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

Short Answer

Expert verified
(d) is not a buffer solution because it involves a strong acid ( HClO_4) and its salt.

Step by step solution

01

Understanding Buffer Solution

A buffer solution is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. It resists changes in pH when small amounts of an acid or base are added.
02

Analyze Option (a)

Option (a) consists of hydrogen sulfide ( H_2S) and its salt potassium hydrogen sulfide ( KHS). H_2S is a weak acid, and KHS can act as its conjugate base. Therefore, this is a buffer solution.
03

Analyze Option (b)

Option (b) consists of aniline ( C_6H_5NH_2), a weak base, and anilinium bromide ( C_6H_5NH_3Br), its conjugate acid. This pair can form a buffer solution.
04

Analyze Option (c)

Option (c) consists of carbonic acid ( H_2CO_3) and potassium hydrogen carbonate ( KHCO_3). Here, H_2CO_3 acts as the weak acid and KHCO_3 as the conjugate base. This combination forms a buffer solution.
05

Analyze Option (d)

Option (d) consists of perchloric acid ( HClO_4), a strong acid, and potassium perchlorate ( KClO_4). A strong acid and its salt cannot resist changes in pH when additional acid or base is added, so this does not form a buffer solution.

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Key Concepts

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

Weak Acid
A weak acid is a type of acid that does not completely dissociate in water. This means that only a small fraction of its molecules release hydrogen ions (\( H^+ \)) into the solution. As a result, weak acids are not as acidic as their strong counterparts, which fully dissociate.

Key characteristics of weak acids include:
  • Partial dissociation in solution, which means a dynamic equilibrium exists between the undissociated acid and the ions that it releases.
  • A relatively higher pH compared to strong acids when at the same concentration.
  • Ability to establish an equilibrium with a conjugate base, making them ideal for buffer solutions.
Buffer solutions often rely on a weak acid to help stabilize the pH against added changes. For example, in the body's blood, carbonic acid (\( H_2CO_3 \)) acts to maintain the pH around the necessary physiological range.
Conjugate Base
The conjugate base is what remains after an acid donates a hydrogen ion. When a weak acid releases its hydrogen ion (\( H^+ \)), the resulting species is the conjugate base. This base can accept hydrogen ions back, re-forming the acid, which is a crucial reaction in buffer systems.

Characteristics of a conjugate base include:
  • Acts as the byproduct when an acid loses a proton.
  • Can react with added acids, helping to neutralize them in buffer solutions, thus maintaining the pH.
  • Often appears in equilibrium equations alongside their acid counterparts, forming conjugate acid-base pairs.
In a buffer solution, for instance, hydrogen sulfide (\( H_2S \)) is paired with its conjugate base, the sulfide ion (\( HS^- \)), to stabilize pH levels effectively.
Weak Base
A weak base is a substance that only partially accepts protons when dissolved in water. This means that when a weak base is added to water, only a fraction of its molecules react to form hydroxide ions (\( OH^- \)). Weak bases are not as efficient at raising pH levels compared to strong bases, which fully dissociate.

Key traits of weak bases include:
  • Partial ionization in solution, leading to a less alkaline solution.
  • The presence of equilibrium between the base molecules and ions formed in the solution.
  • Capability to pair with a conjugate acid to create a buffer solution.
An example of a weak base would be aniline (\( C_6H_5NH_2 \)), which can form a buffer with anilinium bromide (\( C_6H_5NH_3Br \)) by equilibrating with its conjugate acid in solution.
Conjugate Acid
A conjugate acid is what forms when a base accepts a hydrogen ion. In the context of buffer solutions, a weak base interacts with hydrogen protons, becoming a conjugate acid, which then can donate the proton back if needed, creating a balance and preventing drastic pH shifts.

Understanding conjugate acids involves noting:
  • They are created when a base gains a proton, thus reversibly turning into an acid.
  • Conjugate acids in buffer solutions are vital to counteract added bases by releasing hydrogen ions.
  • They maintain an equilibrium with the base from which they were derived, pivotal for a buffering action.
In buffer systems, aniline (\( C_6H_5NH_2 \)) forms its conjugate acid, anilinium ion (\( C_6H_5NH_3^+ \)), which can help regulate pH by serving as a reservoir of hydrogen ions.

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Most popular questions from this chapter

\(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}+22.4 \mathrm{kcal}\) formation of \(\mathrm{NH}_{3}\) by above reaction shows: (a) Cyanamide process (b) Serpeck's process (c) Haber process (d) None of these

The vapour density of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at a certain temperature is 30\. What is the percentage dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at this temperature: (a) \(53.3\) (b) \(106.6\) (c) \(26.7\) (d) None of these

For the reaction: \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) at a given temperature, the equilibrium amount of \(\mathrm{CO}_{2}(\mathrm{~g})\) can be increased by (a) Adding a suitable catalyst (b) Adding an inert gas (c) Decreasing the volume of the container (d) Increasing the amount of \(\mathrm{CO}(\mathrm{g})\)

The equilibrium constant for the reaction: \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is \(\mathrm{K}_{1}\) for reaction \(\mathrm{H}_{2} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_{4}^{2-}\) is \(\mathrm{K}_{2}\) and for reaction \(\mathrm{HPO}_{4}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) is \(\mathrm{K}_{3}\) The equilibrium constant \((\mathrm{K})\) for \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons 3 \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) will be: (a) \(\mathrm{K}_{1}, \times \mathrm{K}_{2} \times \mathrm{K}_{3}\) (b) \(\mathrm{K}_{1} / \mathrm{K}_{2} \mathrm{~K}_{3}\) (c) \(\mathrm{K}_{2} / \mathrm{K}_{1} \mathrm{~K}_{3}\) (d) \(\mathrm{K}_{1},+\mathrm{K}_{2}+\mathrm{K}_{3}\)

In what manner will increase of pressure affect the following equation: \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (a) Shift in the reverse direction (b) Shift in the forward direction (c) Increase in the yield of hydrogen (d) No effect

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