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When a buffer solution of \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) is diluted with water then: (a) \(\mathrm{H}^{+}\) ion concentration increases (b) \(\mathrm{H}^{+}\) ion concentration decreases (c) \(\mathrm{H}^{+}\) ion concentration remains constant (d) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ion concentration increases

Short Answer

Expert verified
The \(\mathrm{H}^+\) ion concentration remains constant upon dilution.

Step by step solution

01

Understanding Buffer Solutions

A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. It resists changes in pH upon dilution or the addition of small amounts of acids or bases.
02

Buffer System Components

In this problem, the buffer solution is composed of acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) which is a weak acid, and sodium acetate (\(\mathrm{CH}_{3} \mathrm{COONa}\)), which provides the conjugate base ion \(\mathrm{CH}_{3} \mathrm{COO}^{-}\).
03

Effect of Dilution on Buffer Solution

Diluting a buffer solution with water means adding more solvent without significantly altering the concentration of hydrogen ions (\(\mathrm{H}^+\)) and acetate ions (\(\mathrm{CH}_{3} \mathrm{COO}^-\)) in relative terms.
04

Impact on Hydrogen Ion Concentration

Although the concentrations of both \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COO}^-\) decrease when a buffer is diluted, the ratio of \(\mathrm{CH}_{3} \mathrm{COOH}\) to \(\mathrm{CH}_{3} \mathrm{COO}^-\) remains constant. Thus, according to the Henderson-Hasselbalch equation, the \(\mathrm{H}^+\) ion concentration stays constant.
05

Conclusion on Option Choices

Given the consistent \(\mathrm{H}^{+}\) ion concentration due to the unchanged ratio, option (c) — \(\mathrm{H}^{+}\) ion concentration remains constant — is the correct choice.

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Key Concepts

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

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a critical formula when dealing with buffer solutions. It allows us to calculate the pH of a buffer solution by relating it to the concentrations of the acid and its conjugate base. The equation is given as:\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]where \([\text{A}^-]\) is the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the acid. This equation assumes that changes in the concentrations of \([\text{A}^-]\) and \([\text{HA}]\) due to dilution do not alter their ratio significantly, thus keeping the pH nearly constant.
  • The equation applies to weak acids and their conjugate bases.
  • It emphasizes the importance of the ratio rather than individual concentrations.
Understanding this relationship helps us comprehend how buffer solutions maintain their pH against dilution and the addition of small amounts of other acids or bases.
Acetic Acid
Acetic acid, with the formula \(\mathrm{CH}_3\mathrm{COOH}\), is an example of a weak acid. It partially ionizes in solution, meaning that only a small fraction of the acid dissociates to release \(\mathrm{H}^+\) ions. This property is essential in buffer solutions because it allows for a reversible reaction with its conjugate base. When acetic acid is in a solution, it exists in equilibrium:\[ \mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \]
  • Acetic acid donates \(\mathrm{H}^+\) ions, which can be captured by its conjugate base, \(\mathrm{CH}_3\mathrm{COO}^-\).
  • This dynamic equilibrium makes acetic acid a crucial agent in resisting pH changes.
Knowing the role of weak acids like acetic acid in buffer solutions can help students understand how these solutions mitigate shifts in pH upon dilution.
Conjugate Base
In the context of buffer solutions, the conjugate base is key to maintaining the pH balance. A conjugate base is formed when an acid donates a proton and becomes a species that can accept protons. For acetic acid, its conjugate base is the acetate ion, \(\mathrm{CH}_3\mathrm{COO}^-\).Let's see why conjugate bases are so vital:
  • They can react with added \(\mathrm{H}^+\) ions, preventing a decrease in pH.
  • They work in tandem with the acid to keep the pH stable even when diluted.
The presence of both a weak acid and its conjugate base in significant amounts is what defines a buffer solution. The ability of the conjugate base to neutralize added acids helps explain why dilution does not significantly affect the Hydrogen ion concentration in a buffer solution.

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Most popular questions from this chapter

In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relationship between \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{2}\) (c) \(K_{p}=K_{c}(R T)^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)

In the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) mole/litre respectively. If the value of \(\mathrm{K}_{\mathrm{c}}\) is \(0.5\), what is the concentration of \(\mathrm{Cl}_{2}\) in mole/litre: (a) \(2.0\) (b) \(1.5\) (c) \(1.0\) (d) \(0.5\)

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4\. What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

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) \(\mathrm{K}_{1},+\mathrm{K}_{2}+\mathrm{K}_{3}\)

Which of these reactions are having homogeneous equilibrium: (a) \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) \(' \mathrm{~g}^{\prime} \quad{ }^{6} \mathrm{~g}^{\prime}\) (b) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl}\) \(' \mathrm{~g}^{\prime} \quad{ }^{2} \mathrm{~g}^{\prime} \quad{ }^{\mathrm{s}} \mathrm{g}^{\prime}\) (c) \(\mathrm{CaCO}_{3} \rightleftharpoons \mathrm{CaO}+\mathrm{CO}_{2}\) 's' \(\quad{ }^{\prime} \mathrm{s}^{\prime} \quad{ }^{\prime} \mathrm{g}\) ' (d) Both (a) and (b)

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