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Chapter 5: Problem 51

Which of the following is not a buffer solution? a. \(0.05 \mathrm{M} \mathrm{KClO}_{4}+0.05 \mathrm{M} \mathrm{HClO}_{4}\) b. \(2 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+2 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5}{ }^{+} \mathrm{NH}_{3} \mathrm{Br}\) c. \(0.8 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}+0.8 \mathrm{M} \mathrm{KHS}\) d. \(3 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}+3 \mathrm{M} \mathrm{KHCO}_{3}\)

Short Answer

Expert verified
Option a is not a buffer solution.

Step by step solution

01

Understand What a Buffer Solution Is

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. This typically involves a weak acid and its conjugate base or a weak base and its conjugate acid.
02

Analyze Option A

Option a describes a mixture of a strong acid, HClO鈧�, and its salt, KClO鈧�. Since both are related to a strong acid, they do not form a buffer solution because strong acids completely ionize in solution.
03

Analyze Option B

Option b involves aniline (C鈧咹鈧匩H鈧�), a weak base, and aniline hydrochloride (C鈧咹鈧匩H鈧冣伜 Br鈦�), its conjugate acid form. This fits the criteria for a buffer solution.
04

Analyze Option C

Option c includes H鈧係, a weak acid, and KHS, its salt form (providing HS鈦�, the conjugate base). This combination forms a buffer solution.
05

Analyze Option D

Option d consists of H鈧侰O鈧� (carbonic acid) and KHCO鈧� (bicarbonate), both of which form a classic buffer system for the carbonate ion. This fits the buffer criteria.
06

Identify the Non-Buffer Solution

Since option a, involving a strong acid and its salt, does not meet the criteria for a buffer (weak acid/base with its conjugate), it is not a buffer solution.

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Key Concepts

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

Strong Acids and Bases
In the world of chemistry, acids and bases play crucial roles. A strong acid is one that completely dissociates in water. This means that when you add it to water, all the acid molecules break apart into ions. For instance, perchloric acid (\(\text{HClO}_4\)) is a strong acid. It dissociates entirely to produce hydrogen ions (\(\text{H}^+\)) and perchlorate ions (\(\text{ClO}_4^-\)). Likewise, strong bases dissociate completely in water. Their primary role is to donate hydroxide ions (\(\text{OH}^-\)).
In buffer solutions, strong acids or bases are generally not used. This is because their complete ionization doesn鈥檛 allow for the pH stability that buffers provide. In the given exercise, the choice involving \(\text{HClO}_4\) as a part of the mixture cannot form a buffer because it is a strong acid and does not have the ability to resist changes in pH.
Conjugate Acid-Base Pairs
Conjugate acid-base pairs are fundamental in understanding how buffers work. When an acid donates a proton (\(\text{H}^+\)), it turns into its conjugate base. Conversely, when a base gains a proton, it becomes its conjugate acid. This pairing is crucial for the mechanism of buffer solutions.
Take, for example, aniline (\(\text{C}_6 ext{H}_5 ext{NH}_2\)) and its conjugate acid, anilinium ion (\(\text{C}_6 ext{H}_5 ext{NH}_3^+\)). When a strong base is added to the buffer solution, the anilinium ion donates a proton to neutralize it. When a strong acid is added, aniline binds with the extra hydrogen ions to maintain stable pH. This dynamic interaction is why conjugate acid-base pairs are essential in buffers.
In the exercise, options \(\text{b}\) and \(\text{d}\) effectively demonstrate the use of conjugate pairs (\(\text{C}_6 ext{H}_5 ext{NH}_3^+\) \& \(\text{C}_6 ext{H}_5 ext{NH}_2\) and \(\text{H}_2 ext{CO}_3\) \& \(\text{HCO}_3^-\)) making them suitable buffer systems.
Weak Acids and Bases
Weak acids and bases are pivotal to forming buffer solutions. Unlike strong counterparts, weak acids do not completely dissociate. This partial ionization is what enables them to react with added bases, maintaining the pH balance. Similarly, weak bases don鈥檛 fully ionize, allowing them to react with added acids.
Hydrogen sulfide (\(\text{H}_2 ext{S}\)) and acetic acid are classic examples of weak acids. They do not completely release hydrogen ions into the solution, making them ideal for buffer solutions. Ammonia is a typical example of a weak base.
In our exercise, \(\text{c}\) utilizes the weak acid \(\text{H}_2 ext{S}\) and its conjugate base \(\text{HS}^-\) to create an effective buffer system, allowing the solution to counteract added acids or bases. It highlights how weak acids and their conjugates play significant roles in buffering capabilities.

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Most popular questions from this chapter

Potassium chromate is slowly added to a solution containing \(0.20 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.20 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) Describe what happens if the Ksp for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(1.1 \times 10^{-12}\) and the \(\mathrm{Ksp}\) of \(\mathrm{BaCrO}_{4}\) is \(1.2 \times 10^{-10}\) a. The \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) precipitates first out of solution and then \(\mathrm{BaCrO}_{4}\) precipitates. b. The \(\mathrm{BaCrO}_{4}\) precipitates first out of solution. c. Both \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) and \(\mathrm{BaCrO}_{4}\) precipitate simultaneously out of solution. d. Neither \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) nor \(\mathrm{BaCrO}_{4}\) precipitates out of solution.

(A): When \(\mathrm{Q}=\mathrm{K}_{\mathrm{c}}\), reaction is at equilibrium (R) At equilibrium \(\Delta \mathrm{G}\) is 0 .

Match the following: Column I Column II A. Law of mass action (p) Guldberg wage B. Active mass of \(\mathrm{CaCO}_{3}\) (q) \(\mathrm{Kp}=\mathrm{Kc}(\mathrm{RT})\) (s) C. \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{SO}_{3}\) (r) 1 D. \(\mathrm{PCl}_{5} \leftrightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (s) zero (t) \(\mathrm{Kp}=\mathrm{Kc}(\mathrm{RT})^{-1}\)

(A): On opening a sealed soda bottle dissolved carbon dioxide gas escapes. (R): Gas escapes to each the new equilibrium condition of lower pressure.

(A): On addition of \(\mathrm{NH}_{4} \mathrm{Cl}\) to \(\mathrm{NH}_{4} \mathrm{OH}\) solution, \(\mathrm{pH}\) decreases but remains greater than \(7 .\) (R): Addition of \(\mathrm{NH}_{4}^{+}\)ion decreases ionization of \(\mathrm{NH}_{4} \mathrm{OH}\) thus, \(\left[\mathrm{OH}^{-}\right]\)is decreased, hence, \(\mathrm{pH}\) decreases.
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