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Chapter 5: Problem 103

The highest \(\left[\mathrm{H}^{+}\right]\) will be found in an aqueous solution that is (a) \(0.10 \mathrm{M} \mathrm{HCl} ;\) (b) \(0.10 \mathrm{M} \mathrm{NH}_{3} ;\) (c) \(0.15 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH} ;(\mathrm{d}) 0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\).

Short Answer

Expert verified
The solution with the highest \(\left[\mathrm{H}^{+}\right]\) concentration is (d) 0.10 M H\( _{2}\)SO\( _{4}\).

Step by step solution

01

Understanding the substances

First, familiarize yourself with each substance and their ability to contribute hydrogen ions to the solution. (a) HCl is a strong acid and dissociates completely in solution to give H\(^{+}\) and Cl\(^{-}\). (b) NH\( _{3}\) is a weak base and it does not donate H\(^{+}\) ions, but instead accepts H\(^{+}\) to form NH\( _{4}^{+}\), decreasing the H\(^{+}\) concentration. (c) CH\( _{3}\)COOH is a weak acid and partially dissociates to give H\(^{+}\) and CH\( _{3}\)COO\(^{-}\). (d) H\( _{2}\)SO\( _{4}\) is a strong acid that can donate two H\(^{+}\) ions, one from each hydrogen atom.
02

Evaluating the \(\left[\mathrm{H}^{+}\right]\) concentration provided by each solution

Understand that the strong acids HCl and H\( _{2}\)SO\( _{4}\) will contribute to higher \(\left[\mathrm{H}^{+}\right]\) concentrations than the weak acid and the weak base. Since H\( _{2}\)SO\( _{4}\) is a strong acid and can donate two H\(^{+}\) ions, it will yield a higher concentration than HCl which only donates one H\(^{+}\) ion. NH\( _{3}\) and CH\( _{3}\)COOH will produce less \(\left[\mathrm{H}^{+}\right]\) than the strong acids.
03

Comparing the concentrations and picking the highest

After understanding that strong acids contribute more to \(\left[\mathrm{H}^{+}\right]\) than weak acids and bases, and that H\( _{2}\)SO\( _{4}\) contributes more H\(^{+}\) than HCl, it is safe to say that the highest concentration of \(\left[\mathrm{H}^{+}\right]\) will be found in a 0.10 M H\( _{2}\)SO\( _{4}\) solution.

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Most popular questions from this chapter

In this chapter, we described an acid as a substance capable of producing \(\mathrm{H}^{+}\) and a salt as the ionic compound formed by the neutralization of an acid by a base. Write ionic equations to show that sodium hydrogen sulfate has the characteristics of both a salt and an acid (sometimes called an acid salt).

Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

A neutralization reaction between an acid and a base is a common method of preparing useful salts. Give net ionic equations showing how the following salts could be prepared in this way: (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4};\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3} ;\) and \((\mathrm{c})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Blood alcohol content (BAC) is often reported in weight-volume percent (w/v\%). For example, a BAC of \(0.10 \%\) corresponds to \(0.10 \mathrm{g} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) per 100 mL of blood. Estimates of BAC can be obtained from breath samples by using a number of commercially available instruments, including the Breathalyzer for which a patent was issued to R. F. Borkenstein in 1958\. The chemistry behind the Breathalyzer is described by the oxidation- reduction reaction below, which occurs in acidic solution: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{g})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq}) \quad(\text { not balanced })\)A Breathalyzer instrument contains two ampules, each of which contains \(0.75 \mathrm{mg} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) dissolved in \(3 \mathrm{mL}\) of \(9 \mathrm{mol} / \mathrm{L} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) .\) One of the ampules is used as reference. When a person exhales into the tube of the Breathalyzer, the breath is directed into one of the ampules, and ethyl alcohol in the breath converts \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) into \(\mathrm{Cr}^{3+} .\) The instrument compares the colors of the solutions in the two ampules to determine the breath alcohol content (BrAC), and then converts this into an estimate of BAC. The conversion of BrAC into BAC rests on the assumption that 2100 mL of air exhaled from the lungs contains the same amount of alcohol as \(1 \mathrm{mL}\) of blood. With the theory and assumptions described in this problem, calculate the molarity of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in the ampules before and after a breath test in which a person with a BAC of \(0.05 \%\) exhales 0.500 Lof his breath into a Breathalyzer instrument.
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