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Which of the following would produce a buffer solution when mixed in equal volume? (a) \(1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.5 \mathrm{M} \mathrm{NaOH}\) (b) \(1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.5 \mathrm{M} \mathrm{HCl}\) (c) \(1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\) and \(0.5 \mathrm{M} \mathrm{NaOH}\) (d) \(1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) and \(0.5 \mathrm{M} \mathrm{HCl}\)

Short Answer

Expert verified
Option (a) produces a buffer solution.

Step by step solution

01

Understanding Buffer Solution

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

Analyze Option (a)

Mixing 1 M acetic acid (CH₃COOH) with 0.5 M NaOH in equal volumes forms a solution with acetic acid and its conjugate base (CH₃COO⁻ after NaOH neutralizes some of the acid). This can be a buffer solution because it has both a weak acid and its conjugate base.
03

Analyze Option (b)

Mixing 1 M acetic acid (CH₃COOH) with 0.5 M HCl does not form a buffer. HCl is a strong acid and does not create a conjugate base with acetic acid. This mixture results in a solution with extra strong acid.
04

Analyze Option (c)

Mixing 1 M ammonium hydroxide (NH₄OH) with 0.5 M NaOH does not create a buffer, as both NH₄OH and NaOH are bases. A buffer requires an acid-base pair.
05

Analyze Option (d)

Mixing 1 M ammonium chloride (NH₄Cl) with 0.5 M HCl does not form a buffer because they both create an acid solution, lacking the necessary weak base and its conjugate acid pair.
06

Conclusion

Option (a) produces a buffer solution because it includes a weak acid and its conjugate base after partial neutralization by NaOH.

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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 an essential component for creating a buffer solution. Unlike strong acids, which completely dissociate in water, a weak acid only partially dissociates. This means not all of its molecules release hydrogen ions into the solution. For example, acetic acid (\(\mathrm{CH_3COOH}\)) is a common weak acid. It only releases some of its hydrogen ions when dissolved in water. In a buffer solution, this partial dissociation is crucial because it allows the weak acid to donate hydrogen ions when the pH starts to rise. It helps maintain the balance of the pH. The presence of a weak acid ensures that the buffer can respond to changes by providing extra hydrogen ions as needed.
  • The acid is weak because it doesn't dissociate completely.
  • Weak acids like acetic acid have a minor ionization in water.
  • They play a key role in stabilizing the pH level of a buffer.
Conjugate Base
When a weak acid partially dissociates, it forms a conjugate base. This conjugate base is the ion left behind after the weak acid donates a hydrogen ion. For the acetic acid example, the conjugate base is the acetate ion (\(\mathrm{CH_3COO^-}\)). The conjugate base is important in a buffer solution because it can react with any added hydrogen ions. This reaction prevents the pH from dropping too much when an acid is added to the solution. By neutralizing added acids, the conjugate base helps maintain the overall balance of the buffer solution.
  • A conjugate base is formed from a weak acid by losing a proton.
  • It acts to neutralize added acids, maintaining the pH.
  • Examples include acetate ion for acetic acid.
pH Resistance
The primary goal of a buffer solution is to resist changes in pH. This property is what makes buffers particularly valuable in many biochemical and industrial processes. When small amounts of acid or base are introduced into a buffer solution, the weak acid and its conjugate base work together to neutralize these additions, preventing large swings in pH. This pH resistance occurs because the weak acid can provide hydrogen ions when there's an addition of a base, while the conjugate base can absorb hydrogen ions when an acid is added. This dual capability allows the buffer to maintain a relatively stable pH level, which is essential in many environments such as in biological systems and laboratories.
  • Buffer solutions maintain a stable pH when small amounts of acids or bases are added.
  • They are used extensively in biological, chemical, and industrial applications.
  • pH resistance is made possible by the interplay between weak acid and conjugate base.

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Most popular questions from this chapter

The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4}\). The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is: (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50

For the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta \mathrm{H}=-93.6 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), the concentration of \(\mathrm{H}_{2}\) at equilibrium can be increased by: (1) Lowering the temperature (2) Increasing the volume of the system (3) Adding \(\mathrm{N}_{2}\) at constant volume (4) Adding \(\mathrm{H}_{2}\) at constant volume (a) (ii) and (iv) are correct (b) Only (ii) is correct (c) (i), (ii) and (iii) are correct (d) (iii) and (iv) are correct

In the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\), when 100 \(\mathrm{mL}\) of \(\mathrm{N}_{2}\) has reacted, the volumes of \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are: (a) \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(300 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (b) \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (c) \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (d) \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(100 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\)

A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves: (a) The equilibrium has been established in the solution (b) Radioactive sugar can displace non-radioactive sugar from its solution.

If equilibrium constant for the reaction: \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(\mathrm{K}_{\mathrm{c}}\), then the equilibrium con- stant for the reaction \(\mathrm{NH}_{3} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2}\) will be. (a) \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}\) (b) \(\frac{1}{\mathrm{~K}^{2}}\) (c) \(\sqrt{K}_{c}\) (d) \(\frac{1}{\sqrt{K}_{c}}\)

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