \mathrm{H}\) (B) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (C) \(\mathrm{NH}_{3}\) (D) \(\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The only compound among the given options that would create a good buffer when it is dissolved in formic acid is \(\mathrm{NaCO}_{2} \mathrm{H}\) (option A).

Step by step solution

01

Analyze the given compounds

We are given four compounds: \(\mathrm{NaCO}_{2} \mathrm{H}\), \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), \(\mathrm{NH}_{3}\), and \(\mathrm{H}_{2} \mathrm{O}\) . We need to see which of these when dissolved in formic acid will produce its conjugate base \(\mathrm{HCO}_{2}^{-}\). Dissolving \(\mathrm{NaCO}_{2} \mathrm{H}\) in water will produce \(\mathrm{HCO}_{2}^{-}\) because \(\mathrm{Na}\)+ ions will dissociate leaving behind \(\mathrm{HCO}_{2}^{-}\) ions.
02

Exclude other compounds

Dissolving \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) , \(\mathrm{NH}_{3}\), and \(\mathrm{H}_{2} \mathrm{O}\) in water will not produce a good buffer with formic acid because they do not produce the conjugate base \(\mathrm{HCO}_{2}^{-}\). Therefore, they can be excluded.

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

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE The phosphorus ion is larger than a neutral phosphorus atom, yet a zinc ion is smaller than a neutral zinc atom. Which of the following statements best explains why? (A) The zinc atom has more protons than the phosphorus atom. (B) The phosphorus atom is paramagnetic, but the zinc atom is diamagnetic. (C) Phosphorus gains electrons when forming an ion, but zinc loses them. (D) The valence electrons in zinc are further from the nucleus than those in phosphorus.

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ How many moles of electrons must be transferred to create 127 g of copper? (A) 1 mole of electrons (B) 2 moles of electrons (C) 3 moles of electrons (D) 4 moles of electrons

\(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{PCl}_{5}(g) \Delta H=-92.5 \mathrm{kJ} / \mathrm{mol}\) In which of the following ways could the reaction above be manipulated to create more product? (A) Decreasing the concentration of \(\mathrm{PCl}_{3}\) (B) Increasing the pressure (C) Increasing the temperature (D) None of the above

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. Based on the given reduction potentials, which of the following would lead to a reaction? (A) Placing some \(\operatorname{Cr}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions (B) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (C) Placing some \(\mathrm{Cr}(s)\) in a solution containing \(\mathrm{Cr}^{3+}\) ions (D) Placing some \(\mathrm{Pb}(s)\) in a solution containing \(\mathrm{Pb}^{2+}\) ions

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

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